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Question 28 of 34

Q.Find x, y, z, if {5[011011]−[213−213]}[21]=[x−1y+12z]\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}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 3mImportance★★★★★
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5[011011]−[213−213]=[−242242]5\begin{bmatrix}0&1\\1&0\\1&1\end{bmatrix}-\begin{bmatrix}2&1\\3&-2\\1&3\end{bmatrix}=\begin{bmatrix}-2&4\\2&2\\4&2\end{bmatrix}; multiplying by [21]\begin{bmatrix}2\\1\end{bmatrix} gives [0610]=[x−1y+12z]\begin{bmatrix}0\\6\\10\end{bmatrix}=\begin{bmatrix}x-1\\y+1\\2z\end{bmatrix}, so x=1, y=5, z=5x=1,\ y=5,\ z=5.

Step 1 — scalar multiply.

5[011011]=[055055]5\begin{bmatrix}0&1\\1&0\\1&1\end{bmatrix}=\begin{bmatrix}0&5\\5&0\\5&5\end{bmatrix}

Step 2 — subtract.

[055055]−[213−213]=[−242242]\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}

Step 3 — multiply by the column [21]\begin{bmatrix}2\\1\end{bmatrix} (a 3×23\times2 times a 2×12\times1 gives a 3×13\times1): …

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