Determinants — Class 12 Maths Notes & Practice

September 11, 2026

Meet Your Tutor

Determinants reward organised working more than fast expansion. I will help you choose the shortest row or column, track cofactor signs, use properties before calculating, and connect determinants with area, inverse matrices and linear equations without losing the logic between steps.

Let us begin with the most reassuring sentence in this whole chapter: a determinant is just a number. That is genuinely all it is. You take a square block of numbers, you follow one fixed recipe, and out comes a single value. There is no hidden trick and no special kind of brain required — only a recipe you have practised often enough that your hand starts moving before your head has finished reading the question.

Determinants live inside Unit II (Algebra) of the Class 12 Maths course. That unit — Matrices and Determinants together — carries 10 marks in the board paper, and Determinants usually takes the larger share because it is where the long, satisfying, full-marks questions sit: find the inverse, solve three equations, find the area of a triangle. These are questions you can finish completely and correctly if you are patient. That is a lovely position to be in.

If matrices felt shaky to you, do not panic. You need surprisingly little from that chapter to start here: what a square matrix is, how to multiply two matrices, and what the identity matrix I looks like. Everything else we will build slowly, from zero, together. Take your time with the early sections. Speed comes later, and it comes on its own.

What You'll Learn

What a Determinant Really Is
Determinant of a 2 × 2 Matrix
Determinant of a 3 × 3 Matrix
Minors and Cofactors
Row and Column Behaviour: Safe Evaluation Shortcuts
Singular and Non-Singular Matrices
Area of a Triangle Using Determinants
Collinearity of Three Points
Adjoint of a Square Matrix
Inverse of a Square Matrix
Solving Linear Equations by the Matrix Method
Consistency, Inconsistency and Number of Solutions
Word Problems Solved by the Matrix Method
Exam Strategy and Common Traps

Your Game Plan

Please do not read this chapter like a story from top to bottom in one sitting. Work it in stages, with a pen in your hand. Here is the order I would sit you down and make you follow.

Day 1 — get the number out. Learn to evaluate 2 × 2 and 3 × 3 determinants until you can do a 3 × 3 in under a minute without second-guessing your signs.
Day 2 — minors and cofactors. These are just determinants of smaller pieces, plus a plus-or-minus sign. Do twenty of them. They are the raw material for everything that follows.
Day 3 — adjoint and inverse. This is one long assembly line, and every step is something you already learned on Day 2. Slow and neat beats fast and messy here.
Day 4 — the applications. Area of a triangle, collinearity of three points, and then the big one: solving equations by the matrix method.
Day 5 — consistency and mixed practice. Learn to decide how many solutions a system has, then do the worksheet at the bottom of this page with a timer running.
Every day — three minutes of revision. Re-derive the inverse of one small matrix from memory. Three minutes. That habit alone is worth several marks.

Exam Tip — What the 2026–27 syllabus actually lists

The official curriculum for this chapter names exactly four things: (1) the determinant of a square matrix up to 3 × 3, together with minors and cofactors, and its use in finding the area of a triangle; (2) the adjoint and inverse of a square matrix; (3) consistency, inconsistency and the number of solutions of a system of linear equations, shown through examples; and (4) solving a system in two or three variables that has a unique solution, using the inverse of a matrix. Note carefully what is not there: Cramer's rule has been removed, and "properties of determinants" is not listed as a separate examinable head. We will still meet the row and column properties, but only in the honest role they now play — as shortcuts that make evaluation faster and safer.

Study Notes

What a Determinant Really Is

Think of a matrix as a filing cabinet: rows and columns of numbers, sitting there, holding information. A determinant is a single summary number squeezed out of that cabinet. It is a bit like a batting average — one number that tells you something useful about a whole season of scores, even though it throws away the details.

Two things must be true before a determinant even exists:

The matrix must be square. Same number of rows as columns. A 2 × 3 matrix simply does not have a determinant — the question is meaningless, like asking for the colour of the number seven.
In Class 12 we only go up to 3 × 3. That is the ceiling for your board paper, so you can relax about anything larger.

Notation matters, and marks are quietly lost here every year. A matrix is written inside square brackets. Its determinant is written inside straight vertical bars, or as det(A), or as |A|.

Matrix A (a box of numbers):

[   4   5 ]
[   2   3 ]

and its determinant |A| (one number):

|   4   5 |
|   2   3 |

Key Rule — Matrix versus determinant

A matrix is an arrangement; a determinant is a number. You can never say "the matrix equals 2", but you can absolutely say "the determinant equals 2". Square brackets = arrangement. Vertical bars = number. Keep those two symbols separate in your handwriting and half the confusion in this chapter disappears.

So what does that number mean? Here is the picture that made it click for me. Take the two rows of a 2 × 2 matrix as two arrows drawn from the origin. Those two arrows stretch a unit square into a slanted parallelogram. The determinant measures how much the area changed — it is a scale factor. And that is why a determinant of zero is such big news: zero area means the two arrows have collapsed onto the same line, the matrix has squashed the plane flat, and nothing that is squashed flat can ever be un-squashed. That single idea explains why a matrix with determinant zero has no inverse, and why a system of equations with determinant zero refuses to give you one clean answer.

Example 1 — Reading the notation correctly

Write the determinant of A =

[   4   5 ]
[   2   3 ]

and evaluate it.

Working. |A| = (4)(3) − (5)(2) = 12 − 10 = 2.

Why it works. The recipe for a 2 × 2 is "main diagonal product minus the other diagonal product". Geometrically, the arrows (4, 5) and (2, 3) stretch the unit square into a parallelogram of area 2. The matrix is a box; the 2 is the number that box produces.

Example 2 — Which of these even has a determinant?

Consider P =

[   1   2   3 ]
[   4   5   6 ]

and Q =

[   1   2 ]
[   3   4 ]

. Find any determinant that exists.

Working. P is a 2 × 3 matrix — two rows, three columns — so it is not square and |P| does not exist. Do not attempt it; simply state that it is not defined. Q is 2 × 2, so |Q| = (1)(4) − (2)(3) = 4 − 6 = −2.

Why it works. The determinant recipe pairs every row with a column of the same length. If the shape is not square, the pairing breaks down and there is nothing to compute. Writing "not defined" is the fully correct answer and earns the mark.

Notice that |Q| came out negative. A determinant is perfectly allowed to be negative — the sign tells you that the transformation flipped the plane over, like turning a page. When we later use determinants for area, we will wrap the answer in a modulus, because area itself is never negative. But the raw determinant keeps its sign, and you must keep it too.

Common Mistake — Dropping the minus sign too early

Students often see a negative determinant, decide it "looks wrong", and quietly change it to positive. Do not. The sign is real information. It only disappears in the one place where we deliberately take a modulus — the area of a triangle. Everywhere else — adjoint, inverse, consistency — a wrong sign wrecks the whole answer.

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Determinant of a 2 × 2 Matrix

This is the atom of the entire chapter. Every 3 × 3 determinant you ever evaluate will be broken down into 2 × 2 determinants, so getting this rock-solid now saves you hours later.

Key Rule — The 2 × 2 determinant

For A =

[   a   b ]
[   c   d ]

the determinant is |A| = ad − bc.

In words: multiply along the main diagonal (top-left to bottom-right), then subtract the product along the other diagonal (top-right to bottom-left). Draw the two diagonals with your pen the first fifty times. Muscle memory is a real thing.

Example 3 — A straightforward one

Evaluate

|   6  -3 |
|   2   7 |

Working. ad − bc = (6)(7) − (−3)(2) = 42 − (−6) = 42 + 6 = 48.

Why it works. The subtraction sign and the negative entry combine into a plus. Write the substitution out with brackets exactly as above — (−3)(2) inside its own brackets — and the sign takes care of itself. Trying to do it in your head is exactly how the sign gets lost.

Example 4 — Two negatives in play

Evaluate

|  -2   5 |
|   4  -1 |

Working. (−2)(−1) − (5)(4) = 2 − 20 = −18.

Why it works. The main diagonal gives a positive 2 because negative times negative is positive; the other diagonal gives 20, which is then subtracted. A negative final answer is completely fine here. Resist the urge to "correct" it.

Example 5 — When the answer is zero, and why that is interesting

Evaluate

|   9  12 |
|   3   4 |

and comment on the result.

Working. (9)(4) − (12)(3) = 36 − 36 = 0.

Why it works. Look at the two rows: (9, 12) is exactly 3 times (3, 4). The rows are multiples of each other, so as arrows they point along the same line, the parallelogram has collapsed to a flat sliver, and the area is zero. Whenever one row (or column) is a multiple of another, you can write down 0 immediately without any multiplication at all. This matrix is called singular — a word we will meet properly a few sections from now.

Exam Tip — Spot the free marks before you calculate

Before touching your pen to a 2 × 2, glance for three freebies: a row of zeros, a column of zeros, or two rows/columns that are multiples of each other. Any of those and the answer is 0 on sight. In a paper where every minute counts, that is a whole question answered in five seconds.

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Determinant of a 3 × 3 Matrix

Here is the good news up front: you already know how to do this. A 3 × 3 determinant is nothing more than three 2 × 2 determinants glued together with signs. There is no new arithmetic to learn — only a new piece of bookkeeping.

The method is called expansion along a row or a column. Pick any single row (or column). For each element in it, cover up that element's own row and column with your finger; the four numbers still visible form a little 2 × 2 determinant. Multiply the element by that little determinant. Do this for all three elements, attach the signs + − +, and add.

Key Rule — Expansion along the first row

For A =

[   a   b   c ]
[   d   e   f ]
[   g   h   i ]

|A| = a(ei − fh) − b(di − fg) + c(dh − eg)

The bracket after each letter is just the 2 × 2 determinant left behind when you cover that letter's row and column. The signs alternate + − +, and that middle minus is the single most common place marks are dropped in this entire chapter.

The sign pattern is not random and you should never memorise it as a list. It is a chessboard:

[   +   –   + ]
[   –   +   – ]
[   +   –   + ]

Top-left is always +, and the sign flips every time you take one step sideways or one step down. Whichever row or column you decide to expand along, read its signs straight off this board. Draw it in the margin of your rough work every single time until you no longer need to.

