If \(\begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix} = \begin{bmatrix}6\end{bmatrix}\), then the values of \(x\) are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2, -3\)
Multiplying from the left, a \(1 \times 2\) matrix times a \(2 \times 2\) matrix gives a \(1 \times 2\) matrix:
\( \displaystyle \begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix} = \begin{bmatrix}x - 3&2x\end{bmatrix} \)
Then a \(1 \times 2\) matrix times a \(2 \times 1\) matrix gives a \(1 \times 1\) matrix:
\( \displaystyle \begin{bmatrix}x - 3&2x\end{bmatrix}\begin{bmatrix}x\\2\end{bmatrix} = \begin{bmatrix}x^2 - 3x + 4x\end{bmatrix} = \begin{bmatrix}x^2 + x\end{bmatrix} \)
By equality of matrices:
\(x^2 + x = 6\)
\(x^2 + x - 6 = 0\)
\((x - 2)(x + 3) = 0\)
\(x = 2 \text{ or } x = -3\)