Let $d \in R $ and
$A= \begin{bmatrix} -2 & 4 + d& sin \theta - 2\\1 & sin \theta + 2 & d\\5 & 2 sin \theta - d& -sin \theta + 2 + 2d \end {bmatrix}, \theta ←[0, 2\pi ]$. If the maximum value of det (A) is 8, then a value of d is
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : -5
$A= \begin{bmatrix} -2 & 4 + d& sin \theta - 2\\1 & sin \theta + 2 & d\\5 & 2 sin \theta - d& -sin \theta + 2 + 2d \end {bmatrix}$
$⇒|A|=\begin{vmatrix}-2 & 4 + d & sin \theta - 2\\1 & sin \theta + 2 & d\\1 & 0 & 0 \end{vmatrix}$
Applying $R_3→R_3-2R_2+R_1$
$⇒|A|=(4 + d) d- (sin^2 \theta - 4) = d^2 + 4d + 4 -sin^2 \theta $
$⇒|A|=(d+2)^2 -sin^2\theta $ ...............(i)
Clearly, |A| is minimum when sin $\theta = \pm 1.$
Substituting $|A|=8 $ and $sin \theta = \pm 1 $ in (i) ,we get
$∴8= (d+2)^2 -1⇒d= -5, 1$