Key Rule — You may expand along ANY row or ANY column

The answer is identical whichever line you choose. So choose deliberately: pick the row or column containing the most zeros. Every zero kills an entire 2 × 2 calculation, because zero times anything is zero. Choosing well can turn a three-minute question into a forty-second one.

Example 6 — Expanding along the first row

Evaluate

|   2   1   3 |
|   1   4  -2 |
|   3  -1   1 |

Working. Expand along row 1, signs + − +.
• Cover the 2: left with

|   4  -2 |
|  -1   1 |

= (4)(1) − (−2)(−1) = 4 − 2 = 2.
• Cover the 1: left with

|   1  -2 |
|   3   1 |

= (1)(1) − (−2)(3) = 1 + 6 = 7.
• Cover the 3: left with

|   1   4 |
|   3  -1 |

= (1)(−1) − (4)(3) = −1 − 12 = −13.

Now assemble with + − + :
|A| = 2(2) − 1(7) + 3(−13) = 4 − 7 − 39 = −42.

Why it works. Each bracket is the determinant of what survives after you delete that element's row and column. The alternating sign is the chessboard doing its job. Notice how the third term, 3(−13), quietly does most of the damage — keep the brackets around negative values and it stays under control.

Example 7 — Choosing a smart row (a zero is a gift)

Evaluate

|   1   3  -2 |
|   4  -1   5 |
|   2   0   3 |

Working. Row 3 contains a zero, so expand along row 3. Its signs, read off the chessboard, are + − +.
• 2 ×

|   3  -2 |
|  -1   5 |

= 2 × [(3)(5) − (−2)(−1)] = 2 × (15 − 2) = 2 × 13 = 26.
• − 0 × (anything) = 0 — skip it entirely.
• + 3 ×

|   1   3 |
|   4  -1 |

= 3 × [(1)(−1) − (3)(4)] = 3 × (−1 − 12) = 3 × (−13) = −39.

|A| = 26 + 0 − 39 = −13.

Why it works. You did two 2 × 2 determinants instead of three. Expanding along row 1 would give exactly the same −13, but with more arithmetic and more chances to slip. Always scan for the zero-richest line first.

Example 8 — The triangular shortcut

Evaluate

|   5   0   0 |
|   2  -3   0 |
|   7   4   6 |

Working. Expand along row 1 (two zeros there):
|A| = 5 ×

|  -3   0 |
|   4   6 |

− 0 + 0 = 5 × [(−3)(6) − (0)(4)] = 5 × (−18) = −90.

Why it works. Every entry above the main diagonal is zero — this is a triangular matrix. For any triangular matrix the determinant is simply the product of the diagonal entries: 5 × (−3) × 6 = −90. Same answer, no expansion needed. Store this pattern; it turns up in the board paper more often than you would expect.

Common Mistake — Forgetting the middle minus

By far the number one error in this chapter is writing |A| = a(ei − fh) + b(di − fg) + c(dh − eg). The middle term is subtracted. A safety habit: before you add anything up, write the three signs + − + above your three brackets and physically tick them. Ten seconds spent there has saved more marks than any clever shortcut.

Exam Tip — Sanity-check with a second expansion

If a determinant carries three or more marks and you have time at the end, re-evaluate it along a different row or column. If the two answers agree, you are safe. If they disagree, you have caught a mistake that would otherwise have cost you every mark that flows from it — and in this chapter, one wrong determinant can poison an entire six-mark question.

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Minors and Cofactors

You have already been computing minors and cofactors for the last two sections — you just did not know their names. Every one of those "cover the row and column with your finger" determinants was a minor. Now we give them proper labels, because the adjoint and the inverse are built entirely out of them.

Key Rule — Minor and cofactor, defined

The minor M_ij of the element a_ij is the determinant you get after deleting row i and column j from the matrix.

The cofactor A_ij is that same minor with a sign attached:
A_ij = (−1)^i+j × M_ij

So if i + j is even, the cofactor equals the minor. If i + j is odd, the cofactor is the minor with its sign flipped. That is the entire difference between the two words.

Read the subscripts carefully: i is the row number, j is the column number, always in that order. M₂₃ means row 2, column 3 — not the other way round. Students who get this backwards produce beautiful, careful, completely wrong answers.

And notice the sign rule (−1)^i+j is exactly the chessboard from the previous section. For M₂₃, i + j = 2 + 3 = 5, which is odd, so A₂₃ = −M₂₃. You do not need to compute a power of −1; you only need to ask "is i + j odd or even?"

Example 9 — All four minors and cofactors of a 2 × 2

Find every minor and cofactor of A =

[   2  -1 ]
[   3   4 ]

and use them to write |A|.

Working. Deleting a row and a column from a 2 × 2 leaves a single number, and the determinant of a single number is that number itself.
• M₁₁ = 4 (delete row 1, column 1). i + j = 2, even → A₁₁ = 4.
• M₁₂ = 3. i + j = 3, odd → A₁₂ = −3.
• M₂₁ = −1. i + j = 3, odd → A₂₁ = 1.
• M₂₂ = 2. i + j = 4, even → A₂₂ = 2.

Expanding along row 1: |A| = a₁₁A₁₁ + a₁₂A₁₂ = (2)(4) + (−1)(−3) = 8 + 3 = 11.
Direct check: ad − bc = (2)(4) − (−1)(3) = 8 + 3 = 11. They agree.

Why it works. The familiar ad − bc formula is the cofactor expansion in disguise. The minus sign in ad − bc is precisely the (−1)¹⁺² attached to M₁₂.

Example 10 — A single named cofactor from a 3 × 3

For A =

[   3   1   2 ]
[   2  -3  -1 ]
[   1   2   4 ]

find M₂₃ and A₂₃.

Working. Row 2, column 3. Delete row 2 (that is 2, −3, −1) and delete column 3 (that is 2, −1, 4). What survives is

|   3   1 |
|   1   2 |

M₂₃ = (3)(2) − (1)(1) = 6 − 1 = 5.
i + j = 2 + 3 = 5, odd, so A₂₃ = −5. A₂₃ = −5.

Why it works. Delete the whole row and the whole column that the element sits in — including the element itself. Four numbers remain, in their original relative positions. Do not shuffle them.

Example 11 — The full cofactor table, and the determinant from it

For the same A =

[   3   1   2 ]
[   2  -3  -1 ]
[   1   2   4 ]

compute every cofactor, then find |A|.

Working — minors first.
M₁₁ = |−3 −1; 2 4| = −12 + 2 = −10 • M₁₂ = |2 −1; 1 4| = 8 + 1 = 9 • M₁₃ = |2 −3; 1 2| = 4 + 3 = 7
M₂₁ = |1 2; 2 4| = 4 − 4 = 0 • M₂₂ = |3 2; 1 4| = 12 − 2 = 10 • M₂₃ = |3 1; 1 2| = 6 − 1 = 5
M₃₁ = |1 2; −3 −1| = −1 + 6 = 5 • M₃₂ = |3 2; 2 −1| = −3 − 4 = −7 • M₃₃ = |3 1; 2 −3| = −9 − 2 = −11

Now apply the chessboard + − + / − + − / + − + to get the cofactor matrix:

[ -10   -9    7 ]
[   0   10   -5 ]
[   5    7  -11 ]

Determinant from row 1: |A| = (3)(−10) + (1)(−9) + (2)(7) = −30 − 9 + 14 = −25.

Why it works. Once the signs are already baked into the cofactors, expansion is a plain dot product — multiply each element by its own cofactor and add. No extra minus signs to remember. Keep this cofactor matrix; we will reuse it for the adjoint and the inverse later, so the work is not wasted.

Example 12 — The zero test (a free accuracy check)

Using the cofactors from Example 11, compute (3)(0) + (1)(10) + (2)(−5).

Working. 0 + 10 − 10 = 0.

Why it works. Here we multiplied the elements of row 1 by the cofactors of row 2 — a mismatched pairing — and the answer is always exactly zero. Match the row with its own cofactors and you get |A|; mismatch them and you get 0. This gives you a free thirty-second check on a cofactor table you are about to build an inverse from. If a mismatched sum does not come out as 0, one of your cofactors is wrong and you have found it before it cost you anything.

Key Rule — The two expansion facts, side by side

• a_i1A_i1 + a_i2A_i2 + a_i3A_i3 = |A| — matched row and cofactors.
• a_i1A_k1 + a_i2A_k2 + a_i3A_k3 = 0 whenever i ≠ k — mismatched.

The same pair of statements holds for columns. Together they are the reason the adjoint formula works at all, as you will see shortly.

Common Mistake — Calling a minor a cofactor

If a question asks for the minor M₃₂ and you hand in the cofactor, you lose the mark even though your arithmetic was perfect. Read the word in the question, underline it, and answer that word. In Example 11, M₃₂ = −7 but A₃₂ = +7 — the difference is the whole mark.

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Row and Column Behaviour: Safe Evaluation Shortcuts

Now a moment of honesty, because it affects how much time you should spend here. In the 2026–27 curriculum, "properties of determinants" is not listed as a separate examinable heading. What is listed is: evaluating determinants up to 3 × 3, minors and cofactors, area of a triangle, adjoint and inverse, and solving systems by the matrix method. So the row and column facts below are not a topic to be examined for their own sake — they are tools. They make your evaluation faster, they catch your errors, and occasionally they turn an intimidating determinant into a one-line answer. Learn them as shortcuts, not as a syllabus unit.

Key Rule — Six behaviours worth knowing

1. Transpose changes nothing. |A′| = |A|. Rows and columns have equal rights, which is exactly why you may expand along either.
2. Swapping two rows (or two columns) flips the sign. One swap turns |A| into −|A|.
3. Two identical rows (or columns) → determinant is 0.
4. One row is a multiple of another → determinant is 0. More generally, if any row is built from the others by adding and scaling, the determinant is 0.
5. Multiplying one whole row by k multiplies the determinant by k. Consequently, for an n × n matrix, |kA| = kⁿ|A| — because you have scaled all n rows.
6. Adding a multiple of one row to another changes nothing at all. R₁ → R₁ − 3R₂ leaves |A| untouched. This is the workhorse: use it to manufacture zeros.

Rule 6 is the one that earns its keep. Zeros are what make a determinant easy, and rule 6 lets you create zeros for free. Rule 5 is the one students misuse: if you factor 3 out of one row, you take out a single 3; if a 3 comes out of all three rows of a 3 × 3, that is 3³ = 27.

