Chapter 7: Matrices

Matrices are rectangular arrays of numbers arranged in rows and columns. They are a powerful mathematical tool used to organise data, solve systems of equations, and are the backbone of computer graphics and data science. For ICSE Class 10, you must master the definition, types, and all four operations on matrices.

1. Introduction to Matrices

What is a Matrix?

A matrix is an ordered rectangular array of numbers (called elements or entries) arranged in horizontal rows and vertical columns, enclosed in square brackets [ ].

A matrix is usually denoted by a capital letter (A, B, C…). Each element is denoted by $a_{ij}$ where $i$ = row number and $j$ = column number.

Definition

A matrix $A$ with $m$ rows and $n$ columns is written as $A = [a_{ij}]_{m \times n}$ and is said to have order $m \times n$ (read as "m by n").

$$A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}$$

Total number of elements in a matrix of order $m \times n$ = $m \times n$
The element in the $i$-th row and $j$-th column is $a_{ij}$
Example: $a_{23}$ means the element in row 2, column 3

Example of a Matrix

Consider: $A = \begin{bmatrix} 3 & -1 & 0 \\ 5 & 2 & 7 \end{bmatrix}$

Order of A = $2 \times 3$ (2 rows, 3 columns)
$a_{11} = 3$, $a_{12} = -1$, $a_{13} = 0$
$a_{21} = 5$, $a_{22} = 2$, $a_{23} = 7$
Total elements = $2 \times 3 = 6$

2. Types of Matrices

Type	Definition	Example
Row Matrix	Only ONE row; order $1 \times n$	$\begin{bmatrix}2 & -3 & 5\end{bmatrix}$ — order $1 \times 3$
Column Matrix	Only ONE column; order $m \times 1$	$\begin{bmatrix}4 \\ -1 \\ 0\end{bmatrix}$ — order $3 \times 1$
Square Matrix	Number of rows = Number of columns; order $n \times n$	$\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$ — order $2 \times 2$
Zero (Null) Matrix	Every element is 0; denoted by $O$	$\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ — zero matrix of order $2 \times 2$
Identity (Unit) Matrix	Square matrix with 1s on main diagonal and 0s elsewhere; denoted $I$	$\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Diagonal Matrix	Square matrix with non-zero elements only on the main diagonal	$\begin{bmatrix}3 & 0 \\ 0 & -5\end{bmatrix}$

Key Points to Remember

The main (principal) diagonal of a square matrix of order $n \times n$ consists of elements $a_{11}, a_{22}, a_{33}, \ldots, a_{nn}$ (where row index = column index).
Every Identity Matrix is a Diagonal Matrix, but not vice versa.
Every Identity Matrix and Zero Matrix is also a Square Matrix (when applicable).
A $1 \times 1$ matrix $[k]$ is both a row matrix and a column matrix.

Practice Problems — Types of Matrices

Q1. State the order of the following matrices and identify their type:
(a) $A = \begin{bmatrix}5 & -2 & 0 & 1\end{bmatrix}$ (b) $B = \begin{bmatrix}3\\7\\-1\end{bmatrix}$ (c) $C = \begin{bmatrix}2&0\\0&-3\end{bmatrix}$ (d) $D = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

Answers:
(a) Order $1 \times 4$ — Row Matrix
(b) Order $3 \times 1$ — Column Matrix
(c) Order $2 \times 2$ — Square Matrix and Diagonal Matrix
(d) Order $3 \times 3$ — Identity (Unit) Matrix

Q2. For matrix $A = \begin{bmatrix}4 & -1 & 0 \\ 3 & 7 & -2 \\ 1 & 5 & 8\end{bmatrix}$, find: (a) $a_{12}$, (b) $a_{31}$, (c) $a_{23}$, (d) $a_{22}$

Answers: (a) $a_{12} = -1$ (b) $a_{31} = 1$ (c) $a_{23} = -2$ (d) $a_{22} = 7$

Q3. Construct a $2 \times 3$ matrix $A = [a_{ij}]$ where each element is defined by $a_{ij} = 2i + j$.

