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Miscellaneous Exercise 4(B) · Q186

Q.Find matrices A and B, where 3A−B=[−121105]3A-B=\begin{bmatrix}-1 & 2 & 1\\1 & 0 & 5\end{bmatrix} and A+5B=[001−100]A+5B=\begin{bmatrix}0 & 0 & 1\\-1 & 0 & 0\end{bmatrix}

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Let M1=[−121105]M_1=\begin{bmatrix}-1 & 2 & 1\\1 & 0 & 5\end{bmatrix}, M2=[001−100]M_2=\begin{bmatrix}0 & 0 & 1\\-1 & 0 & 0\end{bmatrix}. Given: 3A−B=M13A-B=M_1 ... (1); A+5B=M2A+5B=M_2 ... (2).

From (1): B=3A−M1B=3A-M_1. Substitute into (2): A+5(3A−M1)=M2⇒16A=M2+5M1A+5(3A-M_1)=M_2\Rightarrow 16A=M_2+5M_1.

5M1=[−51055025]5M_1=\begin{bmatrix}-5 & 10 & 5\\5 & 0 & 25\end{bmatrix}, so M2+5M1=[−51064025]M_2+5M_1=\begin{bmatrix}-5 & 10 & 6\\4 & 0 & 25\end{bmatrix}.

A=116[−51064025]=[−5/165/83/81/4025/16]A=\tfrac{1}{16}\begin{bmatrix}-5 & 10 & 6\\4 & 0 & 25\end{bmatrix}=\begin{bmatrix}-5/16 & 5/8 & 3/8\\1/4 & 0 & 25/16\end{bmatrix}. …

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