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If A=[31−12]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} then A2−5AA^2 - 5A is equal to (where I is identity matrix of order 2)

Solution

✅ Correct Option: 3

Given A=[31−12]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

Calculate A2A^2:

A2=A×AA^2 = A \times A

=[31−12]×[31−12]= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

=[(3)(3)+(1)(−1)(3)(1)+(1)(2)(−1)(3)+(2)(−1)(−1)(1)+(2)(2)]= \begin{bmatrix} (3)(3)+(1)(-1) & (3)(1)+(1)(2) \\ (-1)(3)+(2)(-1) & (-1)(1)+(2)(2) \end{bmatrix}

=[9−13+2−3−2−1+4]= \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}

=[85−53]= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}


Calculate 5A5A:

5A=5×[31−12]5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

=[155−510]= \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}


Calculate A2−5AA^2 - 5A:

A2−5A=[85−53]−[155−510]A^2 - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}

=[8−155−5−5−(−5)3−10]= \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix}

=[−700−7]= \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}


The identity matrix I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Therefore:

[−700−7]=−7×[1001]=−7I\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} = -7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -7I

The answer is −7I-7I.

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