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Mathematics Question on Matrices

if A=[410 1−22],B=[20−1 31x],C=[1 2 1]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=[15+x 1]D =\begin{bmatrix}15+x\\\ 1\end{bmatrix} such that (2A−3B)C=D(2A -3B)C=D, then xx =

A

33

B

−4-4

C

−6-6

D

66

Answer

−6-6

Explanation

Solution

(2A−3B)C=D(2A - 3B) C =D ⇒[2[410 1−22]−3[20−1 31x]][1 2 1]=[15+x 1]\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} ⇒[223 −7−74−3x][1 2 1]=[15+x 1]\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} ⇒[9 −17−3x]=[15+x 1]⇒x=−6\Rightarrow \begin{bmatrix}9\\\ -17-3x\end{bmatrix} = \begin{bmatrix}15+x\\\ 1\end{bmatrix} \Rightarrow x = -6