Chapter 6 · Algebra · 2 questions in the paper · 12 min read
Matrices and Determinants
What the syllabus asks for 3
- Types, order and operations on matrices
- Determinants, minors and cofactors
- Adjoint, inverse and solution of linear systems
Matrices carry about two questions: the order of a matrix or a product, a 2 × 2 determinant or inverse, and one property question (singular, symmetric, effect of scalar multiplication).
The Dean's Office model asks "how many elements in a single column of a 5 × 3 matrix". Matrices are written here in rows as [[a, b], [c, d]].
- Order and operations: m\times n, when AB exists and its order, transpose rule.
- Determinants: ad-bc, 3 × 3 expansion, row properties, |kA|=k^n|A|.
- Inverse: exists iff |A|\ne0; 2 × 2 formula; |A^{-1}|=\frac{1}{|A|}, |\text{adj}A|=|A|^{n-1}.
- Linear systems: Cramer's rule and the matrix method.
6.1 Types, order and operations on matrices
- A matrix with m rows and n columns has order m\times n and mn elements; each column has m elements.
- Types: row, column, square, diagonal, scalar, identity I, zero (null), symmetric (A=A^T), skew-symmetric (A=-A^T, diagonal zero).
- A+B needs equal order. AB exists only if columns of A = rows of B; order of AB is (rows of A)×(columns of B). In general AB\ne BA; AI=IA=A.
- (A+B)^T=A^T+B^T and (AB)^T=B^TA^T.


6.2 Determinants, minors and cofactors
- For
[[a, b], [c, d]], |A|=ad-bc. For a 3 × 3, expand along any row: a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13} with cofactor C_{ij}=(-1)^{i+j}M_{ij}. - A determinant is 0 if two rows (or columns) are equal or proportional or a row is all zero. Swapping two rows changes the sign. A triangular matrix has |A|= product of the diagonal.
- |AB|=|A||B|, |A^T|=|A|, and |kA|=k^n|A| for order n.
6.3 Adjoint, inverse and linear systems
- A^{-1}=\frac{\text{adj}A}{|A|} exists iff |A|\ne0 (non-singular). For 2 × 2
[[a, b], [c, d]]: \text{adj}A=[[d, -b], [-c, a]]. - AA^{-1}=I, (AB)^{-1}=B^{-1}A^{-1}, |A^{-1}|=\frac{1}{|A|}, |\text{adj}A|=|A|^{n-1}.
- Cramer's rule for a_1x+b_1y=c_1, a_2x+b_2y=c_2: x=\frac{D_x}{D}, y=\frac{D_y}{D}, D=a_1b_2-a_2b_1, with D\ne0.
6.4 Eight traps in matrices and determinants
Matrix questions reward a clear habit: check the order, then the sign, then the rule. These eight traps cover the model-paper style questions on products, determinants and inverses.
- Check the order before you multiply. A_{m\times n}B_{n\times p} is defined only if the columns of A equal the rows of B, and the product is m\times p (Fig 6.1). A 2\times3 times a 3\times2 gives 2\times2, but a 2\times3 times another 2\times3 is undefined. Always write the orders beside the matrices first.
- AB\neq BA in general. Even when both products exist they are usually different, as PQ and QP show in Fig 6.1. So (A+B)^2=A^2+AB+BA+B^2, not A^2+2AB+B^2. The reversal laws are (AB)^T=B^TA^T and (AB)^{-1}=B^{-1}A^{-1}.
- A zero product does not mean a zero factor. AB=0 can happen with A\neq0 and B\neq0, for example \begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=0. Likewise AB=AC does not give B=C unless A is invertible. Do not cancel matrices as you cancel numbers.
- Only square matrices have determinants, and they obey set rules. |kA|=k^n|A| for an n\times n matrix, so doubling a 2\times2 multiplies the determinant by 4, not 2. Also |AB|=|A||B| and |A^T|=|A|. Swapping two rows changes the sign, and two equal rows give zero (Fig 6.2).
- Cofactor signs follow a checkerboard. The cofactor is C_{ij}=(-1)^{i+j}M_{ij}, so the signs start +,-,+ along the first row. The minor M_{ij} has no sign of its own. When expanding, pick the row or column with the most zeros, as the zero in Fig 6.2 saved a product.
- The adjoint is the transpose of the cofactor matrix. Many students forget the transpose. For a 2\times2 matrix the shortcut is to swap the diagonal entries and negate the other two (Fig 6.3). The key identity is A\,(\operatorname{adj}A)=|A|\,I, and |\operatorname{adj}A|=|A|^{n-1}.