Example 13 — Zero on sight

Evaluate

|   1   2   3 |
|   2   4   6 |
|   5   7   9 |

Working. Look at rows 1 and 2: (2, 4, 6) = 2 × (1, 2, 3), so R₂ = 2R₁. By behaviour 4, |A| = 0. No expansion required.

Why it works. Two rows lying along the same direction means the three arrows are squashed into a plane instead of filling space — the volume they enclose is zero. Write one line of justification ("R₂ = 2R₁") and the mark is yours.

Example 14 — Manufacturing zeros with column operations

Evaluate

|   1   1   1 |
|   3   4   5 |
|   9  16  25 |

Working. Apply C₂ → C₂ − C₁ and C₃ → C₃ − C₁. By behaviour 6 the value is unchanged:

|   1   0   0 |
|   3   1   2 |
|   9   7  16 |

Now expand along row 1 — only the first term survives:
|A| = 1 × [(1)(16) − (2)(7)] = 16 − 14 = 2.

Why it works. Two operations bought us two zeros in the top row, which collapsed a three-term expansion into a single 2 × 2. Direct expansion gives 2 as well — try it and confirm. The shortcut is not magic, just less arithmetic and therefore fewer slips.

Example 15 — A disguised multiple

Evaluate

| 102  18  36 |
|   1   3   4 |
|  17   3   6 |

Working. The big numbers look frightening, so look for structure instead. Row 3 is (17, 3, 6). Multiply it by 6: (102, 18, 36). That is exactly row 1. So R₁ = 6R₃, and by behaviour 4, |A| = 0.

Why it works. Large entries in a board question are usually a hint that a pattern is hiding. Before grinding out the arithmetic, spend five seconds asking "is one row a multiple of another?" Here that question saved the entire calculation.

Example 16 — Proving a determinant is zero without expanding

Show, without expanding, that

|     a    a+b   a+2b |
|   a+b   a+2b  a+3b |
|  a+2b   a+3b  a+4b |

= 0.

Working. Add row 1 to row 3:
R₁ + R₃ = (a + a+2b, a+b + a+3b, a+2b + a+4b) = (2a+2b, 2a+4b, 2a+6b) = 2(a+b, a+2b, a+3b) = 2R₂.
So R₁ − 2R₂ + R₃ = 0: the three rows are linearly dependent. Hence the determinant is 0.

Why it works. Every entry moves in an arithmetic progression, so the middle row is exactly the average of the outer two. A row that is built out of the other rows contributes nothing new, the three arrows lie in one plane, and the determinant collapses to zero. Notice this argument never needed a single multiplication.

Exam Tip — The |kA| trap

If |A| = 5 for a 3 × 3 matrix, then |2A| = 2³ × 5 = 40, not 10. The exponent is the order of the matrix. Quick self-test: |A| = −25 for a 3 × 3 gives |2A| = 8 × (−25) = −200. For a 2 × 2 with |A| = 5, |3A| = 9 × 5 = 45. Say the sentence "k to the power of the order" out loud once and it sticks.

Common Mistake — Mixing rows and columns in one step

An operation such as R₁ → R₁ − C₂ is meaningless — you cannot subtract a column from a row. Keep each step purely row-to-row or purely column-to-column, and write the operation down in the margin as you do it. Also, never perform two operations that feed each other in the same step (using the old R₁ and the new R₁ at once); do them one at a time.

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Singular and Non-Singular Matrices

Two vocabulary words, and they carry a lot of weight in the rest of the chapter.

Key Rule — Singular and non-singular

A square matrix A is singular if |A| = 0, and non-singular if |A| ≠ 0.

Everything that follows hangs on this single question. A non-singular matrix has an inverse; a singular matrix does not. A system whose coefficient matrix is non-singular has exactly one solution; a singular one does not. So the very first thing you compute in almost every question of this chapter is |A|, and the very first thing you ask is "is it zero?"

The word "singular" is worth a moment. It suggests something unusual or degenerate — and that is right. A singular matrix has squashed space flat. Information has been permanently lost, so there is no way back, which is precisely why no inverse can exist.

Example 17 — Find k for a singular 2 × 2

Find k so that A =

[   k   3 ]
[   2   5 ]

is singular.

Working. Singular means |A| = 0.
|A| = (k)(5) − (3)(2) = 5k − 6.
Set 5k − 6 = 0 → k = 6/5.

Why it works. "Singular" is not a new calculation — it is just the instruction "set the determinant equal to zero". Translate the word into that equation and the question becomes ordinary algebra. Check: with k = 6/5, the rows are (6/5, 3) and (2, 5), and indeed (6/5, 3) = 0.6 × (2, 5).

Example 18 — A quadratic, so expect two answers

Find all values of k for which

[  k-1   2 ]
[   3   k ]

is singular.

Working. |A| = (k − 1)(k) − (2)(3) = k² − k − 6.
Set k² − k − 6 = 0 → (k − 3)(k + 2) = 0 → k = 3 or k = −2.

Why it works. With k appearing in two positions on the main diagonal, the determinant becomes a quadratic, so there are two answers. Giving only one is a classic half-mark loss. Verify k = 3: rows (2, 2) and (3, 3) — multiples, so determinant 0. Verify k = −2: rows (−3, 2) and (3, −2) — again multiples. Both check out.

Example 19 — A 3 × 3 with an unknown

For which k is A =

[   2   3   1 ]
[   1   k   2 ]
[   3   1   4 ]

singular? For which k does A⁻¹ exist?

Working. Expand along row 1:
|A| = 2[(k)(4) − (2)(1)] − 3[(1)(4) − (2)(3)] + 1[(1)(1) − (k)(3)]
= 2(4k − 2) − 3(4 − 6) + (1 − 3k)
= 8k − 4 + 6 + 1 − 3k = 5k + 3.
Singular when 5k + 3 = 0, i.e. k = −3/5.
A⁻¹ exists for every k except −3/5.

Why it works. Keep k symbolic all the way through the expansion, tidy up at the end, and you are left with a simple linear equation. The second part needs no extra work — "inverse exists" is literally the negation of "singular". Two marks for one calculation.

Exam Tip — The phrase that means "set |A| = 0"

All of these instructions are the same instruction wearing different clothes: "find k so that A is singular", "find k so that A⁻¹ does not exist", "find k so that the system has no unique solution", "find k so that the points are collinear", "find k so that the triangle has zero area". Every one of them means |A| = 0. Recognising the disguise is most of the battle.

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Area of a Triangle Using Determinants

The same triangle, two ways: ½ × base × height = ½ × 6 × 4 = 12, and ½ × |determinant| = ½ × |24| = 12.

This is the first genuine application in the syllabus, and it is a lovely one. Given three points on the coordinate plane, you can find the area of the triangle they form with a single 3 × 3 determinant — no need to find side lengths, no need to hunt for a height, no Heron's formula.

Key Rule — Area from coordinates

The triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) has

Area = ½ × |D| where D =

|  x1  y1   1 |
|  x2  y2   1 |
|  x3  y3   1 |

Two details decide your mark. First, the third column is always a column of 1s — write it before you write anything else. Second, the modulus is compulsory: area can never be negative, so if D comes out as −48 the area is 24, not −24.

Why the 1s? Think of it as the determinant keeping track of the fact that a triangle is a two-dimensional shape sitting in a plane. The column of 1s is what turns a "volume of three arrows" calculation into an "area of three points" calculation. You do not need that justification for the exam, but it is nice to know the formula is not arbitrary.

Notice from the figure above that this agrees perfectly with what you learned years ago. The triangle with vertices A(1, 2), B(7, 2) and C(4, 6) has a horizontal base of length 6 and a vertical height of 4, giving ½ × 6 × 4 = 12. The determinant route gives ½ × |24| = 12 as well. The determinant is not a different geometry — it is the same geometry, packaged so that it works even when no side is conveniently horizontal.

Example 20 — The triangle from the figure

Find the area of the triangle with vertices A(1, 2), B(7, 2) and C(4, 6).

Working. Set up the determinant with the 1s column:
D =

|   1   2   1 |
|   7   2   1 |
|   4   6   1 |

Expand along row 1 with signs + − +:
D = 1[(2)(1) − (1)(6)] − 2[(7)(1) − (1)(4)] + 1[(7)(6) − (2)(4)]
= 1(2 − 6) − 2(7 − 4) + 1(42 − 8)
= −4 − 6 + 34 = 24.
Area = ½ × |24| = 12 square units.

Why it works. The base AB runs from (1, 2) to (7, 2), so it is 6 units long and horizontal; C sits 4 units above that line. The school formula gives ½ × 6 × 4 = 12, exactly matching. Whenever a question hands you a horizontal or vertical side, use it as a free check on your determinant.

Example 21 — Nothing convenient about these points

Find the area of the triangle with vertices (3, 1), (−2, 4) and (5, −3).

Working.
D =

|   3   1   1 |
|  -2   4   1 |
|   5  -3   1 |

= 3[(4)(1) − (1)(−3)] − 1[(−2)(1) − (1)(5)] + 1[(−2)(−3) − (4)(5)]
= 3(4 + 3) − 1(−2 − 5) + 1(6 − 20)
= 21 + 7 − 14 = 14.
Area = ½ × |14| = 7 square units.

Why it works. No side is horizontal, no side is vertical, and the coordinates include negatives — and none of that matters to the determinant. That is exactly why this method is worth having. Keep every substitution inside brackets and the negatives look after themselves.

Example 22 — When the determinant comes out negative

Find the area of the triangle with vertices (0, 4), (6, 0) and (−3, −2).

Working.
D =

|   0   4   1 |
|   6   0   1 |
|  -3  -2   1 |

= 0[(0)(1) − (1)(−2)] − 4[(6)(1) − (1)(−3)] + 1[(6)(−2) − (0)(−3)]
= 0 − 4(6 + 3) + (−12 − 0)
= −36 − 12 = −48.
Area = ½ × |−48| = 24 square units.

Why it works. The determinant is −48 because of the order in which the vertices were listed — write the same three points in a different order and you would get +48. The modulus removes that arbitrariness. Never report a negative area, and never quietly change −48 to 48 before taking the modulus; just take the modulus properly and say so in your working.

