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Answer:

$x=2$

Hint:

Take determinant of matrix on both side and equate it.

Given:

\begin{aligned} &\left|\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right|=\left|\begin{array}{cc} 4 & -1 \\ 1 & 3 \end{array}\right| \end{aligned}

Solution:

\begin{aligned} &\left|\begin{array}{ll} x+1 & x-1 \\ x-3 & x+2 \end{array}\right|=\left|\begin{array}{cc} 4 & -1 \\ 1 & 3 \end{array}\right| \\ &(x+1)(x+2)-(x-3)(x-1)=12+1 \\ &x^{2}+3 x+2-x^{2}+4 x-3=13 \\ &7 x-1=13 \\ &7 x=14 \\ &x=2 \end{aligned}

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