# Compute A B and BA, which ever exists when

Question:

Compute $A B$ and BA, which ever exists when

$A=\left[\begin{array}{ll}-1 & 1 \\ -2 & 2 \\ -3 & 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}3 & -2 & 1 \\ 0 & 1 & 2 \\ -3 & 4 & -5\end{array}\right]$

Solution:

Given : $A=\left[\begin{array}{cc}-1 & 1 \\ -2 & 2 \\ -3 & 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}3 & -2 & 1 \\ 0 & 1 & 2 \\ -3 & 4 & -5\end{array}\right]$

Matrix $A$ is of order $3 \times 2$, and Matrice $B$ is of order $3 \times 3$

To find: matrix $A B$ and $B A$

Formula used:

Where $c_{i j}=a_{i 1} b_{1 j}+a_{i 2} b_{2 j}+a_{i 3} b_{3 j}+\ldots \ldots \ldots \ldots \ldots . . a_{i n} b_{n j}$

If $A$ is a matrix of order $a \times b$ and $B$ is a matrice of order $c \times d$, then matrice $A B$ exists and is of order $a \times d$,if and only if $b$ $=\mathrm{C}$

If $A$ is a matrix of order $a \times b$ and $B$ is a matrice of order $c \times d$, then matrice BA exists and is of order $c \times b$, if and only if $d$ $=\mathrm{a}$

For matrix $A B, a=3, b=2, c=3, d=3$, thus matrix $A B$ does not exist, as $d \neq a$

But $2 \neq 3$

Thus matrix $A B$ does not exist

For matrix $\mathrm{BA}, \mathrm{a}=3, \mathrm{~b}=2, \mathrm{c}=3, \mathrm{~d}=3$, thus matrix $\mathrm{BA}$ is of order $3 \times 2$

as $d=a=3$

Matrix BA $=\left[\begin{array}{cc}-3+4-3 & 3-4+3 \\ 0-2-6 & 0+2+6 \\ 3-8+15 & -3+8-15\end{array}\right]=\left[\begin{array}{cc}-2 & 2 \\ -8 & 8 \\ 10 & -10\end{array}\right]$

Matrix $\mathrm{BA}=\left[\begin{array}{cc}-2 & 2 \\ -8 & 8 \\ 10 & -10\end{array}\right]$

Matrix $B A=\left[\begin{array}{cc}-2 & 2 \\ -8 & 8 \\ 10 & -10\end{array}\right]$