Example 23 — Working backwards from a known area

The triangle with vertices (k, 0), (4, 0) and (0, 3) has area 9 square units. Find k.

Working.
D =

|   k   0   1 |
|   4   0   1 |
|   0   3   1 |

= k[(0)(1) − (1)(3)] − 0[(4)(1) − (1)(0)] + 1[(4)(3) − (0)(0)]
= −3k − 0 + 12 = 12 − 3k.
Area = ½|12 − 3k| = 9, so |12 − 3k| = 18.
Case 1: 12 − 3k = 18 → −3k = 6 → k = −2.
Case 2: 12 − 3k = −18 → −3k = −30 → k = 10.
k = −2 or k = 10.

Why it works. The modulus that protected you in Example 22 now demands payment: |something| = 18 gives two equations, so there are two answers. Geometrically that is obvious — the third vertex can sit on either side and still make a triangle of the same area. Check k = 10: base from (10, 0) to (4, 0) is 6, height 3, area 9. Check k = −2: base from (−2, 0) to (4, 0) is 6, height 3, area 9. Both genuinely work.

Common Mistake — Losing the third column, or losing the ½

Three errors cost marks here, every single year: (1) forgetting the column of 1s and computing a 3 × 3 of coordinates only; (2) forgetting to halve the determinant; (3) reporting a negative area. Build the habit of writing the empty frame with the 1s column first, then filling in the coordinates, and finish every answer with the words "square units".

Exam Tip — Order of the vertices does not matter

You may list the three points in any order you like. Swapping two rows only flips the sign of D, and the modulus erases that. So write them in whatever order keeps the arithmetic simplest — a row containing a zero at the top, for instance, makes the expansion much shorter.

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Collinearity of Three Points

This section is really a corollary of the last one, and it is almost free. Ask yourself: what shape do three points make if they all sit on one straight line? They make a triangle with no thickness — a triangle of zero area. So the test for collinearity is simply the area formula set equal to zero.

Key Rule — The collinearity test

Three points (x₁, y₁), (x₂, y₂), (x₃, y₃) are collinear if and only if

|  x1  y1   1 |
|  x2  y2   1 |
|  x3  y3   1 |

= 0

Same determinant as the area formula, with the ½ and the modulus dropped — because half of zero is zero, and the modulus of zero is zero. One setup, two questions.

Because it is an "if and only if", the test runs in both directions. If you are asked to prove three points are collinear, evaluate the determinant and show it is 0. If you are told they are collinear and asked for an unknown, set the determinant to 0 and solve. That second version is the one boards prefer, because it tests both the geometry and the algebra in one go.

Example 24 — Proving three points are collinear

Show that the points (2, 3), (4, 7) and (6, 11) lie on one straight line.

Working.
D =

|   2   3   1 |
|   4   7   1 |
|   6  11   1 |

= 2[(7)(1) − (1)(11)] − 3[(4)(1) − (1)(6)] + 1[(4)(11) − (7)(6)]
= 2(7 − 11) − 3(4 − 6) + (44 − 42)
= −8 + 6 + 2 = 0.
Since the determinant is 0, the three points are collinear.

Why it works. A sanity check by slope: from (2, 3) to (4, 7) the gradient is (7 − 3)/(4 − 2) = 2; from (4, 7) to (6, 11) it is (11 − 7)/(6 − 4) = 2. Same gradient from a shared point, so one line. The determinant packages that slope comparison into a single calculation that never divides by zero — which is why it also works for vertical lines, where the slope method breaks down.

Example 25 — Finding the missing x-coordinate

If (3, −2), (k, 4) and (9, 10) are collinear, find k.

Working.

|   3  -2   1 |
|   k   4   1 |
|   9  10   1 |

= 0
Expand along row 1:
3[(4)(1) − (1)(10)] − (−2)[(k)(1) − (1)(9)] + 1[(k)(10) − (4)(9)] = 0
3(4 − 10) + 2(k − 9) + (10k − 36) = 0
−18 + 2k − 18 + 10k − 36 = 0
12k − 72 = 0 → k = 6.

Why it works. The unknown appears in two places, both linearly, so the expansion is linear in k and there is exactly one answer. Check it: the points become (3, −2), (6, 4), (9, 10). Gradient from the first to the second is 6/3 = 2; from the second to the third, 6/3 = 2. Confirmed.

Example 26 — Finding the missing y-coordinate

If (1, 4), (3, k) and (−2, −2) are collinear, find k.

Working.

|   1   4   1 |
|   3   k   1 |
|  -2  -2   1 |

= 0
1[(k)(1) − (1)(−2)] − 4[(3)(1) − (1)(−2)] + 1[(3)(−2) − (k)(−2)] = 0
(k + 2) − 4(3 + 2) + (−6 + 2k) = 0
k + 2 − 20 − 6 + 2k = 0
3k − 24 = 0 → k = 8.

Why it works. The unknown sits in the second column this time, and nothing about the method changes. Check: with the points (1, 4), (3, 8), (−2, −2), the gradient from (1, 4) to (3, 8) is 4/2 = 2, and from (1, 4) to (−2, −2) it is (−6)/(−3) = 2. Collinear.

Exam Tip — Always verify with slopes

Every collinearity answer can be checked in fifteen seconds by comparing two gradients. Do it. It is the cheapest insurance in this chapter, and it also protects you against the classic slip of expanding a determinant that you copied down with a coordinate in the wrong column.

Common Mistake — Setting the area, not the determinant, to zero incorrectly

Some students write "½|D| = 0" and then get tangled in the modulus. There is no need. If the area is zero then D itself is zero, so write D = 0 straight away and expand. Introducing the modulus here only creates confusion — save it for genuine area questions.

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Adjoint of a Square Matrix

Five short steps, each one something you can already do. Work left to right and never skip a station.

We now start assembling the biggest machine in the chapter: the inverse of a matrix. The adjoint is the last part we need before that machine runs, and the diagram above is the map. Notice that the first three stations — matrix, minors, cofactors — are exactly what you practised in the minors and cofactors section. Nothing new is being asked of you.

Key Rule — The adjoint

The adjoint of a square matrix A, written adj A, is the transpose of the matrix of cofactors.

Step 1: find every cofactor A_ij.
Step 2: arrange them in a matrix in their natural positions.
Step 3: transpose it — turn rows into columns.

That third step is not optional decoration. Skip it and every downstream answer is wrong.

For a 2 × 2 there is a shortcut worth memorising, because it appears constantly:

Key Rule — The 2 × 2 adjoint shortcut

If A =

[   a   b ]
[   c   d ]

then adj A =

[   d  -b ]
[  -c   a ]

In words: swap the two main-diagonal entries, and change the sign of the other two. Say it out loud a few times — "swap the diagonal, flip the signs of the others". For 2 × 2 matrices you never need to build a cofactor table again.

Example 27 — Adjoint of a 2 × 2, with proof it is right

Find adj A for A =

[   4   3 ]
[   2   5 ]

and verify that A(adj A) = |A| I.

Working. Swap 4 and 5; flip the signs of 3 and 2:
adj A =

[   5  -3 ]
[  -2   4 ]

Also |A| = (4)(5) − (3)(2) = 20 − 6 = 14.

Now multiply:
A(adj A) =

[   4   3 ]
[   2   5 ]

[   5  -3 ]
[  -2   4 ]

Top-left: (4)(5) + (3)(−2) = 20 − 6 = 14.
Top-right: (4)(−3) + (3)(4) = −12 + 12 = 0.
Bottom-left: (2)(5) + (5)(−2) = 10 − 10 = 0.
Bottom-right: (2)(−3) + (5)(4) = −6 + 20 = 14.
So A(adj A) =

[  14   0 ]
[   0  14 ]

= 14 I = |A| I. Verified.

Why it works. Look closely at where the 14s and the 0s came from. The 14s are matched row-and-own-cofactor sums, which give |A|. The 0s are the mismatched sums from Example 12, which are always zero. The whole adjoint identity is just those two facts arranged in a grid.

Key Rule — The adjoint identity

For any square matrix A of order n:

A(adj A) = (adj A)A = |A| I

This single line is the reason the inverse formula exists. Divide both sides by |A| — legal only when |A| ≠ 0 — and you get A × [(1/|A|)adj A] = I, which says exactly that (1/|A|)adj A is the inverse of A.

Two useful consequences, both examinable-friendly, for an n × n matrix:
• |adj A| = |A|ⁿ⁻¹
• adj(A) for a singular matrix still exists — you can always build it — but it will not be an inverse.

Example 28 — Adjoint of a 3 × 3 (the transpose really matters here)

Find adj A for A =

[   3   1   2 ]
[   2  -3  -1 ]
[   1   2   4 ]

and verify A(adj A) = |A| I.

Working. We already built the cofactor matrix in Example 11:
cofactor matrix =

[ -10   -9    7 ]
[   0   10   -5 ]
[   5    7  -11 ]

Now transpose it — row 1 becomes column 1, and so on:
adj A =

[ -10    0    5 ]
[  -9   10    7 ]
[   7   -5  -11 ]

We also know |A| = −25. Multiplying out A(adj A) gives

[ -25    0    0 ]
[   0  -25    0 ]
[   0    0  -25 ]

= −25 I = |A| I. Verified.

(Check one entry yourself so you trust it. Row 1 of A times column 1 of adj A: (3)(−10) + (1)(−9) + (2)(7) = −30 − 9 + 14 = −25. Row 1 of A times column 2 of adj A: (3)(0) + (1)(10) + (2)(−5) = 0 + 10 − 10 = 0.)

Why it works. Compare the cofactor matrix with the adjoint carefully. The −9 that sat in position (1, 2) has moved to position (2, 1); the 0 has moved from (2, 1) to (1, 2). If you forget the transpose, those two entries stay swapped and your inverse silently becomes wrong — the arithmetic will look tidy and the answer will still be worth zero.

Example 29 — Determinant of an adjoint, with no matrix given

(a) If A is a 3 × 3 matrix with |A| = 5, find |adj A|. (b) If B is a 3 × 3 matrix with |B| = −3, find |adj B|. (c) If C is 2 × 2 with |C| = 7, find |adj C|.

Working. Use |adj A| = |A|ⁿ⁻¹.
(a) n = 3, so the power is 2: |adj A| = 5² = 25.
(b) n = 3: |adj B| = (−3)² = 9.
(c) n = 2, so the power is 1: |adj C| = 7¹ = 7.

