Chapter 3: Matrices

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

Order of a Matrix

A matrix having $m$ rows and $n$ columns is called a matrix of order $m \times n$ or simply $m \times n$ matrix (read as an $m$ by $n$ matrix). In general, an $m \times n$ matrix has the following rectangular array:

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

Types of Matrices

• Column Matrix: A matrix is said to be a column matrix if it has only one column. Order $m \times 1$.
• Row Matrix: A matrix is said to be a row matrix if it has only one row. Order $1 \times n$.
• Square Matrix: A matrix in which the number of rows is equal to the number of columns is said to be a square matrix. Thus an $m \times n$ matrix is said to be a square matrix if $m = n$.
• Diagonal Matrix: A square matrix $B = [b_{ij}]{m \times m}$ is said to be a diagonal matrix if all its non-diagonal elements are zero, that is $b{ij} = 0$, when $i \neq j$.
• Scalar Matrix: A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, $B = [b_{ij}]{n \times n}$ is a scalar matrix if $b{ij} = 0$, when $i \neq j$ and $b_{ij} = k$, when $i = j$, for some constant $k$.
• Identity Matrix: A square matrix in which elements in the diagonal are all $1$ and rest are all zero is called an identity matrix. Denoted by $I$.

Operations on Matrices

• Addition: Two matrices can be added if they are of the same order. Addition is element-wise.
• Scalar Multiplication: If $A = [a_{ij}]_{m \times n}$ is a matrix and $k$ is a scalar, then $kA$ is another matrix which is obtained by multiplying each element of $A$ by $k$.
• Multiplication: The product of two matrices $A$ and $B$ is defined if the number of columns of $A$ is equal to the number of rows of $B$. If $A = [a_{ij}]$ is an $m \times n$ matrix and $B = [b_{jk}]$ is an $n \times p$ matrix, then the product $AB$ is an $m \times p$ matrix $C = [c_{ik}]$. Matrix multiplication is not commutative in general, i.e., $AB \neq BA$.

Competency-Based Questions

1. [Sample Paper 2024] Find $X$ and $Y$ if: $$ 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} $$ $$ 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} $$

Solution:

Let Eq 1 be: $2X + 3Y = A$

Let Eq 2 be: $3X + 2Y = B$

Multiply Eq 1 by 3: $6X + 9Y = 3A = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}$

Multiply Eq 2 by 2: $6X + 4Y = 2B = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}$

Subtracting the two equations: $5Y = 3A - 2B = \begin{bmatrix} 6-4 & 9 - (-4) \\ 12 - (-2) & 0 - 10 \end{bmatrix} = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}$

Therefore, $Y = \frac{1}{5} \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix} = \begin{bmatrix} 2/5 & 13/5 \\ 14/5 & -2 \end{bmatrix}$.

Substituting $Y$ in Eq 1:

$2X = A - 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} 6/5 & 39/5 \\ 42/5 & -6 \end{bmatrix} = \begin{bmatrix} 4/5 & -24/5 \\ -22/5 & 6 \end{bmatrix}$

Therefore, $X = \begin{bmatrix} 2/5 & -12/5 \\ -11/5 & 3 \end{bmatrix}$.

2. [CBSE 2022] Express the matrix $A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Solution:

We know that any square matrix $A$ can be expressed as the sum of a symmetric and a skew-symmetric matrix:

$A = \frac{1}{2}(A + A’) + \frac{1}{2}(A - A’)$

where $A’$ is the transpose of $A$.

$A’ = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$

Then, $A + A’ = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix}$. Let $P = \frac{1}{2}(A + A’) = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}$. $P$ is symmetric since $P’ = P$.

Also, $A - A’ = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix}$. Let $Q = \frac{1}{2}(A - A’) = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$. $Q$ is skew-symmetric since $Q’ = -Q$.

Therefore, $A = P + Q = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$.

3. [CBSE 2019] If $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, then show that $A^T A = I$.

Solution:

$A^T = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$

$A^T A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

$= \begin{bmatrix} (\cos\alpha)(\cos\alpha) + (\sin\alpha)(\sin\alpha) & (\cos\alpha)(-\sin\alpha) + (\sin\alpha)(\cos\alpha) \\ (-\sin\alpha)(\cos\alpha) + (\cos\alpha)(\sin\alpha) & (-\sin\alpha)(-\sin\alpha) + (\cos\alpha)(\cos\alpha) \end{bmatrix}$

$= \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & -\sin\alpha\cos\alpha + \sin\alpha\cos\alpha \\ -\sin\alpha\cos\alpha + \sin\alpha\cos\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.

Hence Proved.