If matrix X = [−342−3][2−2] and 2X - 3Y=[10−8]\begin{bmatrix}[r] -3 & 4 \ 2 & -3 \end{bmatrix}\begin{bmatrix}[r] 2 \ -2 \end{bmatrix} \text{ and 2X - 3Y} = \begin{bmatrix}[r] 10 \ -8 \end{bmatrix}[−324−3][2−2] and 2X - 3Y=[10−8], find the matrix 'X' and matrix 'Y'.
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Given,
X=[−342−3][2−2]=[−3×2+4×(−2)2×2+(−3)×(−2)]=[−6+(−8)4+6]=[−1410].X = \begin{bmatrix}[r] -3 & 4 \ 2 & -3 \end{bmatrix}\begin{bmatrix}[r] 2 \ -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -3 \times 2 + 4 \times (-2) \ 2 \times 2 + (-3) \times (-2) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 + (-8) \ 4 + 6 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -14 \ 10 \end{bmatrix}.X=[−324−3][2−2]=[−3×2+4×(−2)2×2+(−3)×(−2)]=[−6+(−8)4+6]=[−1410].
⇒2X−3Y=[10−8]⇒2[−1410]−3Y=[10−8]⇒[−2820]−3Y=[10−8]⇒3Y=[−2820]−[10−8]⇒3Y=[−28−1020−(−8)]⇒3Y=[−3828]⇒Y=13[−3828].\Rightarrow 2X - 3Y = \begin{bmatrix}[r] 10 \ -8 \end{bmatrix} \\[1em] \Rightarrow 2\begin{bmatrix}[r] -14 \ 10 \end{bmatrix} - 3Y = \begin{bmatrix}[r] 10 \ -8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -28 \ 20 \end{bmatrix} - 3Y = \begin{bmatrix}[r] 10 \ -8 \end{bmatrix} \\[1em] \Rightarrow 3Y = \begin{bmatrix}[r] -28 \ 20 \end{bmatrix} - \begin{bmatrix}[r] 10 \ -8 \end{bmatrix} \\[1em] \Rightarrow 3Y = \begin{bmatrix}[r] -28 - 10 \ 20 - (-8) \end{bmatrix} \\[1em] \Rightarrow 3Y = \begin{bmatrix}[r] -38 \ 28 \end{bmatrix} \\[1em] \Rightarrow Y = \dfrac{1}{3}\begin{bmatrix}[r] -38 \ 28 \end{bmatrix}.⇒2X−3Y=[10−8]⇒2[−1410]−3Y=[10−8]⇒[−2820]−3Y=[10−8]⇒3Y=[−2820]−[10−8]⇒3Y=[−28−1020−(−8)]⇒3Y=[−3828]⇒Y=31[−3828].
Hence, X = [−1410] and Y=13[−3828]\begin{bmatrix}[r] -14 \ 10 \end{bmatrix} \text{ and Y} = \dfrac{1}{3}\begin{bmatrix}[r] -38 \ 28 \end{bmatrix}[−1410] and Y=31[−3828].
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