Solution:
$a_{11}=2(1)+1=3$, $a_{12}=2(1)+2=4$, $a_{13}=2(1)+3=5$
$a_{21}=2(2)+1=5$, $a_{22}=2(2)+2=6$, $a_{23}=2(2)+3=7$
$$A = \begin{bmatrix}3&4&5\\5&6&7\end{bmatrix}$$

3. Equality of Matrices

Definition: Equal Matrices

Two matrices $A$ and $B$ are equal (written $A = B$) if and only if:

They have the same order, AND
Every corresponding element is equal: $a_{ij} = b_{ij}$ for all $i$ and $j$.

Worked Example — Using Equality to Find Unknown Values

If $\begin{bmatrix}x+2 & y-1 \\ 3 & 2z\end{bmatrix} = \begin{bmatrix}5 & 4 \\ 3 & 14\end{bmatrix}$, find $x$, $y$, $z$.

Solution: Since the matrices are equal, all corresponding elements are equal:

$x + 2 = 5 \Rightarrow x = 3$
$y - 1 = 4 \Rightarrow y = 5$
$2z = 14 \Rightarrow z = 7$

Practice Problems — Equality of Matrices

Q1. Find $a, b, c, d$ if $\begin{bmatrix}a+b & 2c \\ d-3 & a-b\end{bmatrix} = \begin{bmatrix}7 & 10 \\ 1 & 1\end{bmatrix}$

Solution:
$a+b=7$ … (i) $2c=10 \Rightarrow c=5$
$d-3=1 \Rightarrow d=4$ $a-b=1$ … (ii)
Adding (i)+(ii): $2a=8 \Rightarrow a=4$; From (i): $b=3$
∴ a=4, b=3, c=5, d=4

Q2. If $\begin{bmatrix}x^2 & y \\ 4 & x+y\end{bmatrix} = \begin{bmatrix}9 & 3 \\ 4 & 6\end{bmatrix}$, find $x$ and $y$.

Solution:
$x^2=9 \Rightarrow x=3$ or $x=-3$ $y=3$ $x+y=6 \Rightarrow x=3$
∴ x = 3, y = 3

Q3. Find $p, q, r, s$ if $\begin{bmatrix}p+q & p-q \\ 2r & r+s\end{bmatrix} = \begin{bmatrix}8 & 2 \\ 6 & 5\end{bmatrix}$

Solution:
$p+q=8$, $p-q=2$ → Adding: $2p=10$, $p=5$, $q=3$
$2r=6 \Rightarrow r=3$ $r+s=5 \Rightarrow s=2$
∴ p=5, q=3, r=3, s=2

4. Addition and Subtraction of Matrices

Rule

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

If $A = [a_{ij}]_{m\times n}$ and $B = [b_{ij}]_{m\times n}$, then:

$$A + B = [a_{ij} + b_{ij}]_{m\times n} \qquad A - B = [a_{ij} - b_{ij}]_{m\times n}$$

Properties of Matrix Addition

Commutative: $A + B = B + A$
Associative: $(A + B) + C = A + (B + C)$
Additive Identity: $A + O = A$ (where $O$ is the zero matrix of same order)
Additive Inverse: $A + (-A) = O$ (where $-A$ means each element negated)

Worked Example

If $A = \begin{bmatrix}2 & -1 \\ 3 & 0\end{bmatrix}$ and $B = \begin{bmatrix}1 & 4 \\ -2 & 5\end{bmatrix}$, find $A+B$ and $A-B$.