Why it works. Take determinants on both sides of A(adj A) = |A| I. The left gives |A| × |adj A|; the right gives |A|ⁿ × |I| = |A|ⁿ. Cancel one |A| and you are left with |adj A| = |A|ⁿ⁻¹. Notice part (c): for a 2 × 2, the adjoint has the same determinant as the original. That surprises almost everyone the first time.

Common Mistake — Building the adjoint from minors instead of cofactors

There are two separate steps between the minors and the adjoint: attach the signs, then transpose. Students routinely do one and forget the other. Write the words "signs" and "transpose" as a two-item checklist in your margin and tick them off. Also, only ever transpose once — transposing twice returns you to where you started.

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Inverse of a Square Matrix

Here is the payoff. In ordinary numbers, the inverse of 5 is 1/5, because 5 × (1/5) = 1. Matrices work the same way, except that "1" is replaced by the identity matrix I, and "dividing" is replaced by multiplying by the inverse. There is no such thing as dividing by a matrix — only multiplying by its inverse.

Key Rule — The inverse formula

If A is a square matrix with |A| ≠ 0, then

A⁻¹ = (1 / |A|) × adj A

and it satisfies AA⁻¹ = A⁻¹A = I.

If |A| = 0 the matrix is singular and A⁻¹ does not exist — because you would be dividing by zero. So the very first line of every inverse question you ever write should be the value of |A|.

Think of it as three sentences you write every time, in the same order. "|A| = … , which is not zero, so A⁻¹ exists." Then "adj A = …". Then "A⁻¹ = (1/|A|) adj A = …". Examiners award marks along that path, so even if your arithmetic wobbles at the end, a clearly signposted method keeps most of the marks.

Example 30 — Inverse of a 2 × 2, done in three lines

Find A⁻¹ for A =

[   4   3 ]
[   2   5 ]

and check your answer.

Working.
Step 1: |A| = (4)(5) − (3)(2) = 20 − 6 = 14. Since 14 ≠ 0, A⁻¹ exists.
Step 2: adj A =

[   5  -3 ]
[  -2   4 ]

(swap the diagonal, flip the other signs).
Step 3: A⁻¹ = (1/14)

[   5  -3 ]
[  -2   4 ]

[  5/14  -3/14 ]
[  -1/7    2/7 ]

Check. AA⁻¹ should be I. Top-left: (4)(5/14) + (3)(−1/7) = 20/14 − 3/7 = 10/7 − 3/7 = 1. Top-right: (4)(−3/14) + (3)(2/7) = −6/7 + 6/7 = 0. Good.

Why it works. Leaving the answer as (1/14) times a clean integer matrix is perfectly acceptable and far safer than distributing the fraction into all four entries. Most examiners prefer the factored form, and you will make fewer mistakes.

Example 31 — Inverse of a 3 × 3, the full assembly line

Find A⁻¹ for A =

[   1   2  -1 ]
[   3   0   2 ]
[  -2   1   1 ]

Step 1 — cofactors.
A₁₁ = +|0 2; 1 1| = 0 − 2 = −2 • A₁₂ = −|3 2; −2 1| = −(3 + 4) = −7 • A₁₃ = +|3 0; −2 1| = 3 − 0 = 3
A₂₁ = −|2 −1; 1 1| = −(2 + 1) = −3 • A₂₂ = +|1 −1; −2 1| = 1 − 2 = −1 • A₂₃ = −|1 2; −2 1| = −(1 + 4) = −5
A₃₁ = +|2 −1; 0 2| = 4 − 0 = 4 • A₃₂ = −|1 −1; 3 2| = −(2 + 3) = −5 • A₃₃ = +|1 2; 3 0| = 0 − 6 = −6

Step 2 — determinant (row 1 elements times their own cofactors):
|A| = (1)(−2) + (2)(−7) + (−1)(3) = −2 − 14 − 3 = −19. Not zero, so the inverse exists.

Step 3 — cofactor matrix, then transpose.
cofactor matrix =

[  -2   -7    3 ]
[  -3   -1   -5 ]
[   4   -5   -6 ]

→ adj A =

[  -2   -3    4 ]
[  -7   -1   -5 ]
[   3   -5   -6 ]

Step 4 — divide by the determinant.
A⁻¹ = (1/−19)

[  -2   -3    4 ]
[  -7   -1   -5 ]
[   3   -5   -6 ]

= (1/19)

[   2    3   -4 ]
[   7    1    5 ]
[  -3    5    6 ]

Why it works. Dividing by a negative determinant flips every sign, so it is much tidier to pull the minus inside and present a positive 1/19 outside. Do that flip in one deliberate step and state it, rather than letting stray minus signs wander through nine entries. Quick check: row 1 of A times column 1 of A⁻¹ = (1/19)[(1)(2) + (2)(7) + (−1)(−3)] = (1/19)(2 + 14 + 3) = 1. Correct.

Example 32 — Reusing work you have already done

Find A⁻¹ for A =

[   3   1   2 ]
[   2  -3  -1 ]
[   1   2   4 ]

Working. From Examples 11 and 28 we already have |A| = −25 and adj A =

[ -10    0    5 ]
[  -9   10    7 ]
[   7   -5  -11 ]

So A⁻¹ = (1/−25) × adj A = (1/25)

[  10    0   -5 ]
[   9  -10   -7 ]
[  -7    5   11 ]

Check. Row 1 of A times column 1 of A⁻¹ = (1/25)[(3)(10) + (1)(9) + (2)(−7)] = (1/25)(30 + 9 − 14) = 25/25 = 1. Row 1 of A times column 2 = (1/25)[(3)(0) + (1)(−10) + (2)(5)] = (1/25)(0 − 10 + 10) = 0. Both as they should be.

Why it works. A well-organised answer sheet is a real asset. Because you kept the cofactor table from earlier, this five-mark question collapsed into two lines of writing. In the exam, if one part of a question gives you |A| or adj A, look hard for a later part that reuses it — the paper is usually built that way on purpose.

Key Rule — Four inverse facts worth carrying

• (A⁻¹)⁻¹ = A — undoing the undoing gets you back.
• (AB)⁻¹ = B⁻¹A⁻¹ — the order reverses. Socks then shoes; to undo, shoes off then socks off.
• |A⁻¹| = 1 / |A|.
• The inverse, when it exists, is unique. There is only one right answer, so a verification check is always meaningful.

Common Mistake — Writing A⁻¹ = adj A / A

You divide by the determinant, a single number, never by the matrix itself. "Matrix divided by matrix" has no meaning. Similarly, never write B/A when you mean BA⁻¹ or A⁻¹B — and note those two are generally different, because matrix multiplication does not commute. Which side you multiply on matters enormously in the next section.

Exam Tip — Verify one entry, not nine

A full AA⁻¹ = I check takes far too long under exam pressure. Instead, compute one diagonal entry (it should come out as 1) and one off-diagonal entry (it should come out as 0). Two dot products, about thirty seconds, and it catches the overwhelming majority of arithmetic slips.

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Solving Linear Equations by the Matrix Method

This is the destination the whole chapter has been walking towards, and it is where the biggest marks live. You will be handed two or three simultaneous equations and asked to solve them using the inverse of a matrix. That exact phrase — the matrix method — is what the 2026–27 syllabus names. Cramer's rule has been removed from the course, so do not spend a minute learning it; if a question says "solve using matrices", it wants A⁻¹B.

Key Rule — The matrix method, in four steps

Any system such as
a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃
can be written as AX = B, where A holds the coefficients, X = (x, y, z) as a column, and B holds the right-hand sides as a column.

1. Write A, X and B, and compute |A|.
2. If |A| ≠ 0, say so — a unique solution exists.
3. Find A⁻¹ = (1/|A|) adj A.
4. Compute X = A⁻¹B and read off x, y, z.

Why does step 4 look like that? Start from AX = B and multiply on the left of both sides by A⁻¹. That gives A⁻¹AX = A⁻¹B, and since A⁻¹A = I and IX = X, you are left with X = A⁻¹B. The word "left" matters: BA⁻¹ is not the same thing and, for a column B, is not even defined. Always write A⁻¹B, in that order.

Common Mistake — Equations not in standard form

Before you build A, every equation must be arranged as (terms in x) + (terms in y) + (terms in z) = constant, with the variables in the same order in every row. If a question gives you 3x = 2y + 7, rewrite it as 3x − 2y + 0z = 7 first. And if a variable is missing from an equation, its coefficient is 0, not blank — that zero must appear in A.

Example 33 — Two variables, step by step

Solve 2x + 3y = 13 and 3x − y = 3 by the matrix method.

Step 1. AX = B with A =

[   2   3 ]
[   3  -1 ]

, X =

[ x ]
[ y ]

, B =

[  13 ]
[   3 ]

Step 2. |A| = (2)(−1) − (3)(3) = −2 − 9 = −11 ≠ 0, so a unique solution exists.
Step 3. adj A =

[  -1  -3 ]
[  -3   2 ]

, so A⁻¹ = (1/−11)

[  -1  -3 ]
[  -3   2 ]

= (1/11)

[   1   3 ]
[   3  -2 ]

Step 4. X = A⁻¹B = (1/11)

[   1   3 ]
[   3  -2 ]

[  13 ]
[   3 ]

= (1/11)

[  13 + 9 ]
[  39 – 6 ]

= (1/11)

[  22 ]
[  33 ]

[  2 ]
[  3 ]

Answer: x = 2, y = 3.
Check. 2(2) + 3(3) = 4 + 9 = 13 ✓ and 3(2) − 3 = 6 − 3 = 3 ✓

Why it works. Every step is something you have already practised: a 2 × 2 determinant, the swap-and-flip adjoint, and one matrix multiplication. Substituting back into the original equations costs twenty seconds and turns a hopeful answer into a certain one.

Example 34 — Another two-variable system

Solve 3x + 2y = 7 and 5x − 4y = −3 using matrices.

