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Question

(C) First,multiply the first two matrices: $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 1 \ 0 & 3 & 2 \end{bmatrix}

Vedclass · Accepted Answer

$\pm \sqrt{53}$

Vedclass · Answer

(C) First,multiply the first two matrices: $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 1 \ 0 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1(1)+x(0)+1(0) & 1(3)+x(5)+1(3) & 1(2)+x(1)+1(2) \end{bmatrix} = \begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix}$ Now,multiply this result by the third matrix: $\begin{bmatrix} 1 & 5x+6 & x+4 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ x \end{bmatrix} = 1(1) + (5x+6)(1) + (x+4)(x) = 0$ $1 + 5x + 6 + x^2 + 4x = 0$ $x^2 + 9x + 7 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-9 \pm \sqrt{81 - 4(1)(7)}}{2} = \frac{-9 \pm \sqrt{81 - 28}}{2} = \frac{-9 \pm \sqrt{53}}{2}$ We need to find $2x + 9$: $2x + 9 = 2 \left( \frac{-9 \pm \sqrt{53}}{2} ight) + 9 = -9 \pm \sqrt{53} + 9 = \pm \sqrt{53}$ Thus,the correct option is $C$.

If $\begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix} = 0$,then $2x + 9 =$ . . . . . .

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