Solve this following

Question:

Show that $\mathrm{AB} \neq \mathrm{BA}$ in each of the following cases:

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

Solution:

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

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

To show : matrix $\mathrm{AB} \neq \mathrm{BA}$

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

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

For matrix $A B, a=3, b=c=3, d=3$, thus matrix $A B$ is of order $3 \times 3$

Matrix $\mathrm{AB}=$

Matrix $A B=\left[\begin{array}{ccc}-1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0\end{array}\right]=\left[\begin{array}{ccc}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{array}\right]$

Matrix $A B=\left[\begin{array}{ccc}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{array}\right]$

For matrix $B A, a=3, b=c=3, d=3$, thus matrix $A B$ is of order $3 \times 3$

Matrix $\mathrm{BA}=$

Matrix BA $=\left[\begin{array}{ccc}-1+0+0 & -2+1+0 & -3+0+0 \\ 0-1+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0\end{array}\right]=\left[\begin{array}{ccc}-1 & -1 & -3 \\ 0 & 0 & 0 \\ 6 & 11 & 6\end{array}\right]$

Matrix BA $=\left[\begin{array}{ccc}-1 & -1 & -3 \\ 0 & 0 & 0 \\ 6 & 11 & 6\end{array}\right]$

Matrix $B A=\left[\begin{array}{ccc}-1 & -1 & -3 \\ 0 & 0 & 0 \\ 6 & 11 & 6\end{array}\right]$ and Matrix $A B=\left[\begin{array}{ccc}5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1\end{array}\right]$

Matrix $A B \neq B A$