Working. A =

[   3   2 ]
[   5  -4 ]

, B =

[   7 ]
[  -3 ]

|A| = (3)(−4) − (2)(5) = −12 − 10 = −22 ≠ 0.
adj A =

[  -4  -2 ]
[  -5   3 ]

so A⁻¹ = (1/−22)

[  -4  -2 ]
[  -5   3 ]

= (1/22)

[   4   2 ]
[   5  -3 ]

X = (1/22)

[   4   2 ]
[   5  -3 ]

[   7 ]
[  -3 ]

= (1/22)

[  28 – 6 ]
[  35 + 9 ]

= (1/22)

[  22 ]
[  44 ]

[  1 ]
[  2 ]

Answer: x = 1, y = 2. Check: 3(1) + 2(2) = 7 ✓ and 5(1) − 4(2) = 5 − 8 = −3 ✓

Why it works. Watch the two negatives in step 3 — a negative determinant and a mostly negative adjoint. Pulling the minus sign out and writing a clean positive 1/22 makes the final multiplication far less error-prone. Get in the habit of tidying the sign before you multiply, never during.

Example 35 — Three variables, the classic board question

Solve, by the matrix method:
x + y + z = 6, x − y + z = 2, 2x + y − z = 1.

Step 1. A =

[   1   1   1 ]
[   1  -1   1 ]
[   2   1  -1 ]

, B =

[   6 ]
[   2 ]
[   1 ]

Step 2 — cofactors.
A₁₁ = +|−1 1; 1 −1| = 1 − 1 = 0 • A₁₂ = −|1 1; 2 −1| = −(−1 − 2) = 3 • A₁₃ = +|1 −1; 2 1| = 1 + 2 = 3
A₂₁ = −|1 1; 1 −1| = −(−1 − 1) = 2 • A₂₂ = +|1 1; 2 −1| = −1 − 2 = −3 • A₂₃ = −|1 1; 2 1| = −(1 − 2) = 1
A₃₁ = +|1 1; −1 1| = 1 + 1 = 2 • A₃₂ = −|1 1; 1 1| = −(1 − 1) = 0 • A₃₃ = +|1 1; 1 −1| = −1 − 1 = −2

Step 3. |A| = (1)(0) + (1)(3) + (1)(3) = 6 ≠ 0, so there is a unique solution.
adj A = transpose of

[   0    3    3 ]
[   2   -3    1 ]
[   2    0   -2 ]

[   0    2    2 ]
[   3   -3    0 ]
[   3    1   -2 ]

Step 4. X = (1/6)(adj A)B:
Row 1: (0)(6) + (2)(2) + (2)(1) = 0 + 4 + 2 = 6 → x = 6/6 = 1
Row 2: (3)(6) + (−3)(2) + (0)(1) = 18 − 6 + 0 = 12 → y = 12/6 = 2
Row 3: (3)(6) + (1)(2) + (−2)(1) = 18 + 2 − 2 = 18 → z = 18/6 = 3

Answer: x = 1, y = 2, z = 3.
Check. 1 + 2 + 3 = 6 ✓ 1 − 2 + 3 = 2 ✓ 2(1) + 2 − 3 = 1 ✓

Why it works. A five- or six-mark question, and every single line of it is a skill from an earlier section. Notice the layout trick in step 4: instead of writing out a fraction inside every entry, multiply (adj A) by B first and divide by |A| at the very end. Fewer fractions, fewer mistakes.

Example 36 — Three variables with a negative determinant

Solve: 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4.

Working. A =

[   3  -2   3 ]
[   2   1  -1 ]
[   4  -3   2 ]

, B =

[   8 ]
[   1 ]
[   4 ]

Cofactors: A₁₁ = −1, A₁₂ = −8, A₁₃ = −10, A₂₁ = −5, A₂₂ = −6, A₂₃ = 1, A₃₁ = −1, A₃₂ = 9, A₃₃ = 7.
|A| = (3)(−1) + (−2)(−8) + (3)(−10) = −3 + 16 − 30 = −17 ≠ 0.
adj A =

[   -1   -5   -1 ]
[   -8   -6    9 ]
[  -10    1    7 ]

(adj A)B: Row 1: (−1)(8) + (−5)(1) + (−1)(4) = −8 − 5 − 4 = −17 → x = −17/−17 = 1
Row 2: (−8)(8) + (−6)(1) + (9)(4) = −64 − 6 + 36 = −34 → y = −34/−17 = 2
Row 3: (−10)(8) + (1)(1) + (7)(4) = −80 + 1 + 28 = −51 → z = −51/−17 = 3

Answer: x = 1, y = 2, z = 3. Check in equation 3: 4(1) − 3(2) + 2(3) = 4 − 6 + 6 = 4 ✓

Why it works. The determinant is negative and so is most of the adjoint, yet the answers are clean positive integers because negative divided by negative is positive. Do not let a screen full of minus signs rattle you — work steadily, keep every product in brackets, and divide at the end.

Exam Tip — Always substitute back

Once you have x, y and z, put them into the original equations — all of them, not just one. It takes under a minute and it is the only way to be certain. If a value fails, the fault is nearly always a single cofactor sign, and you now know exactly where to look.

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Consistency, Inconsistency and Number of Solutions

Two questions, three outcomes. Learn this shape and you will never guess at a consistency question again.

Not every system of equations behaves. Two straight lines usually cross at one point — but they might be parallel and never meet, or they might be the very same line lying on top of each other. Those three pictures are exactly the three cases below, and the determinant tells you which one you are looking at.

Key Rule — The vocabulary

A system is consistent if it has at least one solution — that includes both "exactly one" and "infinitely many".
A system is inconsistent if it has no solution at all.

So "consistent" does not mean "one answer". It means "an answer exists". Read the question's wording carefully — asking whether a system is consistent is a different question from asking for the number of solutions.

Test on \|A\|	Test on (adj A)B	Verdict	What you must do
\|A\| ≠ 0	not needed	Consistent — unique solution	Solve it: X = A⁻¹B
\|A\| = 0	(adj A)B ≠ O	Inconsistent — no solution	State it. Do not solve.
\|A\| = 0	(adj A)B = O	No solution OR infinitely many	Inspect the equations to decide which

Key Rule — What the syllabus expects of you here

You are required to solve only the systems that have a unique solution, using the inverse of a matrix. For the other two cases you must be able to identify what is happening — using |A| and (adj A)B — and say so clearly. You are not expected to produce a general solution for an infinitely-many case. Recognising and naming the case is the full answer.

Example 37 — Two parallel lines (no solution)

Examine the consistency of x + 2y = 3 and 2x + 4y = 7.

Working. A =

[   1   2 ]
[   2   4 ]

, B =

[   3 ]
[   7 ]

|A| = (1)(4) − (2)(2) = 4 − 4 = 0. So there is no unique solution — move to the second test.
adj A =

[   4  -2 ]
[  -2   1 ]

(adj A)B =

[  (4)(3) + (-2)(7) ]
[ (-2)(3) +  (1)(7) ]

[  12 – 14 ]
[  -6 +  7 ]

[  -2 ]
[   1 ]

≠ O

Conclusion: the system is inconsistent — it has no solution.

Why it works. Look at the equations directly. Doubling the first gives 2x + 4y = 6, but the second says 2x + 4y = 7. The same quantity cannot be both 6 and 7. Geometrically these are two parallel lines that never meet. The determinant test simply detects that clash without you having to spot it by eye.

Example 38 — The same line twice (infinitely many)

Examine the consistency of x + 2y = 3 and 2x + 4y = 6.

Working. A is the same as before, so |A| = 0 and adj A =

[   4  -2 ]
[  -2   1 ]

, but now B =

[   3 ]
[   6 ]

(adj A)B =

[  12 – 12 ]
[  -6 +  6 ]

[   0 ]
[   0 ]

= O

So we are in the third row of the table — either no solution or infinitely many. Now inspect: the second equation is exactly 2 × the first, so the two equations say the same thing. Every point on the line x + 2y = 3 works.

Conclusion: the system is consistent with infinitely many solutions.

Why it works. Only the constants changed between Examples 37 and 38, and that flipped the answer from "no solution" to "infinitely many". That is precisely why the (adj A)B test is needed — |A| = 0 on its own cannot tell those two situations apart. And notice that even after the test, the last step was plain observation, exactly as the syllabus intends.

Example 39 — A three-variable inconsistent system

Examine the consistency of x + y + z = 1, 2x + 2y + 2z = 5, x − y + z = 0.

Working. A =

[   1   1   1 ]
[   2   2   2 ]
[   1  -1   1 ]

, B =

[   1 ]
[   5 ]
[   0 ]

Row 2 = 2 × Row 1, so |A| = 0 immediately — no calculation needed.
Building the cofactors gives adj A =

[   4   -2    0 ]
[   0    0    0 ]
[  -4    2    0 ]

(adj A)B: Row 1: (4)(1) + (−2)(5) + (0)(0) = 4 − 10 = −6. Row 2: 0. Row 3: (−4)(1) + (2)(5) + 0 = −4 + 10 = 6.
(adj A)B = (−6, 0, 6)′ ≠ O.

Conclusion: inconsistent — no solution.

Why it works. Confirm it by eye: equation 1 says x + y + z = 1, so 2x + 2y + 2z must be 2 — but equation 2 insists it is 5. A flat contradiction. Spotting R₂ = 2R₁ saved you a full 3 × 3 expansion, and it is worth scanning for that pattern before you start any consistency question.

Example 40 — Finding the value that breaks a system

For what value of λ does the system x + y + z = 2, x + 2y + 3z = 5, x + 4y + λz = 10 fail to have a unique solution?

Working. A unique solution exists exactly when |A| ≠ 0, so we want |A| = 0.
|A| =

|   1   1   1 |
|   1   2   3 |
|   1   4   L |

(writing L for λ)
= 1[(2)(L) − (3)(4)] − 1[(1)(L) − (3)(1)] + 1[(1)(4) − (2)(1)]
= (2L − 12) − (L − 3) + (4 − 2)
= 2L − 12 − L + 3 + 2 = L − 7.
So |A| = 0 when λ = 7; for every other value of λ the system has a unique solution.

Why it works. Carry the unknown symbolically right through the expansion and only tidy up at the end — the algebra stays simple because λ appears just once in the matrix. This is one of the most popular one- and two-mark questions in the whole unit, and it takes about ninety seconds once you recognise it.

Exam Tip — "Homogeneous" systems, in one line

If every right-hand side is zero (B = O), the system always has at least the solution x = y = z = 0, called the trivial solution — so it can never be inconsistent. If |A| ≠ 0 that trivial solution is the only one; if |A| = 0 there are infinitely many. Note that (adj A)B = O automatically here, which is why B = O never lands you in the "no solution" box.

