If A =[(3,1)(-1,2)] , show that A2 – 5A + 7I = 0.

Solving A2, Solving 5A, Solving 7I, Now, A2 – 5A + 7I Hence, A2 – 5A + 7I = .

Mathematics

If A = $\begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix}$ , show that A² – 5A + 7I = 0.

Matrices

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Answer

Solving A²,

$\Rightarrow A^2 = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \ -1 \times 3 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - 1 & 3 + 2 \ -3 - 2 & -1 + 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix}.$

Solving 5A,

$\Rightarrow 5A = 5 \times \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix}.$

Solving 7I,

$\Rightarrow 7I = 7 \times \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix}.$

Now, A² – 5A + 7I

$\Rightarrow \begin{bmatrix} 8 & 5 \ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 8 - (15) + 7 & 5 - (5) + 0 \ -5 - 5 + 0 & 3 - (10) + 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}.$

Hence, A² – 5A + 7I = $\begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$ .

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Matrices

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