$A + B$ $= \begin{bmatrix}2+1 & -1+4 \\ 3+(-2) & 0+5\end{bmatrix} = \begin{bmatrix}3 & 3 \\ 1 & 5\end{bmatrix}$

$A - B$ $= \begin{bmatrix}2-1 & -1-4 \\ 3-(-2) & 0-5\end{bmatrix} = \begin{bmatrix}1 & -5 \\ 5 & -5\end{bmatrix}$

Practice Problems — Addition & Subtraction

Q1. If $A = \begin{bmatrix}3 & 1 \\ -1 & 2\end{bmatrix}$, $B = \begin{bmatrix}-1 & 3 \\ 4 & -2\end{bmatrix}$, find $2A + 3B$.

Solution:
$2A = \begin{bmatrix}6&2\\-2&4\end{bmatrix}$, $3B = \begin{bmatrix}-3&9\\12&-6\end{bmatrix}$
$2A+3B = \begin{bmatrix}6-3 & 2+9\\-2+12 & 4-6\end{bmatrix} = \begin{bmatrix}3&11\\10&-2\end{bmatrix}$

Q2. Find matrix $X$ if $2X + \begin{bmatrix}1&2\\3&4\end{bmatrix} = \begin{bmatrix}7&8\\9&10\end{bmatrix}$.

Solution:
$2X = \begin{bmatrix}7-1&8-2\\9-3&10-4\end{bmatrix} = \begin{bmatrix}6&6\\6&6\end{bmatrix}$
$X = \begin{bmatrix}3&3\\3&3\end{bmatrix}$

Q3. If $A + B = \begin{bmatrix}5&6\\8&2\end{bmatrix}$ and $A - B = \begin{bmatrix}1&2\\4&0\end{bmatrix}$, find matrices $A$ and $B$.

Solution:
Adding: $2A = \begin{bmatrix}6&8\\12&2\end{bmatrix} \Rightarrow A = \begin{bmatrix}3&4\\6&1\end{bmatrix}$
Subtracting: $2B = \begin{bmatrix}4&4\\4&2\end{bmatrix} \Rightarrow B = \begin{bmatrix}2&2\\2&1\end{bmatrix}$

Q4. If $A = \begin{bmatrix}1&-1\\2&3\end{bmatrix}$, $B = \begin{bmatrix}-1&0\\1&-2\end{bmatrix}$ and $C = \begin{bmatrix}3&2\\-1&0\end{bmatrix}$, find $A - B + C$.

Solution:
$A-B = \begin{bmatrix}2&-1\\1&5\end{bmatrix}$
$(A-B)+C = \begin{bmatrix}2+3&-1+2\\1-1&5+0\end{bmatrix} = \begin{bmatrix}5&1\\0&5\end{bmatrix}$

5. Scalar Multiplication

Rule: Scalar Multiplication

To multiply a matrix $A$ by a scalar (real number) $k$, multiply every element of the matrix by $k$.

$$kA = k[a_{ij}]_{m\times n} = [k \cdot a_{ij}]_{m\times n}$$

Properties of Scalar Multiplication

$k(A + B) = kA + kB$
$(k + l)A = kA + lA$
$(kl)A = k(lA)$
$1 \cdot A = A$ and $0 \cdot A = O$ (zero matrix)

Worked Example

If $A = \begin{bmatrix}4 & -2 \\ 6 & 0\end{bmatrix}$, find $3A$ and $-2A$.

$3A = \begin{bmatrix}12 & -6 \\ 18 & 0\end{bmatrix}$ $-2A = \begin{bmatrix}-8 & 4 \\ -12 & 0\end{bmatrix}$

Practice Problems — Scalar Multiplication

Q1. If $A = \begin{bmatrix}2&-1\\0&3\end{bmatrix}$ and $B = \begin{bmatrix}1&0\\4&-2\end{bmatrix}$, find $3A - 2B$.

Solution:
$3A = \begin{bmatrix}6&-3\\0&9\end{bmatrix}$, $2B = \begin{bmatrix}2&0\\8&-4\end{bmatrix}$
$3A-2B = \begin{bmatrix}4&-3\\-8&13\end{bmatrix}$

Q2. Find matrix $X$ such that $5X - \begin{bmatrix}10&15\\-5&20\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$.

