ICSE Class 10 Mathematics Question 7 of 21

Solved 2024 Specimen Paper ICSE Class 10 Mathematics — Question 1

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Question

Question 4(i)

If A = [4−4−44]\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}, find A2. If A2 = pA, then find the value of p.

Answer

Given,

⇒ A2 = pA

⇒[4−4−44][4−4−44]=p[4−4−44]⇒[4×4+(−4)×(−4)4×(−4)+(−4)×4−4×4+4×(−4)(−4)×(−4)+4×4]=p[4−4−44]⇒[16+16−16−16−16−1616+16]=p[4−4−44]⇒[32−32−3232]=p[4−4−44]⇒32[1−1−11]=4p[1−1−11]⇒32=4p⇒p=324=8.\Rightarrow \begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*}\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 4 \times 4 + (-4) \times (-4) & 4 \times (-4) + (-4) \times 4 \\ -4 \times 4 + 4 \times (-4) & (-4) \times (-4) + 4 \times 4 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 16 + 16 & -16 - 16 \\ -16 - 16 & 16 + 16 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] 32 & -32 \\ -32 & 32 \end{bmatrix*} = p\begin{bmatrix*}[r] 4 & -4 \\ -4 & 4 \end{bmatrix*} \\[1em] \Rightarrow 32\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} = 4p\begin{bmatrix*}[r] 1 & -1 \\ -1 & 1 \end{bmatrix*} \\[1em] \Rightarrow 32 = 4p \\[1em] \Rightarrow p = \dfrac{32}{4} = 8.

Hence, p = 8.

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