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Class 12 Maths · Chapter 3 · Bihar Board

Matrices & Operations

Master matrix fundamentals with visual diagrams, step-by-step solutions with marking scheme, MCQs, Assertion-Reasoning, and previous year questions.

9 Types of Matrices 6 Visual Diagrams Step Marking
9Matrix Types
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

What is a matrix? Let's visualize it.

3.1 Introduction to Matrices

A matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called elements or entries of the matrix.

Definition — Matrix

A matrix is an ordered rectangular array of numbers (or functions) enclosed in brackets $[ \ ]$ or $( \ )$.

We denote matrices by capital letters like A, B, C, etc. A matrix with $m$ rows and $n$ columns is called an $m \times n$ matrix.

3.2 Order of a Matrix

If a matrix has $m$ rows and $n$ columns, its order is $m \times n$ (read as "m by n").

$$\text{Order} = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n$$

Total number of elements in an $m \times n$ matrix = $m \times n$

Memory Trick — R × C

Always write order as Rows × Columns (R × C).

  • A $\mathbf{2 \times 3}$ matrix has 2 rows and 3 columns = 6 elements
  • A $\mathbf{3 \times 3}$ matrix has 3 rows and 3 columns = 9 elements

Elements are denoted by $a_{ij}$ where $i$ = row number, $j$ = column number. So $a_{23}$ means the element in 2nd row and 3rd column.

Section 02

🎨 Visual Diagrams

See how matrices look and behave

V1 Anatomy of a Matrix

Rows, Columns & Elements ← Row 1 ← Row 2 ← Row 3 Col 1 Col 2 Col 3 a₁₁ a₁₂ a₁₃ a₂₁ a₂₂ a₂₃ a₃₁ a₃₂ a₃₃ Order: 3 × 3 | Total Elements: 9
Each element is identified by its position $a_{ij}$ — where $i$ is the row number and $j$ is the column number.

V2 Six Common Matrix Types — Visualized

Shape & Structure Comparison Row Matrix (1×4) Column Matrix (4×1) Square Matrix (3×3) Diagonal (3×3) Identity Matrix (3×3) 1 1 1 Zero Matrix (3×3) 0 0 0 0 0 0 0 0 0 Scalar Matrix (3×3) k k k
Different types of matrices based on shape and structure. Green/cyan cells highlight non-zero elements.

V3 Matrix Multiplication — Row × Column

How to compute $c_{ik}$ A (2×3) 1 -1 2 0 3 4 × B (3×2) 2 7 5 -1 1 2 = C (2×2) 15 13 -2 17 -13 c₁₁ = (1)(2) + (-1)(-1) + (2)(5) = 2 + 1 + 10 = 13 c₁₂ = (1)(7) + (-1)(1) + (2)(-4) = 7 - 1 - 8 = -2 ⚠ Rule: Columns of A must equal Rows of B for product to exist
To find the $(i, k)$ element of the product $AB$, take the $i$-th row of A and $k$-th column of B, multiply element-wise, and sum them up.

V4 Transpose — Rows become Columns

$A^T$ — Flipping across the diagonal A (2×3) 1 2 3 4 5 6 → Transpose Aᵀ (3×2) 1 4 — 2 5 — 3 6 — Row 1 of A becomes Column 1 of Aᵀ, Row 2 becomes Column 2, etc.
The transpose of a matrix is obtained by interchanging its rows and columns. Order changes from $m \times n$ to $n \times m$.

V5 Symmetric vs Skew-Symmetric

Mirror properties across diagonal Symmetric: A = Aᵀ 1 2 3 2 5 4 3 4 6 Skew-Symmetric: A = -Aᵀ 0 2 3 -2 0 4 -3 -4 0
Symmetric: mirror elements across diagonal are equal. Skew-Symmetric: mirror elements are negatives, and diagonal is always zero.

V6 Matrix Addition — Same Order Required

Element-wise addition A 1 3 2 4 + B 5 2 1 3 = A + B 6 5 3 7 A + B = [aᵢⱼ + bᵢⱼ] — corresponding elements added ⚠ Only possible if A and B have SAME order!
Matrix addition is done element-by-element. Both matrices MUST have the same order, otherwise addition is undefined.
Section 03

