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### Answers (1)

Hence proved $AB \neq BA$

$\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \neq\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right]$

Hint: matrix multiplication is only possible, when number of columns of first matrix is equal to the number of rows of second matrix.

Given: $A =\begin{bmatrix} -1 &1 &0 \\ 0& -1 &1 \\ 2& 3 &4 \end{bmatrix}$ and $B=\begin{bmatrix} 1 &2 &3 \\ 0& 1 &0 \\ 1& 1 &0 \end{bmatrix}$

Consider,

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

\begin{aligned} &A B=\left[\begin{array}{ccc} (-1) \times 1+1 \times 0+0 \times 1 & -1 \times 2+1 \times 1+0 \times 1 & -1 \times 3+1 \times 0+0 \times 0 \\ 0 \times 1+(-1) \times 0+1 \times 1 & 0 \times 2+(-1) \times 1+1 \times 1 & 0 \times 3+(-1) \times 0+1 \times 0 \\ 2 \times 1+3 \times 0+4 \times 1 & 2 \times 2+3 \times 1+4 \times 1 & 2 \times 3+3 \times 0+4 \times 0 \end{array}\right] \\ &A B=\left[\begin{array}{ccc} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0 \end{array}\right] \end{aligned}

$AB =\begin{bmatrix} -1 & -1 &-3 \\ 1& 0 &0 \\ 6 &1 1 &6 \end{bmatrix}$                                        ...(i)

Now again consider,

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

\begin{aligned} &B A=\left[\begin{array}{lll} 1 \times(-1)+2 \times 0+3 \times 2 & 1 \times 1+2 \times(-1)+3 \times 3 & 1 \times 0+2 \times 1+3 \times 4 \\ 0 \times(-1)+1 \times 0+0 \times 2 & 0 \times 1+1 \times(-1)+0 \times 3 & 0 \times 0+1 \times 1+0 \times 4 \\ 1 \times(-1)+1 \times 0+0 \times 2 & 1 \times 1+1 \times(-1)+0 \times 3 & 1 \times 0+1 \times 1+0 \times 4 \end{array}\right] \\ B A & =\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] \end{aligned}

$BA =\begin{bmatrix} 5 &8 &14 \\ 0 & -1 & 1\\ -1 & 0 & 1 \end{bmatrix}$                                         ...(ii)

From equation (i) and (ii), it is clear that $AB \neq BA$

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