Question
Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$.
If $A^{2} - 4A + I = O$ and $B^{2} - 5B - 6I = O$, then among the two statements :
(S1) : $\lbrack(B - A)(B + A)\rbrack^{T} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2) : $det(adj(A + B)) = - 5$,
Options
Solution
Sol. [Key Concept: Use the trace (Cayley–Hamilton) to find $\alpha,\beta$, then test each statement with the adj/det identities.]
$${\ A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix} }{A^{2} - 4\text{ }A + I = 0 }$$trace of matrix
$$\alpha + 1 = 4 $$
$$\alpha = 3 $$
$$B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix} $$
$$B^{2} - 5\text{ }B - 6I = 0 $$
$$det.(B) = 6 - 3\beta = - 6 $$
$$3\beta = 12 $$
$$\beta = 4 $$
$$B - A = \begin{bmatrix} 2 & 1 \\ 3 & - 1 \end{bmatrix};B + A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} $$
$$(B - A)(B + A) = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix} $$
$$adj(A) = \begin{bmatrix} 3 & - 2 \\ - 1 & 1 \end{bmatrix},adjB = \begin{bmatrix} 2 & - 3 \\ - 4 & 3 \end{bmatrix} $$
$$adj(A + B) = \begin{bmatrix} 5 & - 5 \\ - 5 & 4 \end{bmatrix} $$
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