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(B) To find the value of $A$,we can use the method of substitution since the given equation holds for all values of $x$.Let $x = 1$.Substituting $x =

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$24$

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(B) To find the value of $A$,we can use the method of substitution since the given equation holds for all values of $x$.Let $x = 1$.Substituting $x = 1$ into the determinant:$\left| {\begin{array}{*{20}{c}}{{1^2} + 1}&{1 + 1}&{1 - 2}\ {2(1)^2 + 3(1) - 1}&{3(1)}&{3(1) - 3}\ {{1^2} + 2(1) + 3}&{2(1) - 1}&{2(1) - 1}\end{array}} ight| = A(1) - 12$$\left| {\begin{array}{*{20}{c}}2&2&{ - 1}\ 4&3&0\ 6&1&1\end{array}} ight| = A - 12$Now,evaluate the determinant:$2(3 	imes 1 - 0 	imes 1) - 2(4 	imes 1 - 0 	imes 6) + (-1)(4 	imes 1 - 3 	imes 6) = A - 12$$2(3) - 2(4) - 1(4 - 18) = A - 12$$6 - 8 + 14 = A - 12$$12 = A - 12$$A = 24$.

If $\left| {\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\ {2{x^2} + 3x - 1}&{3x}&{3x - 3}\\ {{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}} \right| = Ax - 12$,then the value of $A$ is

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