If $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$,then show that $|2A| = 4|A|$.
(A) The given matrix is $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$.
First,calculate $2A$:
$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$.
Now,calculate the determinant of $2A$ $(LHS)$:
$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2 \times 4) - (4 \times 8) = 8 - 32 = -24$.
Next,calculate the determinant of $A$:
$|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = (1 \times 2) - (2 \times 4) = 2 - 8 = -6$.
Now,calculate $4|A|$ $(RHS)$:
$4|A| = 4 \times (-6) = -24$.
Since $LHS$ = $-24$ and $RHS$ = $-24$,we have shown that $|2A| = 4|A|$.
First,calculate $2A$:
$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$.
Now,calculate the determinant of $2A$ $(LHS)$:
$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2 \times 4) - (4 \times 8) = 8 - 32 = -24$.
Next,calculate the determinant of $A$:
$|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = (1 \times 2) - (2 \times 4) = 2 - 8 = -6$.
Now,calculate $4|A|$ $(RHS)$:
$4|A| = 4 \times (-6) = -24$.
Since $LHS$ = $-24$ and $RHS$ = $-24$,we have shown that $|2A| = 4|A|$.
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