- An inverse exists only when |A|\neq0. Then A^{-1}=\frac{\operatorname{adj}A}{|A|}, and |A^{-1}|=\frac{1}{|A|}. A matrix with determinant zero is singular and has no inverse, as when two rows are proportional. In Fig 6.2 a zero determinant would collapse the square onto a line.
- Solve AX=B with X=A^{-1}B in that order. Writing BA^{-1} is wrong because the product is not commutative. If |A|=0 the system has no unique solution: it has none or infinitely many. Cramer's rule gives each unknown as a ratio of determinants, x=\frac{D_x}{D}.
6.4.1 Worked checks
Q. If A=\begin{pmatrix}1&2&3\\4&5&6\end{pmatrix} and B=\begin{pmatrix}2&1\\0&3\\1&2\end{pmatrix}, find AB.
Answer. AB is 2\times2. Row 1 by column 1: 1\cdot2+2\cdot0+3\cdot1=5. Row 1 by column 2: 1+6+6=13. Row 2 by column 1: 8+0+6=14. Row 2 by column 2: 4+15+12=31. So AB=\begin{pmatrix}5&13\\14&31\end{pmatrix}.
Q. Evaluate \begin{vmatrix}2&0&1\\1&3&2\\1&1&4\end{vmatrix} and say whether the matrix is invertible.
Answer. Expand along row 1: 2(12-2)-0+1(1-3)=20-2=18. Since 18\neq0 the matrix is non-singular, so the inverse exists. Sarrus' rule gives the same value, 24+0+1-3-4-0=18.
Q. Solve 2x+y=7 and x-y=2 using the inverse.
Answer. A=\begin{pmatrix}2&1\\1&-1\end{pmatrix}, |A|=-3, A^{-1}=\frac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}.
Then X=A^{-1}\begin{pmatrix}7\\2\end{pmatrix}=\frac{1}{-3}\begin{pmatrix}-9\\-3\end{pmatrix}=\begin{pmatrix}3\\1\end{pmatrix}. So x=3,y=1, which agrees with Fig 6.3.
Formula and fact sheet
| Item | Result |
|---|---|
| 2\times2 determinant | ad-bc |
| Product order | (m\times n)(n\times p)=m\times p |
| (AB)^T, (AB)^{-1} | B^TA^T, B^{-1}A^{-1} |
| \lvert kA\rvert | k^n\lvert A\rvert |
| \lvert \text{adj}A\rvert | \lvert A\rvert^{n-1} |
Chapter summary
- Order m\times n: m rows, n columns; a column holds m elements.
- AB exists only if inner orders match, and AB\ne BA in general.
- A zero determinant means singular: no inverse, and a row dependency.
- |kA|=k^n|A|: the factor is raised to the order.
- Reverse the order for (AB)^T and (AB)^{-1}.
- Use Cramer's rule when D\ne0.
Practice questions
Set A — Recall (20 questions)
- A matrix of order 5 × 3 has how many elements in a single column?a5b3c4d15
- The determinant of
[[2, 3], [1, 4]]isa11b7c5d1 - The product of a 2 × 3 matrix and a 3 × 4 matrix has ordera3 × 3b2 × 4c4 × 2dnot defined
- A square matrix whose determinant is zero is calledaidentitybsymmetriccorthogonaldsingular
- For a square matrix A of order n, AI equalsaAb0cIdA^2
- The transpose of a 2 × 3 matrix has ordera2 × 3b3 × 2c2 × 2d3 × 3
- (AB)^T equalsaA^TB^TbABcB^TA^TdBA
- The inverse of a square matrix A exists if and only ifaA is symmetricb|A|=0cA is nulld|A|\ne0
- If A=
[[1, 2], [3, 4]]and B=[[0, 1], [1, 0]], the element in row 2, column 1 of A+B isa3b4c5d1 - The determinant of the 3 × 3 identity matrix isa1b0c3d9
- A square matrix is symmetric ifaA=-A^TbA=A^{-1}c|A|=1dA=A^T
- If A is 3 × 3 and |A|=2, then |2A| isa4b8c16d6
- The determinant of a matrix with two identical rows isa0b1c-1d2
- The inverse of the 2 × 2 matrix 2I isa2Ib\frac{1}{2}IcId-2I
- A matrix all of whose elements are zero is called aaidentity matrixbscalar matrixcdiagonal matrixdnull matrix
- The number of elements in a 3 × 4 matrix isa7b10c12d34
- The determinant of
[[1, 2], [2, 4]]isa4b0c8d-4 - In the adjoint of
[[a, b], [c, d]], the element in row 1, column 2 isa-bbbc-cdc - The diagonal elements of a skew-symmetric matrix area1b-1call equaldzero
- |A^{-1}| equalsa|A|b-|A|c\frac{1}{|A|}d|A|^2
Answers and explanations — Set A
1. (a) — a 5 × 3 matrix has 5 rows, so each column holds 5 entries.