Types of Matrices

Learn all 9 types with conditions and examples

T1 Complete Classification

TypeConditionExample
Row MatrixOnly 1 row ($1 \times n$)$[1 \ \ 2 \ \ 3]$
Column MatrixOnly 1 column ($m \times 1$)$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
Square MatrixRows = Columns ($m = n$)$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
Diagonal MatrixNon-diagonal elements = 0$\begin{bmatrix} 2 & 0 \\ 0 & 5 \end{bmatrix}$
Scalar MatrixDiagonal has same value $k \neq 0$$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
Identity MatrixDiagonal = 1, rest = 0$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Zero MatrixAll elements = 0$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Symmetric$A^T = A$$\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
Skew-Symmetric$A^T = -A$ (diagonal = 0)$\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$
Section 04

Matrix Operations

Addition, Scalar Multiplication, Multiplication, Transpose

O1 Addition of Matrices

Two matrices can be added only if they have the same order. The result is obtained by adding corresponding elements.

$$(A + B)_{ij} = a_{ij} + b_{ij}$$
Key Properties
  • Commutative: $A + B = B + A$
  • Associative: $(A + B) + C = A + (B + C)$
  • Additive Identity: $A + O = O + A = A$ (O = zero matrix)
  • Additive Inverse: $A + (-A) = O$

O2 Scalar Multiplication

If $k$ is a scalar (real number), then $kA$ is obtained by multiplying every element of A by $k$.

$$(kA)_{ij} = k \cdot a_{ij}$$

Negative of a matrix: $-A = (-1)A$

O3 Matrix Multiplication

The product $AB$ is defined only when columns of A = rows of B.

$$\text{If } A_{m \times n} \text{ and } B_{n \times p} \text{ then } (AB)_{m \times p}$$

The $(i,k)$ element of $AB$ is:

$$c_{ik} = \sum_{j=1}^{n} a_{ij} \cdot b_{jk}$$
⚠ Key Points
  • Not commutative: $AB \neq BA$ (in general)
  • Associative: $(AB)C = A(BC)$
  • Distributive: $A(B + C) = AB + AC$
  • Non-zero product: $AB = O$ doesn't mean $A = O$ or $B = O$
  • Identity: $AI = IA = A$

O4 Transpose of a Matrix

If $A$ is $m \times n$, then $A^T$ is $n \times m$ obtained by interchanging rows and columns.

$$(A^T)_{ij} = a_{ji}$$
Properties of Transpose
  • $(A^T)^T = A$
  • $(A + B)^T = A^T + B^T$
  • $(kA)^T = kA^T$
  • $(AB)^T = B^T A^T$ (reverse order!)

O5 Symmetric & Skew-Symmetric

Symmetric Matrix

$A^T = A$, i.e., $a_{ij} = a_{ji}$ for all $i, j$.

Skew-Symmetric Matrix

$A^T = -A$, i.e., $a_{ij} = -a_{ji}$ for all $i, j$. All diagonal elements = 0.

Important Theorems
  • $A + A^T$ is always symmetric
  • $A - A^T$ is always skew-symmetric
  • Any square matrix can be written as: $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

O6 Invertible Matrices

A square matrix $A$ of order $n$ is invertible if there exists a matrix $B$ such that:

$$AB = BA = I_n$$

Then $B = A^{-1}$ (called the inverse of A).

Key Results
  • Inverse, if it exists, is unique
  • $(AB)^{-1} = B^{-1} A^{-1}$ (reverse order!)
  • $(A^{-1})^{-1} = A$
  • If $A$ is invertible, $A^T$ is also invertible
Quick Reference

📐 All Important Formulas

Memorize these for board exams

F1 Basic Operations

$(A + B)_{ij} = a_{ij} + b_{ij}$
$(A - B)_{ij} = a_{ij} - b_{ij}$
$(kA)_{ij} = k \cdot a_{ij}$
$(AB)_{ik} = \sum_{j} a_{ij} \cdot b_{jk}$

F2 Transpose Properties

$(A^T)^T = A$
$(A + B)^T = A^T + B^T$
$(AB)^T = B^T A^T$
$(kA)^T = kA^T$

F3 Symmetric & Skew

Symmetric: $A^T = A$
Skew-Symmetric: $A^T = -A$
$A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

F4 Inverse Properties

$(AB)^{-1} = B^{-1} A^{-1}$
$(A^{-1})^{-1} = A$
$(A^T)^{-1} = (A^{-1})^T$
$AA^{-1} = A^{-1}A = I$

F5 Counting Formulas

Total elements in $m \times n$ matrix = $mn$
Number of elements in $n \times n$ square matrix = $n^2$
Number of possible $3 \times 3$ matrices with entries 0 or 1 = $2^9 = 512$
Shortcuts

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — Order Memory

"Row first, Column second"

  • A $\mathbf{2 \times 3}$ matrix has 2 rows, 3 columns
  • Order is always written as m × n (rows × columns)
  • Total elements = $m \times n$
Trick 2 — Multiplication Defined?

