The maximum value of the determinant of the matrix $\begin{bmatrix}1&1&1\\1&1+\sin x&1\\1+\cos x&1&1\end{bmatrix}$ is: (where x is real)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2}$
Given matrix:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{bmatrix}$
Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Rightarrow \begin{bmatrix} 1 & 1 & 1 \\ 0 & \sin x & 0 \\ \cos x & 0 & 0 \end{bmatrix}$
Now, expand along the third column:
$\det(A) = 1\begin{vmatrix} 0 & \sin x \\ \cos x & 0 \end{vmatrix} = -( \sin x \cos x ) = -\sin x \cos x$
So $|\det(A)| = |\sin x \cos x| = \frac{1}{2}|\sin 2x|$
Maximum value of $|\sin 2x| = 1$
Hence, maximum value of determinant = $\frac{1}{2}$