Solution:
$5X = \begin{bmatrix}10&15\\-5&20\end{bmatrix} \Rightarrow X = \begin{bmatrix}2&3\\-1&4\end{bmatrix}$

6. Matrix Multiplication

Condition for Multiplication

Matrix $A$ (of order $m \times n$) can be multiplied by matrix $B$ (of order $n \times p$) only if the number of columns of A equals the number of rows of B.

The resulting product $AB$ will have order $m \times p$.

$$\underbrace{A}_{m \times n} \times \underbrace{B}_{n \times p} = \underbrace{AB}_{m \times p}$$

Memory Aid: $(m \times \mathbf{n}) \times (\mathbf{n} \times p) = m \times p$ — the inner dimensions must match.

How to Multiply — The Dot Product Method

Each element $c_{ij}$ of the product $AB$ is computed as the dot product of row $i$ of $A$ with column $j$ of $B$:

$$c_{ij} = \sum_{k=1}^{n} a_{ik} \cdot b_{kj} = a_{i1}b_{1j} + a_{i2}b_{2j} + \cdots + a_{in}b_{nj}$$

Worked Example 1 — $2 \times 2$ matrices

$A = \begin{bmatrix}1&2\\3&4\end{bmatrix}$, $B = \begin{bmatrix}5&6\\7&8\end{bmatrix}$. Find $AB$.

$c_{11} = (1)(5)+(2)(7) = 5+14 = 19$

$c_{12} = (1)(6)+(2)(8) = 6+16 = 22$

$c_{21} = (3)(5)+(4)(7) = 15+28 = 43$

$c_{22} = (3)(6)+(4)(8) = 18+32 = 50$

$$AB = \begin{bmatrix}19&22\\43&50\end{bmatrix}$$

Worked Example 2 — $2\times3$ times $3\times2$

$A = \begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}_{2\times3}$, $B = \begin{bmatrix}7&8\\9&10\\11&12\end{bmatrix}_{3\times2}$. Find $AB$ (order will be $2\times2$).

$c_{11} = 1(7)+2(9)+3(11) = 7+18+33=58$

$c_{12} = 1(8)+2(10)+3(12) = 8+20+36=64$

$c_{21} = 4(7)+5(9)+6(11) = 28+45+66=139$

$c_{22} = 4(8)+5(10)+6(12) = 32+50+72=154$

$$AB = \begin{bmatrix}58&64\\139&154\end{bmatrix}$$

Important Properties of Matrix Multiplication

NOT Commutative: In general, $AB \neq BA$ (order matters!)
Associative: $(AB)C = A(BC)$
Distributive: $A(B+C) = AB + AC$ and $(A+B)C = AC + BC$
Identity: $AI = IA = A$ (where $I$ is the identity matrix of appropriate order)
Zero: $AO = O$ and $OA = O$
Cancellation LAW FAILS: $AB = AC$ does NOT necessarily imply $B = C$
Zero Product Property FAILS: $AB = O$ does NOT necessarily imply $A = O$ or $B = O$

Practice Problems — Matrix Multiplication

Q1. Find $AB$ and $BA$ if $A = \begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $B = \begin{bmatrix}0&1\\1&0\end{bmatrix}$. Verify $AB \neq BA$.

Solution:
$AB = \begin{bmatrix}0+2&1+0\\0+4&3+0\end{bmatrix} = \begin{bmatrix}2&1\\4&3\end{bmatrix}$
$BA = \begin{bmatrix}0+3&0+4\\1+0&2+0\end{bmatrix} = \begin{bmatrix}3&4\\1&2\end{bmatrix}$
Since $AB \neq BA$, matrix multiplication is not commutative.

Q2. If $A = \begin{bmatrix}2&1\\1&3\end{bmatrix}$, find $A^2$ (i.e., $A \times A$).