"Inner dimensions must match"

  • $A_{m \times n} \times B_{n \times p}$ → Defined, result $m \times p$
  • $A_{m \times n} \times B_{k \times p}$ with $n \neq k$ → Not defined
  • Result order = outer dimensions (m × p)
Trick 3 — Transpose Reversal

Whenever you take transpose of a product, REVERSE the order:

  • $(AB)^T = B^T A^T$ (NOT $A^T B^T$!)
  • $(ABC)^T = C^T B^T A^T$
  • Same for inverse: $(AB)^{-1} = B^{-1} A^{-1}$
Trick 4 — Symmetric/Skew Quick Check

"Mirror check" — Look at elements across the main diagonal:

  • Equal values → Symmetric
  • Negative values → Skew-Symmetric
  • Mixed → Neither
  • Skew-symmetric matrix always has zeros on diagonal
Trick 5 — Reverse Order Rule

For any product-based operation:

  • Transpose: $(AB)^T = B^T A^T$
  • Inverse: $(AB)^{-1} = B^{-1} A^{-1}$
  • Product of 3: $(ABC)^T = C^T B^T A^T$

Remember: "Flip and Reverse"

Trick 6 — Identity Matrix Power

For any positive integer $n$: $I^n = I$

  • $I \cdot A = A \cdot I = A$
  • Identity behaves like "1" in multiplication
Trick 7 — Zero Matrix Caution

Unlike real numbers:

  • $AB = O$ does NOT imply $A = O$ or $B = O$
  • Both A and B could be non-zero
  • Example: $\begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Trick 8 — Diagonal Matrix Multiplication

Multiplication of diagonal matrices of same order is commutative:

  • $D_1 \cdot D_2 = D_2 \cdot D_1$
  • Result is also a diagonal matrix
Trick 9 — Symmetric Decomposition

Any square matrix A can be split into:

  • Symmetric part: $P = \frac{1}{2}(A + A^T)$
  • Skew-Symmetric part: $Q = \frac{1}{2}(A - A^T)$
  • Then $A = P + Q$
Trick 10 — Power of 2x2 Matrix

For rotational matrices $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$:

  • $A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}$
  • Easy to remember using rotation angle doubling
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. If a matrix has 8 elements, what are the possible orders it can have?

Step 1 — Use Formula0.5 Mark

If a matrix has order $m \times n$, then total elements = $mn$.

Given: $mn = 8$.

Step 2 — Find factor pairs1 Mark

Factor pairs of 8: $(1, 8), (2, 4), (4, 2), (8, 1)$.

Step 3 — List orders0.5 Mark

Possible orders: $1 \times 8, \ 2 \times 4, \ 4 \times 2, \ 8 \times 1$.

Possible orders are $1 \times 8, 8 \times 1, 4 \times 2, 2 \times 4$
📋 Total: 3 Marks

Q2. Construct a $3 \times 2$ matrix whose elements are given by $a_{ij} = \frac{1}{2}|i - 3j|$.

Step 1 — Write general form0.5 Mark

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$ with $i = 1, 2, 3$ and $j = 1, 2$.

Step 2 — Compute each element2 Marks

$a_{11} = \frac{1}{2}|1 - 3| = \frac{1}{2}(2) = 1$

$a_{12} = \frac{1}{2}|1 - 6| = \frac{1}{2}(5) = \frac{5}{2}$

$a_{21} = \frac{1}{2}|2 - 3| = \frac{1}{2}(1) = \frac{1}{2}$

$a_{22} = \frac{1}{2}|2 - 6| = \frac{1}{2}(4) = 2$

$a_{31} = \frac{1}{2}|3 - 3| = 0$

$a_{32} = \frac{1}{2}|3 - 6| = \frac{1}{2}(3) = \frac{3}{2}$

Step 3 — Write final matrix0.5 Mark

$A = \begin{bmatrix} 1 & 5/2 \\ 1/2 & 2 \\ 0 & 3/2 \end{bmatrix}$

Matrix constructed successfully
📋 Total: 3 Marks

Q3. If $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}$, find $2A - B$.

Step 1 — Compute 2A1 Mark

$2A = \begin{bmatrix} 2 & 4 & 6 \\ 4 & 6 & 2 \end{bmatrix}$

Step 2 — Compute 2A - B1.5 Marks

$2A - B = \begin{bmatrix} 2-3 & 4-(-1) & 6-3 \\ 4-(-1) & 6-0 & 2-2 \end{bmatrix}$

$= \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$

$2A - B = \begin{bmatrix} -1 & 5 & 3 \\ 5 & 6 & 0 \end{bmatrix}$
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

1. A matrix having $m$ rows and $n$ columns is of order:

Correct: B — Order is always written as rows × columns = $m \times n$.

