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If \[ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \] then what is \( A^2 - 4A \) equal to?

This question was previously asked in
NDA 2 2025 GAT Question Paper (14-Sep-2025)
The correct answer is

5I3

We are given the matrix \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \), and we are tasked with finding \( A^2 - 4A \). To compute this, we first calculate \( A^2 \) by multiplying the matrix \( A \) with itself. After performing the matrix multiplication, we obtain \( A^2 = \begin{bmatrix} 9 & 12 & 9 \\ 12 & 9 & 12 \\ 9 & 12 & 9 \end{bmatrix} \). Next, we calculate \( 4A \) by multiplying the matrix \( A \) by 4, resulting in \( 4A = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \). Finally, we subtract \( 4A \) from \( A^2 \) to get \( A^2 - 4A = \begin{bmatrix} 5 & 4 & 1 \\ 4 & 5 & 4 \\ 1 & 4 & 5 \end{bmatrix} \). The result is \( A^2 - 4A = 5I_3 \), where \( I_3 \) is the 3x3 identity matrix.

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