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Mathematics

Find, x and y from the following equations :

[52−1y−1]−[1x−12−3]=[47−32]\begin{bmatrix} 5 & 2 \ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \ -3 & 2 \end{bmatrix}

Matrices

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Answer

Given,

⇒[52−1y−1]−[1x−12−3]=[47−32]⇒[5−12−(x−1)−1−2y−1−(−3)]=[47−32]⇒[43−x−3y+2]=[47−32]\Rightarrow \begin{bmatrix} 5 & 2 \ -1 & y - 1 \end{bmatrix} - \begin{bmatrix} 1 & x - 1 \ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5 - 1 & 2 - (x - 1) \ -1 - 2 & y - 1 - (-3) \end{bmatrix} = \begin{bmatrix} 4 & 7 \ -3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 3 - x \ -3 & y + 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \ -3 & 2 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

⇒ 3 - x = 7
⇒ x = 3 - 7 = -4.

⇒ y + 2 = 2 ⇒ y = 0.

Hence, x = -4 and y = 0.

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