2. The number of elements in a $4 \times 5$ matrix is:

Correct: C — Total elements = $m \times n = 4 \times 5 = 20$.

3. A matrix with only one column is called:

Correct: A — Column matrix has only 1 column ($m \times 1$).

4. A square matrix $A$ is symmetric if:

Correct: D — Symmetric: $A = A^T$ (transpose equals original).

5. In a skew-symmetric matrix, all diagonal elements are:

Correct: B — For skew-symmetric, $a_{ii} = -a_{ii}$, so $a_{ii} = 0$.

6. If $A$ is $2 \times 3$ and $B$ is $3 \times 4$, then $AB$ is:

Correct: C — Inner dims match (3 = 3), result outer dims: $2 \times 4$.

7. $(A + B)^T$ equals:

Correct: A — Transpose of sum = sum of transposes.

8. $(AB)^T$ equals:

Correct: D — Reverse order rule: $(AB)^T = B^T A^T$.

9. Identity matrix $I$ of order 3 has:

Correct: B — Identity matrix has 1's on the diagonal, 0's elsewhere.

10. If $A$ and $B$ are both symmetric of same order, then $AB - BA$ is:

Correct: A — $(AB - BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB - BA)$.

11. If $A$ is $3 \times 3$ matrix, then order of $2A$ is:

Correct: C — Scalar multiplication doesn't change the order.

12. If $AB = O$ (zero matrix), then:

Correct: D — Matrix product can be zero even when neither factor is zero.

13. $(A^T)^T$ equals:

Correct: A — Transpose of transpose gives back the original matrix.

14. Number of possible $2 \times 2$ matrices with entries 0 or 1 is:

Correct: B — $2^4 = 16$ possible matrices.

15. Scalar matrix is a special case of:

Correct: C — Scalar matrix is a diagonal matrix with equal diagonal entries.

16. If $A$ is a square matrix, then $A + A^T$ is:

Correct: A — $(A + A^T)^T = A^T + A = A + A^T$, so it's symmetric.

17. If $A$ is a square matrix, then $A - A^T$ is:

Correct: D — $(A - A^T)^T = A^T - A = -(A - A^T)$, so it's skew-symmetric.

18. If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, then $A \cdot A^T$ equals:

Correct: B — This is an orthogonal matrix, so $AA^T = I$.

19. If $A^2 = A$, then $(I + A)^3 - 7A$ equals:

Correct: C — Expand using $A^2 = A$: $(I + A)^3 = I + 3A + 3A^2 + A^3 = I + 7A$. So $(I+A)^3 - 7A = I$.

20. If the matrix is both symmetric and skew-symmetric, then it is:

Correct: B — $A = A^T$ and $A = -A^T$ imply $A = -A$, so $A = O$.
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) A matrix of order $3 \times 4$ has 12 elements.
Reason (R) Total number of elements in $m \times n$ matrix is $mn$.
Correct: A — Both true, R correctly explains A. Elements = $3 \times 4 = 12$.
2
Assertion (A) Every square matrix is symmetric.
Reason (R) A square matrix has equal number of rows and columns.
Correct: D — A is false (not every square matrix is symmetric). R is true (definition of square matrix).
3
Assertion (A) If $A = [a_{ij}]_{m \times n}$ and $B = [b_{ij}]_{m \times n}$, then $A + B$ is defined.
Reason (R) Addition of matrices is defined only when both matrices have the same order.
Correct: A — Both A and R are true. R correctly explains why A + B is defined.
4
Assertion (A) For any two matrices A and B, $AB = BA$.
Reason (R) Matrix multiplication is generally not commutative.
Correct: D — A is false ($AB \neq BA$ in general). R is true (matrix multiplication is non-commutative).
5
Assertion (A) If $A$ and $B$ are symmetric matrices, then $AB - BA$ is skew-symmetric.
Reason (R) $(AB - BA)^T = -(AB - BA)$ when A, B are symmetric.
Correct: A — Both true, and R directly proves A.
6
Assertion (A) If $AB = O$ (zero matrix), then either $A = O$ or $B = O$.
Reason (R) There exist non-zero matrices $A$ and $B$ such that $AB = O$.
Correct: D — A is false: for example, $A = \begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $B = \begin{bmatrix}0&0\\1&0\end{bmatrix}$ are both non-zero, yet $AB = O$. R is true (such matrices do exist).
7
Assertion (A) For any square matrix A, $A + A^T$ is symmetric.
Reason (R) $(A + A^T)^T = A^T + A = A + A^T$.
Correct: A — Both true, R is the formal proof of A.
8
Assertion (A) A skew-symmetric matrix has all non-zero diagonal entries.
Reason (R) For skew-symmetric, $a_{ij} = -a_{ji}$.
Correct: D — A is false (diagonal is always ZERO for skew-symmetric). R is true (defining condition).
9
Assertion (A) If $A$ is a square matrix, then $A - A^T$ is skew-symmetric.
Reason (R) $(A - A^T)^T = A^T - A = -(A - A^T)$.
Correct: A — Both true, R is the formal proof.
10
Assertion (A) $(AB)^T = A^T B^T$ for all matrices A, B where AB is defined.
Reason (R) $(AB)^T = B^T A^T$ always.
Correct: D — A is false (order must be reversed: $B^T A^T$). R is true (this is the correct rule).
Score: 0 / 10
Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

