Q : 3         If   A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}  , then show that  | 2 A |=4|A|

Answers (1)

Given determinant A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} then we have to show that | 2 A |=4|A|,

So, A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} then, 2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}

Hence we have \left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24

So, L.H.S. = |2A| = -24

then calculating R.H.S. 4\left | A \right |

We have,

\left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6

hence R.H.S becomes 4\left | A \right | = 4\times(-6) = -24

Therefore L.H.S. =R.H.S.

Hence proved.

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