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  • Verify that A^2 = I when A equals [0 1 -1]

Verify that A^2 = I when A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]

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We need to prove that

\begin{array}{l} A^{2}=I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \because A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \\ \therefore A^{2}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \end{array}

According to the rule of matrix multiplication we have-

\begin{aligned} &A^{2}=\left[\begin{array}{ccc} 0 \times 0+1 \times 4+(-1) \times 3 & 0 \times 1+1 \times(-3)+(-1) \times(-3) & 0 \times(-1)+1 \times 4+(-1) \times 4 \\ 4 \times 0+(-3) \times 4+4 \times 3 & 4 \times 1+(-3) \times(-3)+4 \times(-3) & 4 \times(-1)+(-3) \times 4+4 \times 4 \\ 3 \times 0+(-3) \times 4+4 \times 3 & 3 \times 1+(-3) \times(-3)+4 \times(-3) & 3 \times(-1)+(-3) \times 4+4 \times 4 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ccc} 4-3 & -3+3 & 4-4 \\ -12+12 & 4+9-12 & -4-12+16 \\ -12+12 & 3+9-12 & -3+16-12 \end{array}\right]\\ &\therefore \mathrm{A}^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I\\ &\text { Hence Proved } \end{aligned}

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