If \(z=\begin{bmatrix}x & x^2\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}\begin{bmatrix}x\\x^2\end{bmatrix}\), then \(\left(\displaystyle\frac{d^2z}{dx^2}\right)_{x=-1}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 32
Multiplying the row matrix by the square matrix first:
\( \displaystyle \begin{bmatrix}x & x^2\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix} = \begin{bmatrix}x + 2x^2 & -x + 3x^2\end{bmatrix} \)
Then by the column matrix:
\( z = x(x + 2x^2) + x^2(-x + 3x^2) \)
\( z = x^2 + 2x^3 - x^3 + 3x^4 \)
\( z = 3x^4 + x^3 + x^2 \)
Differentiating twice:
\( \displaystyle \frac{dz}{dx} = 12x^3 + 3x^2 + 2x \)
\( \displaystyle \frac{d^2z}{dx^2} = 36x^2 + 6x + 2 \)
At x = −1:
\( 36(-1)^2 + 6(-1) + 2 = 36 - 6 + 2 = 32 \)