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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right], B=\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$ and $2 \mathrm{~A}+3 \mathrm{~B}-5 \mathrm{C}=0$, then $\mathrm{C}=$
MathematicsMatricesAP EAMCETAP EAMCET 2023 (19 May Shift 1)
Options:
  • A $\left[\begin{array}{cccc}2 & 1 & 6 / 5 & 7 / 5 \\ 1 & 7 / 5 & 2 & 3 / 5\end{array}\right]$
  • B $\left[\begin{array}{cccc}-2 & 1 & 6 / 5 & 7 / 5 \\ 1 & -7 / 5 & 2 & 3 / 5\end{array}\right]$
  • C $\left[\begin{array}{cccc}-2 & 1 & 6 / 5 & 7 / 5 \\ 1 & 7 / 5 & 2 & 3 / 5\end{array}\right]$
  • D $\left[\begin{array}{cccc}2 & 1 & 6 / 5 & 7 / 5 \\ 1 & -7 / 5 & 2 & 3 / 5\end{array}\right]$
Solution:
1961 Upvotes Verified Answer
The correct answer is: $\left[\begin{array}{cccc}2 & 1 & 6 / 5 & 7 / 5 \\ 1 & -7 / 5 & 2 & 3 / 5\end{array}\right]$
$\begin{aligned} & \text { Since } 2 \mathrm{~A}+3 \mathrm{~B}-5 \mathrm{C}=0 \\ & \Rightarrow 5 \mathrm{C}=2 \mathrm{~A}+3 \mathrm{~B} \\ & \Rightarrow 5 \mathrm{C}=2\left[\begin{array}{llll}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right]+3\left[\begin{array}{llll}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right] \\ & \Rightarrow \quad 5 \mathrm{C}=\left[\begin{array}{cccc}4+6 & 2+3 & 6+0 & -2+9 \\ 2+3 & -4-3 & 4+6 & -6+9\end{array}\right] \\ & \Rightarrow 5 \mathrm{C}=\left[\begin{array}{cccc}10 & 5 & 6 & 7 \\ 5 & -7 & 10 & 3\end{array}\right] \\ & \end{aligned}$
$\Rightarrow C=\left[\begin{array}{cccc}2 & 1 & 6 / 5 & 7 / 5 \\ 1 & -7 / 5 & 2 & 3 / 5\end{array}\right]$

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