Solution:
$A^2 = \begin{bmatrix}2&1\\1&3\end{bmatrix}\begin{bmatrix}2&1\\1&3\end{bmatrix} = \begin{bmatrix}4+1&2+3\\2+3&1+9\end{bmatrix} = \begin{bmatrix}5&5\\5&10\end{bmatrix}$

Q3. If $A = \begin{bmatrix}1&0\\2&1\end{bmatrix}$ and $B = \begin{bmatrix}3&1\\-1&2\end{bmatrix}$, find $2AB - 3I$ where $I$ is the $2\times2$ identity matrix.

Solution:
$AB = \begin{bmatrix}3+0&1+0\\6-1&2+2\end{bmatrix} = \begin{bmatrix}3&1\\5&4\end{bmatrix}$
$2AB = \begin{bmatrix}6&2\\10&8\end{bmatrix}$
$3I = \begin{bmatrix}3&0\\0&3\end{bmatrix}$
$2AB-3I = \begin{bmatrix}3&2\\10&5\end{bmatrix}$

Q4. Determine the order of the product $PQ$ given $P$ is $3\times4$ and $Q$ is $4\times2$. Also state whether $QP$ is possible.

Solution:
$PQ$: Inner dimensions $4=4$ ✓ → Order of $PQ$ = $3\times2$
$QP$: $Q$ is $4\times2$, $P$ is $3\times4$. Inner dimensions $2\neq3$ ✗ → $QP$ is NOT possible.

Q5 (ICSE Type). If $A = \begin{bmatrix}3&x\\0&1\end{bmatrix}$ and $B = \begin{bmatrix}9&16\\0&-y\end{bmatrix}$, find $x$ and $y$ given that $A^2 = B$.

Solution:
$A^2 = \begin{bmatrix}3&x\\0&1\end{bmatrix}\begin{bmatrix}3&x\\0&1\end{bmatrix} = \begin{bmatrix}9+0&3x+x\\0&0+1\end{bmatrix} = \begin{bmatrix}9&4x\\0&1\end{bmatrix}$
Comparing with $B = \begin{bmatrix}9&16\\0&-y\end{bmatrix}$:
$4x = 16 \Rightarrow x = 4$ and $1 = -y \Rightarrow y = -1$

Q6. Verify that $A(B+C) = AB+AC$ for $A = \begin{bmatrix}1&2\\0&1\end{bmatrix}$, $B = \begin{bmatrix}1&0\\2&1\end{bmatrix}$, $C = \begin{bmatrix}0&1\\1&0\end{bmatrix}$.

Solution:
$B+C = \begin{bmatrix}1&1\\3&1\end{bmatrix}$. $A(B+C) = \begin{bmatrix}1+6&1+2\\0+3&0+1\end{bmatrix} = \begin{bmatrix}7&3\\3&1\end{bmatrix}$
$AB = \begin{bmatrix}1+4&0+2\\0+2&0+1\end{bmatrix} = \begin{bmatrix}5&2\\2&1\end{bmatrix}$
$AC = \begin{bmatrix}0+2&1+0\\0+1&0+0\end{bmatrix} = \begin{bmatrix}2&1\\1&0\end{bmatrix}$
$AB+AC = \begin{bmatrix}7&3\\3&1\end{bmatrix}$ ✓ Both sides are equal.

7. Transpose of a Matrix

Definition: Transpose

The transpose of a matrix $A$, denoted by $A^T$ (or $A'$), is obtained by interchanging its rows and columns.

If $A = [a_{ij}]_{m\times n}$, then $A^T = [a_{ji}]_{n\times m}$.

The element at position $(i,j)$ in $A$ becomes the element at position $(j,i)$ in $A^T$.

Order changes: If $A$ is $m \times n$, then $A^T$ is $n \times m$.

Example

$A = \begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}_{2\times3}$ → $A^T = \begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}_{3\times2}$

Properties of Transpose

$(A^T)^T = A$ (transpose of transpose = original)
$(A+B)^T = A^T + B^T$
$(kA)^T = kA^T$
$(AB)^T = B^T A^T$ (note the reversal of order!)

