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If $ A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} $ then show that | 2A | = 4| A |

1 Answer

Toolbox:
  • If $A= \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
  • Then $|A|=a_{11} \times a_{22}- a_{12} \times a_{21}$
Step 1:
Given $A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$
The value of the determinent A is
$|A|= 1 \times 2 - 4 \times 2$
$=2-8$
$=-6$
Step 2:
$2A=2\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
Therefore $|2A|= 2 \times 4 - 8 \times 4$
$=8-32$
$=-24$
$4|A|=-6 \times 4$
$=-24$
Therefore $|2A|=4|A|$
Solution:Hence proved
answered Apr 3, 2013 by meena.p
 

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