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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
For some $\alpha, \beta \in R$, let $A = \begin{bmatrix} \alpha & 2 \ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \ 1 & \beta \end{bmatrix}$ be such that $A^2 - 4A + 2I = B^2 - 3B + I = O$. Then $(\det(\text{adj}(A^3 - B^3)))$ is equal to ______
Solution
$\text{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$
$\text{Tr}(B) = 3 \Rightarrow \beta + 1 = 3 \Rightarrow \beta = 2$
$A^2 - 4A + 2I = 0$
$A^3 = 4A^2 - 2A = 16A - 8I - 2A = 14A - 8I$
$= \begin{bmatrix} 28 & 28 \ 14 & 20 \end{bmatrix}$
$B^2 - 3B + I = 0 \Rightarrow B^2 = 3B - I$
$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$
$= \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix}$
$A^3 - B^3 = \begin{bmatrix} 15 & 20 \ 6 & 7 \end{bmatrix}$
$\Rightarrow |A^3 - B^3| = 105 - 120 = -15$
$\Rightarrow \det(\text{adj}(A^3 - B^3)) = |A^3 - B^3|^2 = 225$
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