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Q. if $A = \begin{bmatrix}4&1&0\\ 1&-2&2\end{bmatrix} , B = \begin{bmatrix}2&0&-1\\ 3&1&x\end{bmatrix} , C = \begin{bmatrix}1\\ 2\\ 1\end{bmatrix}$ and $D =\begin{bmatrix}15+x\\ 1\end{bmatrix}$ such that $(2A -3B)C=D$, then $x$ =

Matrices

Solution:

$(2A - 3B) C =D$
$\Rightarrow \left[2\begin{bmatrix}4&1&0\\ 1&-2&2\end{bmatrix}-3 \begin{bmatrix}2&0&-1\\ 3&1&x\end{bmatrix}\right] \begin{bmatrix}1\\ 2\\ 1\end{bmatrix} =\begin{bmatrix}15+x\\ 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2&2&3\\ -7&-7&4-3x\end{bmatrix} \begin{bmatrix}1\\ 2\\ 1\end{bmatrix} =\begin{bmatrix}15+x\\ 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}9\\ -17-3x\end{bmatrix} = \begin{bmatrix}15+x\\ 1\end{bmatrix} \Rightarrow x = -6$