Find x and y, if [2xxy3y][32]=[169].\begin{bmatrix}[r] 2x & x \ y & 3y \end{bmatrix} \begin{bmatrix}[r] 3 \ 2 \end{bmatrix} = \begin{bmatrix}[r] 16 \ 9 \end{bmatrix}.[2xyx3y][32]=[169].
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Given,
[2xxy3y][32]=[169]L.H.S.=[2xxy3y][32]=[2x×3+x×2y×3+3y×2]=[6x+2x3y+6y]=[8x9y]\begin{bmatrix}[r] 2x & x \ y & 3y \end{bmatrix} \begin{bmatrix}[r] 3 \ 2 \end{bmatrix} = \begin{bmatrix}[r] 16 \ 9 \end{bmatrix} \\[1em] \text{L.H.S.} = \begin{bmatrix}[r] 2x & x \ y & 3y \end{bmatrix} \begin{bmatrix}[r] 3 \ 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2x \times 3 + x \times 2 \ y \times 3 + 3y \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 6x + 2x \ 3y + 6y \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 8x \ 9y \end{bmatrix}[2xyx3y][32]=[169]L.H.S.=[2xyx3y][32]=[2x×3+x×2y×3+3y×2]=[6x+2x3y+6y]=[8x9y]
Comparing L.H.S. with R.H.S. we get,
⇒[8x9y]=[169]⇒8x=16 and 9y=9∴x=2 and y=1.\Rightarrow \begin{bmatrix}[r] 8x \ 9y \end{bmatrix} = \begin{bmatrix}[r] 16 \ 9 \end{bmatrix} \\[1em] \Rightarrow 8x = 16 \text{ and } 9y = 9 \\[1em] \therefore x = 2 \text{ and } y = 1.⇒[8x9y]=[169]⇒8x=16 and 9y=9∴x=2 and y=1.
Hence, the value of x = 2 and y = 1.
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