Let $X=\left[\begin{array}{l}\mathrm{a} \\ \mathrm{b} \\ \mathrm{c}\end{array}\right],…

Let $X=\left[\begin{array}{l}\mathrm{a} \\ \mathrm{b} \\ \mathrm{c}\end{array}\right], \mathrm{A}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A X=B$, then the value of $2 a-3 b+4 c$ will be
  1. 0
  2. -4
  3. 6
  4. 4

Solution

$\begin{aligned} & \mathrm{AX}=\mathrm{B} \\ \therefore \quad & {\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{a} \\ \mathrm{b} \\ \mathrm{c}\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right] }\end{aligned}$ $\begin{array}{ll} \therefore \quad & a-b+2 c=3 ...(i)\\ & 2 a+c=1...(ii) \\ & 3 a+2 b+c=4 ...(iii)\end{array}$ Solving (i), (ii) and (iii), we get $\begin{aligned} & a=-1, b=2, c=3 \\ \therefore \quad 2 a-3 b+4 c & =2(-1)-3(2)+4(3) \\ & =4 \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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