2. (c) — 2·4 − 3·1 = 5.
3. (b) — (2×3)(3×4) gives 2 × 4.
4. (d) — zero determinant means singular.
5. (a) — I is the multiplicative identity.
6. (b) — rows and columns swap.
7. (c) — the order reverses.
8. (d) — |A| = 0 means no inverse.
9. (b) — row 2, column 1: 3 + 1 = 4.
10. (a) — diagonal product 1·1·1.
11. (d) — (a) defines skew-symmetric.
12. (c) — 2³ × 2 = 16, not 2 × 2 = 4.
13. (a) — equal rows make the determinant vanish.
14. (b) — (2I)(½I) = I.
15. (d) — null or zero matrix.
16. (c) — 3 × 4 = 12.
17. (b) — 1·4 − 2·2 = 0, rows proportional.
18. (a) — adj has −b in the top right.
19. (d) — aᵢᵢ = −aᵢᵢ forces 0.
20. (c) — |A||A⁻¹| = |I| = 1.
Set B — Understanding (12 questions)
- If the determinant of
[[x, 2], [3, 4]]is -2, then x isa3b2c-1d1 - For A=
[[1, 2], [3, 4]], the element in row 1, column 1 of A^2 isa1b7c10d15 - For A=
[[1, 2], [0, 1]]and B=[[1, 0], [3, 1]], the element in row 2, column 1 of AB isa3b7c1d0 - In the inverse of
[[3, 1], [5, 2]], the element in row 1, column 2 isa1b5c-1d3 - The matrix
[[k, 2], [3, 6]]is singular when k equalsa0b2c3d1 - If A is 3 × 3 and |A|=5, then |\text{adj}A| isa25b5c125d15
- If A is 3 × 2 and B is 2 × 3, the order of AB isa2 × 2b3 × 2c3 × 3d2 × 3
- For the system x+y=5, x-y=1, the value of x isa2b3c4d1
- The determinant of
[[2, 5, 1], [0, 3, 7], [0, 0, 4]]isa24b14c9d0 - If A is 2 × 2 with |A|=3, then |3A| isa9b81c6d27
- The cofactor of the element in row 1, column 2 of
[[1, 2], [3, 4]]isa3b-3c4d-4 - If
[[a, 0], [0, a]]squared equals I, a possible value of a isa0b2c-1d3
Answers and explanations — Set B
1. (d) — 4x − 6 = −2, x = 1.
2. (b) — 1·1 + 2·3 = 7.
3. (a) — row 2 of A is (0, 1); column 1 of B is (1, 3); 0 + 3 = 3.
4. (c) — |A| = 1; inverse is [[2, -1], [-5, 3]].
5. (d) — 6k − 6 = 0, k = 1.
6. (a) — |A|^(3−1) = 25.
7. (c) — (3×2)(2×3) = 3 × 3, while BA would be 2 × 2.
8. (b) — adding the equations gives 2x = 6.
9. (a) — triangular: 2·3·4 = 24.
10. (d) — 3² × 3 = 27.
11. (b) — C₁₂ = −(minor 3) = −3.
12. (c) — a² = 1, so a = ±1; only −1 is listed.
Set C — Application (8 questions)
- Solving 2x+3y=8 and x+y=3 by Cramer's rule gives x+2y equal toa4b5c6d3
- If A is a 3 × 3 matrix with |A|=-2, then |A^{-1}| isa2b-2c\frac{1}{2}d-\frac{1}{2}
- If A=
[[1, 1], [0, 1]], the element in row 1, column 2 of A^3 isa3b1c2d6 - The determinant of
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]isa1b45c0d-6 - For A=
[[2, 1], [1, 1]], the element in row 1, column 1 of A^{-1} isa2b-1c\frac{1}{2}d1 - The product of
[[1, 2], [3, 4]]and the column[[1], [1]]is a column whose entries add up toa7b10c3d14 - If the determinant of
[[x, 2], [3, x]]is 3, a possible value of x isa9b6c3d0 - If A is the 2 × 2 matrix with a_{ij}=i+j, then |A| isa-1b1c0d17
Answers and explanations — Set C
1. (b) — D = −1, Dₓ = −1, so x = 1; y = 2 and x + 2y = 5.
2. (d) — 1/|A| = −1/2.
3. (a) — each multiplication adds 1 to the corner entry.
4. (c) — rows are in AP, so row 3 − row 2 = row 2 − row 1.
5. (d) — |A| = 1 and adj has d = 1 at the top left.
6. (b) — the column is (3, 7).
7. (c) — x² − 6 = 3 gives x = ±3.
8. (a) — [[2, 3], [3, 4]]: 8 − 9 = −1.