If A=(4101−22),B=(20−1314),C=(12−1) and (3B−2A)C+2X=O then 'X'=

(3B−2A)C+2X=0⇒2X=(2A−3B)C 2A−3B=2[4101−22]−3[20−1314] =[8202−44]−[60−39312]=[223−7−7−8] 2X=(2A−3B)C=[223−7−7−8][12−1] =[3−13]⇒X=[32−132]=12[3−13]

If A=(4101−22),B=(20−1314),C=(12−1) and (3B−2A)C+2X=O then 'X'=

(3B−2A)C+2X=0⇒2X=(2A−3B)C 2A−3B=2[4101−22]−3[20−1314] =[8202−44]−[60−39312]=[223−7−7−8] 2X=(2A−3B)C=[223−7−7−8][12−1] =[3−13]⇒X=[32−132]=12[3−13]

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If $A = (\begin{matrix} 4 & 1 & 0 \\ 1 & - 2 & 2 \end{matrix}), B = (\begin{matrix} 2 & 0 & - 1 \\ 3 & 1 & 4 \end{matrix}), C = (\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix})$ and $(3 B - 2 A) C + 2 X = O$ then $' X' =$

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\frac{1}{2} [\begin{matrix} 3 \\ 13 \end{matrix}]

\frac{1}{2} [\begin{matrix} 3 \\ - 13 \end{matrix}]

\frac{1}{2} [\begin{matrix} - 3 \\ 13 \end{matrix}]

[\begin{matrix} 3 \\ - 13 \end{matrix}]

answer is D.

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Detailed Solution

(3 B - 2 A) C + 2 X = 0 \Rightarrow 2 X = (2 A - 3 B) C

$2 A - 3 B = 2 [\begin{matrix} 4 & 1 & 0 \\ 1 & - 2 & 2 \end{matrix}] - 3 [\begin{matrix} 2 & 0 & - 1 \\ 3 & 1 & 4 \end{matrix}]$

$= [\begin{matrix} 8 & 2 & 0 \\ 2 & - 4 & 4 \end{matrix}] - [\begin{matrix} 6 & 0 & - 3 \\ 9 & 3 & 12 \end{matrix}] = [\begin{matrix} 2 & 2 & 3 \\ - 7 & - 7 & - 8 \end{matrix}]$

$2 X = (2 A - 3 B) C = [\begin{matrix} 2 & 2 & 3 \\ - 7 & - 7 & - 8 \end{matrix}] [\begin{matrix} 1 \\ 2 \\ - 1 \end{matrix}]$

$= [\begin{matrix} 3 \\ - 13 \end{matrix}] \Rightarrow X = [\begin{matrix} \frac{3}{2} \\ - \frac{13}{2} \end{matrix}] = \frac{1}{2} [\begin{matrix} 3 \\ - 13 \end{matrix}]$