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IPMAT 2025 Exam Solved Papers

Question : 7 of 90

Marks: +1, -0

A=\left[\begin{array}{cc}2 & n \\ 4 & 1\end{array}\right]

A^{3}=27\left[\begin{array}{cc}4 & q \\ p & r\end{array}\right]

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Solution:

\begin{aligned} & A=\left[\begin{array}{ll} 2 & n \\ 4 & 1 \end{array}\right] \\ & \Rightarrow A^{2}=\left[\begin{array}{ll} 2 & n \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 2 & n \\ 4 & 1 \end{array}\right] \\ & \Rightarrow A^{2}=\left[\begin{array}{cc} 4+4 n & 3 n \\ 12 & 4 n+1 \end{array}\right] \end{aligned}

And,

A^{3}=\left[\begin{array}{cc}4+4 n & 3 n \\ 12 & 4 n+1\end{array}\right]\left[\begin{array}{cc}2 & n \\ 4 & 1\end{array}\right]

\Rightarrow A^{3}=\left[\begin{array}{cc} 8+20 n & 4 n^{2}+7 n \\ 16 n+28 & 16 n+1 \end{array}\right] \rightarrow 1

And, it is given that

A^{3}=27\left[\begin{array}{ll}4 & q \\ p & r\end{array}\right]

\Rightarrow A^{3}=\left[\begin{array}{ll} 108 & 27 q \\ 27 p & 27 r \end{array}\right] \rightarrow 2

Comparing each cell value in eq. 1 and eq. 2 -

\begin{aligned} & = > 8+20 n=108 \Rightarrow n=5 \\ & \Rightarrow 4 n^{2}+7 n=27 q \Rightarrow 135=27 q \Rightarrow q=5 \\ & \Rightarrow 16 n+28=27 p \Rightarrow 108=27 p \Rightarrow p=4 \\ & \Rightarrow 16 n+1=27 r \Rightarrow 81=27 r \Rightarrow r=3 \end{aligned}

Thus, the value of

p+q+r=4+5+3=12

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