If {left| {begin{array}{cc} 4 & 1 2 & 1 end{ | vedclass

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(C) Given the equation: ${\left| {\begin{array}{cc} 4 & 1 \ 2 & 1 \end{array}} ight|^2} = \left| {\begin{array}{cc} 3 & 2 \ 1 & x \end{array}} i

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$6$

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(C) Given the equation: ${\left| {\begin{array}{cc} 4 & 1 \ 2 & 1 \end{array}} ight|^2} = \left| {\begin{array}{cc} 3 & 2 \ 1 & x \end{array}} ight| - \left| {\begin{array}{cc} x & 3 \ -2 & 1 \end{array}} ight|$First,calculate the determinant on the left side:$\left| {\begin{array}{cc} 4 & 1 \ 2 & 1 \end{array}} ight| = (4 	imes 1) - (1 	imes 2) = 4 - 2 = 2$So,the left side is $2^2 = 4$.Next,calculate the determinants on the right side:$\left| {\begin{array}{cc} 3 & 2 \ 1 & x \end{array}} ight| = (3 	imes x) - (2 	imes 1) = 3x - 2$$\left| {\begin{array}{cc} x & 3 \ -2 & 1 \end{array}} ight| = (x 	imes 1) - (3 	imes -2) = x + 6$Substitute these values back into the equation:$4 = (3x - 2) - (x + 6)$$4 = 3x - 2 - x - 6$$4 = 2x - 8$$2x = 4 + 8$$2x = 12$$x = 6$

If ${\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right|^2} = \left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| - \left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right|$,then $x =$

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