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Exercise 4.5 · Q114

Q.Solve the following equations for XX and YY, if 3X−Y=[1−1−11]3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} and X−3Y=[0−10−1]X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix}.

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Given 3X−Y=[1−1−11]3X - Y = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} ...(1) and X−3Y=[0−10−1]X - 3Y = \begin{bmatrix} 0 & -1 \\ 0 & -1 \end{bmatrix} ...(2).

Multiply (1) by 3: 9X−3Y=[3−3−33]9X - 3Y = \begin{bmatrix} 3 & -3 \\ -3 & 3 \end{bmatrix} ...(3)

Subtract (2) from (3): (9X−3Y)−(X−3Y)=8X=[3−0−3−(−1)−3−03−(−1)]=[3−2−34](9X-3Y)-(X-3Y) = 8X = \begin{bmatrix} 3-0 & -3-(-1) \\ -3-0 & 3-(-1) \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix}

X=18[3−2−34]=[38−14−3812]X = \frac{1}{8}\begin{bmatrix} 3 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & -\frac{1}{4} \\ -\frac{3}{8} & \frac{1}{2} \end{bmatrix}

Substitute into (1): Y=3X−[1−1−11]=[98−1−34−(−1)−98−(−1)32−1]=[1814−1812]Y = 3X - \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{9}{8}-1 & -\frac{3}{4}-(-1) \\ -\frac{9}{8}-(-1) & \frac{3}{2}-1 \end{bmatrix} = \begin{bmatrix} \frac{1}{8} & \frac{1}{4} \\ -\frac{1}{8} & \frac{1}{2} \end{bmatrix} …

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