Question
If \[ A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \] show that \( A^2 = A \).
Solution
We compute \( A^2 = A \times A \)
\[ A^2 = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \] —Direct Multiplication (Single Matrix Form)
\[ A^2 = \begin{bmatrix} 2\cdot2 + (-3)(-1) + (-5)(1) & 2(-3) + (-3)(4) + (-5)(-3) & 2(-5) + (-3)(5) + (-5)(-4) \\ (-1)(2) + 4(-1) + 5(1) & (-1)(-3) + 4(4) + 5(-3) & (-1)(-5) + 4(5) + 5(-4) \\ 1\cdot2 + (-3)(-1) + (-4)(1) & 1(-3) + (-3)(4) + (-4)(-3) & 1(-5) + (-3)(5) + (-4)(-4) \end{bmatrix} \] —Simplifying
\[ A^2 = \begin{bmatrix} 4 + 3 – 5 & -6 – 12 + 15 & -10 – 15 + 20 \\ -2 – 4 + 5 & 3 + 16 – 15 & 5 + 20 – 20 \\ 2 + 3 – 4 & -3 – 12 + 12 & -5 – 15 + 16 \end{bmatrix} \] —Final Result
\[
A^2 =
\begin{bmatrix}
2 & -3 & -5 \\
-1 & 4 & 5 \\
1 & -3 & -4
\end{bmatrix}
= A
\]
—
Conclusion
\[
A^2 = A
\]
Hence proved.