If for AX = B, B = [ l 9 52 0 ] and A^ - 1 = [ * 20 c 3& - 1 2 & … - Vidyadip

MCQ

If for $AX = B,$ $B = \left[ \begin{array}{l}9\\52\\0\end{array} \right]$ and ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}3&{ - \frac{1}{2}}&{ - \frac{1}{2}}\\{ - 4}&{\,\,\,\frac{3}{4}}&{\,\,\,\,\frac{5}{4}}\\2&{ - \frac{1}{4}}&{ - \frac{3}{4}}\end{array}} \right]$, then $X$ is equal to

✓
$\left[ \begin{array}{l}1\\3\\5\end{array} \right]$
B
$\left[ \begin{array}{l} - \frac{1}{2}\\ - \frac{1}{2}\\\,\,\,2\end{array} \right]$
C
$\left[ \begin{array}{l} - 4\\\,\,2\\\,\,3\end{array} \right]$
D
$\left[ \begin{array}{l}\,\,\,\,3\\\begin{array}{*{20}{c}}{\,\,\,\frac{3}{4}}\\{ - \,\frac{3}{4}}\end{array}\end{array} \right]$

✓

Answer

Correct option: A.

$\left[ \begin{array}{l}1\\3\\5\end{array} \right]$

a
(a) $AX = B \Rightarrow {A^{ - 1}}.AX = {A^{ - 1}}B \Rightarrow X = {A^{ - 1}}B = \left[ \begin{array}{l}1\\3\\5\end{array} \right]$.

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