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Question

(C) First,multiply the first two matrices: $\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & x & 3 \ 2 & 4 & 5 \ 3 & 2 & x \end{bmatrix}

Vedclass · Accepted Answer

$-\frac{5}{3}$

Vedclass · Answer

(C) First,multiply the first two matrices: $\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & x & 3 \ 2 & 4 & 5 \ 3 & 2 & x \end{bmatrix} = \begin{bmatrix} (2(1)+3(2)+4(3)) & (2(x)+3(4)+4(2)) & (2(3)+3(5)+4(x)) \end{bmatrix} = \begin{bmatrix} 20 & 2x+20 & 21+4x \end{bmatrix}$.Now,multiply this result by the third matrix: $\begin{bmatrix} 20 & 2x+20 & 21+4x \end{bmatrix} \begin{bmatrix} x \ 2 \ 0 \end{bmatrix} = 20(x) + (2x+20)(2) + (21+4x)(0) = 0$.This simplifies to: $20x + 4x + 40 = 0$.$24x = -40$.$x = -\frac{40}{24} = -\frac{5}{3}$.

If $\begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & x & 3 \\ 2 & 4 & 5 \\ 3 & 2 & x \end{bmatrix} \begin{bmatrix} x \\ 2 \\ 0 \end{bmatrix} = O$,then $x = $ . . . . . .

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