Common Mistake — Trying to invert a singular matrix

If |A| = 0, then A⁻¹ does not exist and writing X = A⁻¹B is meaningless. Every year students compute |A| = 0, notice nothing, and carry on dividing by zero for another page. The instant you write |A| = 0, stop, switch to the (adj A)B test, and answer the consistency question instead.

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Word Problems Solved by the Matrix Method

Boards love dressing the matrix method up in a story: shopping bills, ages, three unknown numbers, marks in three subjects. The mathematics is identical to the last two sections. The only new skill is translation — turning English into three equations without losing anything on the way.

Key Rule — Translating a word problem

1. Name your variables first, in writing, with units: "Let x be the price of one pen in rupees." Do not skip this — it is worth a mark and it stops you confusing yourself later.
2. Turn each sentence into one equation, keeping x, y, z in the same order every time.
3. Build A, X, B; check |A| ≠ 0; solve X = A⁻¹B.
4. Answer the actual question in words, with units. A bare column of numbers is an incomplete answer.

Example 41 — The stationery shop

Three students buy from the same shop. The first buys 2 pens, 3 notebooks and 1 eraser for Rs 115. The second buys 3 pens, 1 notebook and 2 erasers for Rs 70. The third buys 1 pen, 2 notebooks and 3 erasers for Rs 85. Find the price of each item using matrices.

Step 1 — name the variables. Let x, y, z be the price in rupees of one pen, one notebook and one eraser.
2x + 3y + z = 115 • 3x + y + 2z = 70 • x + 2y + 3z = 85

Step 2 — matrices. A =

[   2   3   1 ]
[   3   1   2 ]
[   1   2   3 ]

, B =

[  115 ]
[   70 ]
[   85 ]

Step 3 — cofactors and determinant.
A₁₁ = 3 − 4 = −1 • A₁₂ = −(9 − 2) = −7 • A₁₃ = 6 − 1 = 5
A₂₁ = −(9 − 2) = −7 • A₂₂ = 6 − 1 = 5 • A₂₃ = −(4 − 3) = −1
A₃₁ = 6 − 1 = 5 • A₃₂ = −(4 − 3) = −1 • A₃₃ = 2 − 9 = −7
|A| = (2)(−1) + (3)(−7) + (1)(5) = −2 − 21 + 5 = −18 ≠ 0.
The cofactor matrix happens to be symmetric here, so adj A =

[  -1  -7   5 ]
[  -7   5  -1 ]
[   5  -1  -7 ]

Step 4 — solve. (adj A)B:
Row 1: (−1)(115) + (−7)(70) + (5)(85) = −115 − 490 + 425 = −180 → x = −180/−18 = 10
Row 2: (−7)(115) + (5)(70) + (−1)(85) = −805 + 350 − 85 = −540 → y = −540/−18 = 30
Row 3: (5)(115) + (−1)(70) + (−7)(85) = 575 − 70 − 595 = −90 → z = −90/−18 = 5

Answer: a pen costs Rs 10, a notebook Rs 30 and an eraser Rs 5.
Check. 2(10) + 3(30) + 5 = 115 ✓ 3(10) + 30 + 2(5) = 70 ✓ 10 + 2(30) + 3(5) = 85 ✓

Why it works. The numbers are large but the method never changes. Two practical habits: keep the items in the same order (pen, notebook, eraser) in every equation, and do the (adj A)B multiplication before dividing by |A| — that way you handle integers all the way to the last line.

Example 42 — Three unknown numbers

The sum of three numbers is 12. Twice the first, minus the second, plus three times the third is 17. The first plus twice the second, minus the third, is 6. Find the numbers using matrices.

Working. Let the numbers be x, y and z.
x + y + z = 12 • 2x − y + 3z = 17 • x + 2y − z = 6
A =

[   1   1   1 ]
[   2  -1   3 ]
[   1   2  -1 ]

, B =

[  12 ]
[  17 ]
[   6 ]

Cofactors: A₁₁ = 1 − 6 = −5, A₁₂ = −(−2 − 3) = 5, A₁₃ = 4 + 1 = 5, A₂₁ = −(−1 − 2) = 3, A₂₂ = −1 − 1 = −2, A₂₃ = −(2 − 1) = −1, A₃₁ = 3 + 1 = 4, A₃₂ = −(3 − 2) = −1, A₃₃ = −1 − 2 = −3.
|A| = (1)(−5) + (1)(5) + (1)(5) = 5 ≠ 0.
adj A =

[  -5   3   4 ]
[   5  -2  -1 ]
[   5  -1  -3 ]

(adj A)B: Row 1: (−5)(12) + (3)(17) + (4)(6) = −60 + 51 + 24 = 15 → x = 15/5 = 3
Row 2: (5)(12) + (−2)(17) + (−1)(6) = 60 − 34 − 6 = 20 → y = 20/5 = 4
Row 3: (5)(12) + (−1)(17) + (−3)(6) = 60 − 17 − 18 = 25 → z = 25/5 = 5

Answer: the three numbers are 3, 4 and 5.
Check. 3 + 4 + 5 = 12 ✓ 2(3) − 4 + 3(5) = 6 − 4 + 15 = 17 ✓ 3 + 2(4) − 5 = 6 ✓

Why it works. The translation is the only place you can go wrong. Read "twice the first, minus the second, plus three times the third" slowly and write 2x − y + 3z, one phrase at a time. Once the three equations are down correctly, the rest is the machine you have already built.

Exam Tip — The value-based ending

Word problems in this unit often close with a short line about honesty, cooperation or thrift. If the question asks you to comment, write one genuine sentence — it usually carries a mark and costs you fifteen seconds. Do not leave it blank because it "is not maths".

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Exam Strategy and Common Traps

You now know all the mathematics in this chapter. This last section is about converting that knowledge into marks under time pressure, which is a separate skill and worth practising on its own.

Key Rule — The order of operations for any determinants question

1. Read the question twice and underline the exact word: minor or cofactor? area or collinear? consistency or solve?
2. Compute |A| before anything else. It decides whether the rest of the question is even possible.
3. Show every step on its own line. Method marks are awarded even when the final number is wrong — but only if the examiner can see the method.
4. Check: substitute the solution back, or verify one entry of AA⁻¹, or compare two gradients.
5. Finish the sentence: units, "square units", or a clear statement such as "the system is inconsistent".

Example 43 — Find the mistake

A student is asked to find A⁻¹ for A =

[   2   5 ]
[   1   3 ]

and writes:
"|A| = 6 − 5 = 1. adj A =

[   3   5 ]
[   1   2 ]

. So A⁻¹ =

[   3   5 ]
[   1   2 ]

."
Identify the error and give the correct inverse.

Working. The determinant is right: (2)(3) − (5)(1) = 6 − 5 = 1. The adjoint is wrong — the student swapped the diagonal entries but forgot to change the signs of the other two. Correctly, adj A =

[   3  -5 ]
[  -1   2 ]

, so
A⁻¹ = (1/1)

[   3  -5 ]
[  -1   2 ]

[   3  -5 ]
[  -1   2 ]

Check. AA⁻¹: top-left (2)(3) + (5)(−1) = 6 − 5 = 1 ✓; top-right (2)(−5) + (5)(2) = −10 + 10 = 0 ✓. The student's version would have given top-left (2)(3) + (5)(1) = 11, nowhere near 1.

Why it works. This is exactly the check described earlier — two dot products — and it would have caught the error in under thirty seconds. Build the check into your routine and errors like this stop reaching the answer line.

Common Mistake — The seven that cost the most marks

1. Forgetting the middle minus in a 3 × 3 expansion.
2. Confusing a minor with a cofactor.
3. Forgetting to transpose the cofactor matrix when finding adj A.
4. Dropping the column of 1s in the area formula, or forgetting the ½.
5. Reporting a negative area.
6. Carrying on to find A⁻¹ after discovering |A| = 0.
7. Writing X = BA⁻¹ instead of X = A⁻¹B.

Read that list before every practice session for a week and it will stop being a list of your mistakes.

Exam Tip — Where the marks actually are

Unit II (Matrices and Determinants together) is worth 10 marks. Within Determinants, the reliable earners are: a short evaluation or a minor/cofactor question, an area or collinearity question, and one long question that is either "find A⁻¹" or "solve the system by the matrix method". That long question is the one to protect — it is completely predictable, entirely mechanical, and fully within your reach if you practise the four steps until they are automatic.

One last thing before the worksheet. If a section above still feels wobbly, go back to it now rather than pressing on. Determinants is a chapter where every idea sits on the one before it — shaky cofactors means a shaky adjoint, which means a shaky inverse, which means a lost six-mark question. Do not move on until minors and cofactors feel genuinely comfortable. Everything after that is assembly.

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Practice Worksheet

Twelve original questions, arranged easy to hard, covering every sub-topic above. Work each one fully on paper before you open the answer — reading a solution feels like learning but is not. Every answer here has been checked, so if yours disagrees, hunt for the difference; that hunt is where the real learning happens.

Q1. Evaluate

|   7  -3 |
|   4   5 |

Show Answer

|A| = (7)(5) − (−3)(4) = 35 − (−12) = 35 + 12 = 47.

The subtraction of a negative product turns into an addition. Writing (−3)(4) inside brackets is what keeps that safe.

Q2. Find all values of x for which

|   x   2 |
|   8   x |

= 0.

Show Answer

(x)(x) − (2)(8) = 0 → x² − 16 = 0 → x² = 16 → x = 4 or x = −4.

Both roots count. Writing only x = 4 loses half the mark. Check x = −4: rows (−4, 2) and (8, −4), and (−4, 2) = −½ × (8, −4), so the rows are multiples and the determinant is indeed 0.

Q3. Evaluate

|   2  -1   3 |
|   1   4   0 |
|   3   2  -2 |

by expanding along the row or column of your choice.

Show Answer

Row 2 contains a zero, so expand along row 2. Its signs from the chessboard are − + −.
= −1 × |−1 3; 2 −2| + 4 × |2 3; 3 −2| − 0 × |2 −1; 3 2|
= −1[(−1)(−2) − (3)(2)] + 4[(2)(−2) − (3)(3)] − 0
= −1(2 − 6) + 4(−4 − 9)
= −1(−4) + 4(−13) = 4 − 52 = −48.

