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Find x, y, z if \(\left\{5\begin{bmatrix}0 & 1\\ 1 & 0\\ 1 & 1\end{bmatrix}-\begin{bmatrix}2 & 1\\ 3 & -2\\ 1 & 3\end{bmatrix}\right\}\begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}x-1\\ y+1\\ 2z\end{bmatrix}\)

Q.2 (A) (ii) Find x, y, z if \(\left\{5\begin{bmatrix}0 & 1\\ 1 & 0\\ 1 & 1\end{bmatrix}-\begin{bmatrix}2 & 1\\ 3 & -2\\ 1 & 3\end{bmatrix}\right\}\begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}x-1\\ y+1\\ 2z\end{bmatrix}\)

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Answer: \(x=1,\ y=5,\ z=5\)

\(5\begin{bmatrix}0 & 1\\ 1 & 0\\ 1 & 1\end{bmatrix}-\begin{bmatrix}2 & 1\\ 3 & -2\\ 1 & 3\end{bmatrix}=\begin{bmatrix}0 & 5\\ 5 & 0\\ 5 & 5\end{bmatrix}-\begin{bmatrix}2 & 1\\ 3 & -2\\ 1 & 3\end{bmatrix}=\begin{bmatrix}-2 & 4\\ 2 & 2\\ 4 & 2\end{bmatrix}\)

\(\begin{bmatrix}-2 & 4\\ 2 & 2\\ 4 & 2\end{bmatrix}\begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}-4+4\\ 4+2\\ 8+2\end{bmatrix}=\begin{bmatrix}0\\ 6\\ 10\end{bmatrix}\)

Equating with \(\begin{bmatrix}x-1\\ y+1\\ 2z\end{bmatrix}\): \(x-1=0,\ y+1=6,\ 2z=10\). Hence \(x=1,\ y=5,\ z=5\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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