Chapter 3: Matrices

Matrix A: $A = \begin{bmatrix} 5 & -2 \\ 0 & 5 \\ 3 & 6 \end{bmatrix}$ (3 rows, 2 columns)

Matrix B: $B = \begin{bmatrix} 1+i & 2 \\ 3.5 & -1 \\ 3 & 5/7 \end{bmatrix}$ (3 rows, 3 columns)

Matrix C: $C = \begin{bmatrix} 1+x & 3 \\ \cos x & \tan x \\ \sin 2x & x^3 \end{bmatrix}$ (2 rows, 3 columns)

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

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

Solution:

A $3 \times 2$ matrix has the form: $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$

Given: $a_{ij} = \frac{1}{2}|i - 3j|$

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

$a_{12} = \frac{1}{2}|1 - 3(2)| = \frac{1}{2}|-5| = \frac{5}{2}$

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

$a_{22} = \frac{1}{2}|2 - 3(2)| = \frac{1}{2}|-4| = 2$

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

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

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

Example 2: 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 $2A - B$

Solution:

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

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

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

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

Example 3: If $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$, show that $A^2 - 2A + I = O$

Solution:

First, find $A^2$:

$A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$

Now, $A^2 - 2A + I$:

$= \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} - 2\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} - \begin{bmatrix} 6 & -8 \\ 2 & -2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 5-6+1 & -8+8+0 \\ 2-2+0 & -3+2+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

Hence proved.

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

Solution:

We know: $B = \frac{1}{2}(B + B') + \frac{1}{2}(B - B')$

First, find $B'$: $B' = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$

$P = \frac{1}{2}(B + B') = \frac{1}{2}\begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} = \begin{bmatrix} 2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$

$Q = \frac{1}{2}(B - B') = \frac{1}{2}\begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$

Verify: $P' = P$ (symmetric) and $Q' = -Q$ (skew symmetric)

Therefore: $B = P + Q$ where P is symmetric and Q is skew symmetric

Q1. If a matrix has 18 elements, what are the possible orders it can have?

Q2. Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = \frac{(i+j)^2}{2}$

Q3. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$, find $A + B$ and $A - B$

Q4. If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ 0 & 2 \end{bmatrix}$, find $AB$ and $BA$. Are they equal?

Q5. For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that $(A + A')$ is symmetric and $(A - A')$ is skew symmetric.

Q6. If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, verify that $A'A = I$

Q7. Express the matrix $A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

Q8. If $A$ and $B$ are symmetric matrices of the same order, show that $AB - BA$ is a skew symmetric matrix.

Q9. If $A = \begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 - 5A + 7I = O$

• A matrix is an ordered rectangular array of numbers or functions

• A matrix having $m$ rows and $n$ columns is of order $m \times n$

• $A = [a_{ij}]_{m \times n}$ is a column matrix if $n = 1$, row matrix if $m = 1$

• A square matrix has $m = n$

• $A = [a_{ij}]_{n \times n}$ is diagonal if $a_{ij} = 0$ when $i \neq j$

• $A = [a_{ij}]_{n \times n}$ is scalar if $a_{ij} = 0$ when $i \neq j$ and $a_{ij} = k$ when $i = j$

• Identity matrix has $a_{ij} = 1$ when $i = j$ and $a_{ij} = 0$ when $i \neq j$

• Zero matrix has all elements equal to zero

• Matrices are equal if they have same order and corresponding elements are equal

• Matrix addition: $A + B = [a_{ij} + b_{ij}]$ (same order required)

• Scalar multiplication: $kA = [ka_{ij}]$

• Matrix multiplication $AB$ is defined only when columns of $A$ = rows of $B$

• Properties: $(AB)C = A(BC)$, $A(B+C) = AB + AC$, $(A+B)C = AC + BC$

• Transpose: If $A = [a_{ij}]_{m \times n}$, then $A' = [a_{ji}]_{n \times m}$

• $(A')' = A$, $(kA)' = kA'$, $(A+B)' = A' + B'$, $(AB)' = B'A'$

• Symmetric matrix: $A' = A$

• Skew symmetric matrix: $A' = -A$ (all diagonal elements are 0)

• Any square matrix can be expressed as sum of symmetric and skew symmetric matrix

• If $AB = BA = I$, then $B = A^{-1}$ and $A = B^{-1}$

• Inverse of a matrix, if it exists, is unique

• $(AB)^{-1} = B^{-1}A^{-1}$