Special Matrices via Transpose

Symmetric Matrix: $A^T = A$ (i.e., $a_{ij} = a_{ji}$ for all $i, j$). Example: $\begin{bmatrix}1&2\\2&5\end{bmatrix}$
Skew-Symmetric Matrix: $A^T = -A$ (i.e., $a_{ij} = -a_{ji}$; all diagonal elements must be 0). Example: $\begin{bmatrix}0&3\\-3&0\end{bmatrix}$

Practice Problems — Transpose

Q1. Find the transpose of $A = \begin{bmatrix}2&-1&0\\3&4&-5\end{bmatrix}$ and state the order of $A^T$.

Solution:
$A^T = \begin{bmatrix}2&3\\-1&4\\0&-5\end{bmatrix}_{3\times2}$ Order of $A^T$ = $3\times2$

Q2. If $A = \begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $B = \begin{bmatrix}0&1\\2&3\end{bmatrix}$, verify $(AB)^T = B^T A^T$.

Solution:
$AB = \begin{bmatrix}4&7\\8&15\end{bmatrix}$, $(AB)^T = \begin{bmatrix}4&8\\7&15\end{bmatrix}$
$B^T = \begin{bmatrix}0&2\\1&3\end{bmatrix}$, $A^T = \begin{bmatrix}1&3\\2&4\end{bmatrix}$
$B^T A^T = \begin{bmatrix}0\cdot1+2\cdot2&0\cdot3+2\cdot4\\1\cdot1+3\cdot2&1\cdot3+3\cdot4\end{bmatrix} = \begin{bmatrix}4&8\\7&15\end{bmatrix}$ ✓

Q3. Check whether $A = \begin{bmatrix}2&-3&1\\-3&0&4\\1&4&-1\end{bmatrix}$ is symmetric, skew-symmetric, or neither.

Solution:
$A^T = \begin{bmatrix}2&-3&1\\-3&0&4\\1&4&-1\end{bmatrix} = A$
Since $A^T = A$, matrix $A$ is Symmetric.

8. Mixed Practice Problems (ICSE Board Style)

Board Level Practice

Q1 (ICSE 2019 Style). If $A = \begin{bmatrix}1&2\\3&4\end{bmatrix}$, $B = \begin{bmatrix}6&1\\1&1\end{bmatrix}$ and $C = \begin{bmatrix}4&1\\2&-1\end{bmatrix}$, find $A(B-C)$. Verify your answer by computing $AB - AC$.

Solution:
$B-C = \begin{bmatrix}2&0\\-1&2\end{bmatrix}$
$A(B-C) = \begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}2&0\\-1&2\end{bmatrix} = \begin{bmatrix}2-2&0+4\\6-4&0+8\end{bmatrix} = \begin{bmatrix}0&4\\2&8\end{bmatrix}$
Verification: $AB = \begin{bmatrix}8&3\\22&7\end{bmatrix}$, $AC = \begin{bmatrix}8&-1\\20&-1\end{bmatrix}$
$AB-AC = \begin{bmatrix}0&4\\2&8\end{bmatrix}$ ✓

Q2. Given $\begin{bmatrix}2&1\\-3&4\end{bmatrix} \cdot X = \begin{bmatrix}7\\2\end{bmatrix}$, find the matrix $X$, given $X$ is a $2\times1$ matrix. Let $X = \begin{bmatrix}a\\b\end{bmatrix}$.

Solution:
$\begin{bmatrix}2a+b\\-3a+4b\end{bmatrix} = \begin{bmatrix}7\\2\end{bmatrix}$
$2a+b=7$ … (i) $-3a+4b=2$ … (ii)
From (i): $b=7-2a$. Sub in (ii): $-3a+4(7-2a)=2 \Rightarrow -3a+28-8a=2 \Rightarrow -11a=-26 \Rightarrow a=\frac{26}{11}$
This gives non-integer values. Let's try: Multiply (i) by 4: $8a+4b=28$. Add to (ii): $5a=30 \Rightarrow a=6$; $b=7-12=-5$
$X = \begin{bmatrix}6\\-5\end{bmatrix}$

Q3. If $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ satisfies $A^2 - (a+d)A = (bc-ad)I$, verify this for $A = \begin{bmatrix}2&3\\1&4\end{bmatrix}$.

