Question
In Problems $51-56,$ solve for $x .$$$\left|\begin{array}{ll}{x} & {1} \\{3} & {x}\end{array}\right|=-2$$
Step 1
The determinant of a 2x2 matrix is calculated as follows: $$ \left|\begin{array}{ll} {a} & {b} \\ {c} & {d} \end{array}\right| = ad - bc $$ So, in our case, the determinant of the matrix is $x \cdot x - 1 \cdot 3 = x^2 - 3$. Show more…
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In Problems $51-56,$ solve for $x$. $$\left|\begin{array}{ll} x & 1 \\ 3 & x \end{array}\right|=-2$$
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In Problems $51-56,$ solve for $x$. $$\left|\begin{array}{lll} x & 1 & 2 \\ 1 & x & 3 \\ 0 & 1 & 2 \end{array}\right|=-4 x$$
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