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Exercise 4.6 · Q147

Q.Find x, y, z if {3[200222]−4[11−1231]}[12]=[x−3y−12z]\left\{3\begin{bmatrix}2&0\\0&2\\2&2\end{bmatrix}-4\begin{bmatrix}1&1\\-1&2\\3&1\end{bmatrix}\right\}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}x-3\\y-1\\2z\end{bmatrix}.

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Given {3[200222]−4[11−1231]}[12]=[x−3y−12z]\left\{3\begin{bmatrix}2&0\\0&2\\2&2\end{bmatrix}-4\begin{bmatrix}1&1\\-1&2\\3&1\end{bmatrix}\right\}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}x-3\\y-1\\2z\end{bmatrix}.

Step 1 — scale each matrix:

3[200222]=[600666]3\begin{bmatrix}2&0\\0&2\\2&2\end{bmatrix}=\begin{bmatrix}6&0\\0&6\\6&6\end{bmatrix}, 4[11−1231]=[44−48124]4\begin{bmatrix}1&1\\-1&2\\3&1\end{bmatrix}=\begin{bmatrix}4&4\\-4&8\\12&4\end{bmatrix}

Step 2 — subtract:

[6−40−40−(−4)6−86−126−4]=[2−44−2−62]\begin{bmatrix}6-4&0-4\\0-(-4)&6-8\\6-12&6-4\end{bmatrix}=\begin{bmatrix}2&-4\\4&-2\\-6&2\end{bmatrix}

Step 3 — multiply by [12]\begin{bmatrix}1\\2\end{bmatrix}:

Row 1: 2(1)+(−4)(2)=2−8=−62(1)+(-4)(2)=2-8=-6 …

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