If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$, then the value of $\det(A^{4}) + \det(A^{10} -…

If $A = \begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$, then the value of $\det(A^{4}) + \det(A^{10} - (\text{Adj}(2A))^{10})$ is equal to ________.

Solution

Given, $A=\begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}$ $\Rightarrow A^2=\begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}=\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix}$ $\Rightarrow A^4=\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 16 & 15 \\ 0 & 1 \end{bmatrix}$ $\Rightarrow A^5=\begin{bmatrix} 16 & 15 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 0 & -1 \end{bmatrix}=\begin{bmatrix} 32 & 33 \\ 0 & -1 \end{bmatrix}$ $\Rightarrow A^{10}=\begin{bmatrix} 32 & 33 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} 32 & 33 \\ 0 & -1 \end{bmatrix}=\begin{bmatrix} 1024 & 1023 \\ 0 & 1 \end{bmatrix}$ Now, $2A=\begin{bmatrix} 4 & 6 \\ 0 & -2 \end{bmatrix}$ $\Rightarrow \text{Adj}(2A)=\begin{bmatrix} -2 & 0 \\ -6 & 4 \end{bmatrix}^T$ $\Rightarrow \text{Adj}(2A)=\begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}$ $\Rightarrow (\text{Adj}(2A))^2=\begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} 4 & -12 \\ 0 & 16 \end{bmatrix}$ $\Rightarrow (\text{Adj}(2A))^4=\begin{bmatrix} 4 & -12 \\ 0 & 16 \end{bmatrix}\begin{bmatrix} 4 & -12 \\ 0 & 16 \end{bmatrix}=\begin{bmatrix} 16 & -240 \\ 0 & 256 \end{bmatrix}$ $\Rightarrow (\text{Adj}(2A))^5=\begin{bmatrix} 16 & -240 \\ 0 & 256 \end{bmatrix}\begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} -32 & -1056 \\ 0 & 1024 \end{bmatrix}$ $\Rightarrow (\text{Adj}(2A))^{10}=\begin{bmatrix} -32 & -1056 \\ 0 & 1024 \end{bmatrix}\begin{bmatrix} -32 & -1056 \\ 0 & 1024 \end{bmatrix}=\begin{bmatrix} 1024 & -1047552 \\ 0 & 1048576 \end{bmatrix}$ From equation (i) & (ii) $\Rightarrow A^{10}-(\text{Adj}(2A))^{10}=\begin{bmatrix} 1024 & 1023 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1024 & -1047552 \\ 0 & 1048576 \end{bmatrix}=\begin{bmatrix} 0 & 1048575 \\ 0 & -1048575 \end{bmatrix}$ $\Rightarrow \text{det}(A^{10}-(\text{Adj}(2A))^{10})=\begin{vmatrix} 0 & 1048575 \\ 0 & -1048575 \end{vmatrix}=0$ Let $E=\text{det}(A^4)+\text{det}(A^{10}-(\text{Adj}(2A))^{10})$ $\Rightarrow E=\text{det}(A)^4+0$ $\Rightarrow E=(-2-0)^4$ $\Rightarrow E=16$

Asked in: JEE Main 2021 (17 Mar Shift 1)

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