Solution:
$a=2, b=3, c=1, d=4$. $a+d=6$, $bc-ad=3-8=-5$.
$A^2 = \begin{bmatrix}4+3&6+12\\2+4&3+16\end{bmatrix} = \begin{bmatrix}7&18\\6&19\end{bmatrix}$
$6A = \begin{bmatrix}12&18\\6&24\end{bmatrix}$
$A^2-6A = \begin{bmatrix}-5&0\\0&-5\end{bmatrix} = -5\begin{bmatrix}1&0\\0&1\end{bmatrix} = -5I = (bc-ad)I$ ✓

Q4. Find matrix $B$ if $3A + 2B = \begin{bmatrix}10&-4\\8&6\end{bmatrix}$ where $A = \begin{bmatrix}2&0\\2&-2\end{bmatrix}$.

Solution:
$3A = \begin{bmatrix}6&0\\6&-6\end{bmatrix}$
$2B = \begin{bmatrix}10-6&-4-0\\8-6&6+6\end{bmatrix} = \begin{bmatrix}4&-4\\2&12\end{bmatrix}$
$B = \begin{bmatrix}2&-2\\1&6\end{bmatrix}$

Q5. Show that $\begin{bmatrix}1&2\\3&4\end{bmatrix} \cdot \begin{bmatrix}4&-2\\-3&1\end{bmatrix} = \begin{bmatrix}-2&0\\0&-2\end{bmatrix}$. Hence or otherwise, write the product as a scalar multiple of $I$.

Solution:
Product $= \begin{bmatrix}4-6&-2+2\\12-12&-6+4\end{bmatrix} = \begin{bmatrix}-2&0\\0&-2\end{bmatrix} = -2\begin{bmatrix}1&0\\0&1\end{bmatrix} = -2I$ ✓

Q6 (ICSE 2022 Pattern). If $\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}25\end{bmatrix}$ and $x - y = 1$, find $x$ and $y$.

Solution:
$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = [x^2+y^2] = [25] \Rightarrow x^2+y^2=25$
Also $x-y=1 \Rightarrow x=y+1$
$(y+1)^2+y^2=25 \Rightarrow y^2+2y+1+y^2=25 \Rightarrow 2y^2+2y-24=0 \Rightarrow y^2+y-12=0$
$(y+4)(y-3)=0 \Rightarrow y=3$ (taking positive) $\Rightarrow x=4$
∴ x=4, y=3 (or x=-3, y=-4)

9. Quick Formula Summary

Concept	Rule / Formula
Order of Matrix	$m \times n$ means $m$ rows, $n$ columns; total elements = $mn$
Equality: $A = B$	Same order AND all corresponding elements equal
Addition $A+B$	Add corresponding elements; requires same order
Scalar multiplication $kA$	Multiply every element by $k$
Matrix product $AB$	Columns of $A$ must = Rows of $B$; $c_{ij} = \sum a_{ik}b_{kj}$
Transpose $A^T$	Rows ↔ Columns; order $m\times n \to n\times m$
Identity Matrix $I$	$AI = IA = A$
$AB \neq BA$	Multiplication is NOT commutative
$(AB)^T$	$= B^T A^T$ (reverse order!)

Examiner's Favourite Questions

Find unknown values using matrix equality — common every year
Compute $A^2$ or $A^2 - kA + I$ type questions
Show $A^2 = kI$ or $A^2 = A$ (idempotent matrix)
Given $AB = C$, find matrix $X$ (the unknown matrix)
Verify properties: $(AB)^T = B^T A^T$, $A(B+C) = AB + AC$
State whether multiplication $AB$ is possible and find the order