Q. Find the values of $x$ and $y$ if $\begin{bmatrix} x+2 & 3 \\ 5 & 2y-1 \end{bmatrix} = \begin{bmatrix} 5 & 3 \\ 5 & 7 \end{bmatrix}$.

Step 1 — Equate corresponding1 Mark

By equality of matrices: $x + 2 = 5$ and $2y - 1 = 7$.

Step 2 — Solve1 Mark

$x = 3$ and $2y = 8 \Rightarrow y = 4$.

$x = 3, \ y = 4$
2024

Q. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$, find $AB$.

Step 1 — Setup0.5 Mark

$AB = \begin{bmatrix} 1(2)+2(4) & 1(1)+2(3) \\ 3(2)+4(4) & 3(1)+4(3) \end{bmatrix}$

Step 2 — Compute1.5 Marks

$= \begin{bmatrix} 2+8 & 1+6 \\ 6+16 & 3+12 \end{bmatrix} = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$

$AB = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$
2023

Q. Show that the matrix $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$ is skew-symmetric.

Step 1 — Find transpose1 Mark

$A^T = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$

Step 2 — Check A^T = -A1 Mark

$-A = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix} = A^T$ ✓

Also diagonal = 0 ✓

A is skew-symmetric
2023

Q. Express the matrix $B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$ as sum of a symmetric and a skew-symmetric matrix.

Step 1 — Find B^T1 Mark

$B^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$

Step 2 — Symmetric part P1 Mark

$P = \frac{1}{2}(B + B^T) = \begin{bmatrix} 2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$

Step 3 — Skew-symmetric part Q1 Mark

$Q = \frac{1}{2}(B - B^T) = \begin{bmatrix} 0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$

$B = P + Q$ ✓
2022

Q. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$, then verify $(AB)^T = B^T A^T$.

Step 1 — Compute AB1 Mark

$AB = \begin{bmatrix} 10 & 7 \\ 22 & 15 \end{bmatrix}$

Step 2 — Compute LHS (AB)^T0.5 Mark

$(AB)^T = \begin{bmatrix} 10 & 22 \\ 7 & 15 \end{bmatrix}$

Step 3 — Compute RHS B^T A^T1 Mark

$B^T A^T = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 10 & 22 \\ 7 & 15 \end{bmatrix}$

LHS = RHS ✓

Verified: $(AB)^T = B^T A^T$
2022

Q. Find $x$, $y$, $z$ if $2\begin{bmatrix} x & 3 \\ 4 & y \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ z & 8 \end{bmatrix}$.

Step 1 — Simplify LHS0.5 Mark

$\begin{bmatrix} 2x & 6 \\ 8 & 2y \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ z & 8 \end{bmatrix}$

Step 2 — Equate1 Mark

$2x = 4 \Rightarrow x = 2$; $2y = 8 \Rightarrow y = 4$; $z = 8$.

$x = 2, \ y = 4, \ z = 8$
2021

Q. If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find $k$ so that $A^2 = kA - 2I$.

Step 1 — Compute A²1 Mark

$A^2 = \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

Step 2 — Set up equation1 Mark

$kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}$

Equate: $3k - 2 = 1 \Rightarrow k = 1$. Check others: $-2k = -2 \Rightarrow k = 1$ ✓

$k = 1$
2020

Q. If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$, show that $A^3 - 23A - 40I = O$.

Step 1 — Find A²1 Mark

$A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}$

Step 2 — Find A³ = A²·A1 Mark

$A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}$

Step 3 — Verify1 Mark

$23A + 40I = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix} = A^3$

So $A^3 - 23A - 40I = O$ ✓

Verified ✓