Expanding along row 1 gives −48 as well — try it as a check. Choosing row 2 saved one whole 2 × 2 evaluation.

Q4. For A =

[   3   1  -2 ]
[   4   0   5 ]
[  -1   2   6 ]

write down the minor M₂₃ and the cofactor A₂₃.

Show Answer

Delete row 2 and column 3. What remains is

|   3   1 |
|  -1   2 |

M₂₃ = (3)(2) − (1)(−1) = 6 + 1 = 7.
i + j = 2 + 3 = 5, which is odd, so A₂₃ = (−1)⁵ × 7 = −7.

Two different answers to two different words. Read which one the question asked for.

Q5. Find the area of the triangle whose vertices are (2, −2), (5, 4) and (−1, 3).

Show Answer

D =

|   2  -2   1 |
|   5   4   1 |
|  -1   3   1 |

= 2[(4)(1) − (1)(3)] − (−2)[(5)(1) − (1)(−1)] + 1[(5)(3) − (4)(−1)]
= 2(4 − 3) + 2(5 + 1) + (15 + 4)
= 2 + 12 + 19 = 33.
Area = ½ × |33| = 33/2 = 16.5 square units.

Do not forget the halving, and finish with the words "square units".

Q6. If the points (2, 5), (k, −3) and (8, 1) are collinear, find k.

Show Answer

Set the determinant to zero:

|   2   5   1 |
|   k  -3   1 |
|   8   1   1 |

= 0
2[(−3)(1) − (1)(1)] − 5[(k)(1) − (1)(8)] + 1[(k)(1) − (−3)(8)] = 0
2(−3 − 1) − 5(k − 8) + (k + 24) = 0
−8 − 5k + 40 + k + 24 = 0
−4k + 56 = 0 → k = 14.

Check by slopes with the points (2, 5), (14, −3), (8, 1): from (2, 5) to (14, −3) the gradient is −8/12 = −2/3; from (2, 5) to (8, 1) it is −4/6 = −2/3. Same gradient, so collinear.

Q7. For A =

[   5   2 ]
[   3   4 ]

find adj A and A⁻¹, and verify that A(adj A) = |A| I.

Show Answer

|A| = (5)(4) − (2)(3) = 20 − 6 = 14 ≠ 0.
adj A =

[   4  -2 ]
[  -3   5 ]

(swap the diagonal, flip the other two signs).
A⁻¹ = (1/14)

[   4  -2 ]
[  -3   5 ]

Verification: A(adj A) — top-left (5)(4) + (2)(−3) = 20 − 6 = 14; top-right (5)(−2) + (2)(5) = −10 + 10 = 0; bottom-left (3)(4) + (4)(−3) = 12 − 12 = 0; bottom-right (3)(−2) + (4)(5) = −6 + 20 = 14. So A(adj A) = 14 I = |A| I. Verified.

Q8. For A =

[   1  -1   2 ]
[   3   0   1 ]
[   2   4  -3 ]

find |A| and A⁻¹.

Show Answer

Cofactors:
A₁₁ = +|0 1; 4 −3| = 0 − 4 = −4 • A₁₂ = −|3 1; 2 −3| = −(−9 − 2) = 11 • A₁₃ = +|3 0; 2 4| = 12 − 0 = 12
A₂₁ = −|−1 2; 4 −3| = −(3 − 8) = 5 • A₂₂ = +|1 2; 2 −3| = −3 − 4 = −7 • A₂₃ = −|1 −1; 2 4| = −(4 + 2) = −6
A₃₁ = +|−1 2; 0 1| = −1 − 0 = −1 • A₃₂ = −|1 2; 3 1| = −(1 − 6) = 5 • A₃₃ = +|1 −1; 3 0| = 0 + 3 = 3

|A| = (1)(−4) + (−1)(11) + (2)(12) = −4 − 11 + 24 = 9 ≠ 0.
Cofactor matrix

[  -4   11   12 ]
[   5   -7   -6 ]
[  -1    5    3 ]

→ transpose → adj A =

[  -4    5   -1 ]
[  11   -7    5 ]
[  12   -6    3 ]

A⁻¹ = (1/9)

[  -4    5   -1 ]
[  11   -7    5 ]
[  12   -6    3 ]

Spot check: row 1 of A times column 1 of A⁻¹ = (1/9)[(1)(−4) + (−1)(11) + (2)(12)] = (1/9)(−4 − 11 + 24) = 9/9 = 1. Correct.

Q9. Solve 2x − 3y = −4 and 4x + y = 20 by the matrix method.

Show Answer

A =

[   2  -3 ]
[   4   1 ]

, B =

[  -4 ]
[  20 ]

|A| = (2)(1) − (−3)(4) = 2 + 12 = 14 ≠ 0, so a unique solution exists.
adj A =

[   1   3 ]
[  -4   2 ]

, so A⁻¹ = (1/14)

[   1   3 ]
[  -4   2 ]

X = A⁻¹B = (1/14)

[  -4 + 60 ]
[  16 + 40 ]

= (1/14)

[  56 ]
[  56 ]

[  4 ]
[  4 ]

x = 4, y = 4. Check: 2(4) − 3(4) = 8 − 12 = −4 ✓ and 4(4) + 4 = 20 ✓

Q10. Examine the consistency of x + y + z = 3, 2x + 2y + 2z = 6, x − y + z = 1. If it is consistent, state how many solutions it has.

Show Answer

A =

[   1   1   1 ]
[   2   2   2 ]
[   1  -1   1 ]

, B =

[   3 ]
[   6 ]
[   1 ]

Row 2 = 2 × Row 1, so |A| = 0 at once. No unique solution — go to the second test.
The cofactor work gives adj A =

[   4   -2    0 ]
[   0    0    0 ]
[  -4    2    0 ]

(adj A)B: Row 1: (4)(3) + (−2)(6) + (0)(1) = 12 − 12 = 0. Row 2: 0. Row 3: (−4)(3) + (2)(6) + 0 = −12 + 12 = 0.
So (adj A)B = O — we are in the third row of the decision table, and must inspect the equations.

Equation 2 is exactly twice equation 1, so it adds nothing. That leaves x + y + z = 3 and x − y + z = 1. Subtracting gives 2y = 2, so y = 1, and then x + z = 2, which has endlessly many solutions (for instance x = 0, z = 2, or x = 5, z = −3).

The system is consistent and has infinitely many solutions. You are not required to write a general solution — identifying the case is the complete answer.

Q11. A is a square matrix of order 3 with |A| = −4. Find (a) |adj A|, (b) |3A|, (c) |A⁻¹|.

Show Answer

(a) |adj A| = |A|ⁿ⁻¹ = (−4)² = 16.
(b) |3A| = 3ⁿ|A| = 3³ × (−4) = 27 × (−4) = −108.
(c) |A⁻¹| = 1/|A| = −1/4.

All three exponents are governed by n = 3. The commonest slip is writing |3A| = 3 × (−4) = −12; the scalar multiplies every one of the three rows, so it appears three times.

Q12. Solve by the matrix method: x + 2y + 3z = 14, 2x − y + z = 3, 3x + y − z = 2.

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A =

[   1   2   3 ]
[   2  -1   1 ]
[   3   1  -1 ]

, B =

[  14 ]
[   3 ]
[   2 ]

Cofactors.
A₁₁ = +|−1 1; 1 −1| = 1 − 1 = 0 • A₁₂ = −|2 1; 3 −1| = −(−2 − 3) = 5 • A₁₃ = +|2 −1; 3 1| = 2 + 3 = 5
A₂₁ = −|2 3; 1 −1| = −(−2 − 3) = 5 • A₂₂ = +|1 3; 3 −1| = −1 − 9 = −10 • A₂₃ = −|1 2; 3 1| = −(1 − 6) = 5
A₃₁ = +|2 3; −1 1| = 2 + 3 = 5 • A₃₂ = −|1 3; 2 1| = −(1 − 6) = 5 • A₃₃ = +|1 2; 2 −1| = −1 − 4 = −5

|A| = (1)(0) + (2)(5) + (3)(5) = 0 + 10 + 15 = 25 ≠ 0, so a unique solution exists.
The cofactor matrix is symmetric here, so adj A =

[   0    5    5 ]
[   5  -10    5 ]
[   5    5   -5 ]

(adj A)B: Row 1: (0)(14) + (5)(3) + (5)(2) = 0 + 15 + 10 = 25 → x = 25/25 = 1
Row 2: (5)(14) + (−10)(3) + (5)(2) = 70 − 30 + 10 = 50 → y = 50/25 = 2
Row 3: (5)(14) + (5)(3) + (−5)(2) = 70 + 15 − 10 = 75 → z = 75/25 = 3

x = 1, y = 2, z = 3.
Check: 1 + 2(2) + 3(3) = 1 + 4 + 9 = 14 ✓ 2(1) − 2 + 3 = 3 ✓ 3(1) + 2 − 3 = 2 ✓

That is the whole chapter, tested. If you scored well, wonderful — now redo the two questions that took you longest, because those are the ones that will slow you down in the exam hall. If you struggled, that is genuinely fine and completely normal: go back to the section that tripped you, work through its examples with the answers covered, and come back tomorrow.

And here is the only progress target that has ever really worked: aim for one more correct question tomorrow than you managed today. Not ten more. One. That is a rise you can actually deliver, every single day, without dread — and a month of one-more-a-day quietly turns a shaky chapter into your safest one. Small, steady, repeated. That is the whole secret.

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Determinants — Class 12 Maths Notes & Practice

Meet Your Tutor

What You'll Learn

Your Game Plan

Study Notes

What a Determinant Really Is

Determinant of a 2 × 2 Matrix

Determinant of a 3 × 3 Matrix

Minors and Cofactors

Row and Column Behaviour: Safe Evaluation Shortcuts

Singular and Non-Singular Matrices

Area of a Triangle Using Determinants

Collinearity of Three Points

Adjoint of a Square Matrix

Inverse of a Square Matrix

Solving Linear Equations by the Matrix Method

Consistency, Inconsistency and Number of Solutions

Word Problems Solved by the Matrix Method

Exam Strategy and Common Traps

Practice Worksheet

Kaizen Close

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