Question

If \[ A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \] show that \( A^2 = I_3 \).

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Solution

Compute \( A^2 = A \times A \)

\[ A^2 = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \] —

Multiplying directly:

\[ A^2 = \begin{bmatrix} 4\cdot4 + (-1)\cdot3 + (-4)\cdot3 & 4(-1) + (-1)\cdot0 + (-4)(-1) & 4(-4) + (-1)(-4) + (-4)(-3) \\ 3\cdot4 + 0\cdot3 + (-4)\cdot3 & 3(-1) + 0\cdot0 + (-4)(-1) & 3(-4) + 0(-4) + (-4)(-3) \\ 3\cdot4 + (-1)\cdot3 + (-3)\cdot3 & 3(-1) + (-1)\cdot0 + (-3)(-1) & 3(-4) + (-1)(-4) + (-3)(-3) \end{bmatrix} \] —

Simplifying:

\[ A^2 = \begin{bmatrix} 16 – 3 – 12 & -4 + 0 + 4 & -16 + 4 + 12 \\ 12 + 0 – 12 & -3 + 0 + 4 & -12 + 0 + 12 \\ 12 – 3 – 9 & -3 + 0 + 3 & -12 + 4 + 9 \end{bmatrix} \] —

Final Result:

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Conclusion

Hence proved.

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