MHT CET · Maths · Determinants
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
- A 0
- B -4
- C 6
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{AX}=\mathrm{B} \\ \therefore \quad & {\left[\begin{array}{ccc}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{a} \\ \mathrm{b} \\ \mathrm{c}\end{array}\right]=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right] }\end{aligned}\)
\(\begin{array}{ll}
\therefore \quad & a-b+2 c=3 ...(i)\\
& 2 a+c=1...(ii) \\
& 3 a+2 b+c=4
...(iii)\end{array}\)
Solving (i), (ii) and (iii), we get
\(\begin{aligned}
& a=-1, b=2, c=3 \\
\therefore \quad 2 a-3 b+4 c & =2(-1)-3(2)+4(3) \\
& =4
\end{aligned}\)
\(\begin{array}{ll}
\therefore \quad & a-b+2 c=3 ...(i)\\
& 2 a+c=1...(ii) \\
& 3 a+2 b+c=4
...(iii)\end{array}\)
Solving (i), (ii) and (iii), we get
\(\begin{aligned}
& a=-1, b=2, c=3 \\
\therefore \quad 2 a-3 b+4 c & =2(-1)-3(2)+4(3) \\
& =4
\end{aligned}\)
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