MHT CET · Maths · Matrices
If \(\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 3 \\ 4\end{array}\right]\), then \(2 x-y+z=\)
- A 3
- B 2
- C 1
- D \(-3\)
Answer & Solution
Correct Answer
(D) \(-3\)
Step-by-step Solution
Detailed explanation
From the given equation we have
\(x+y+z=0 \quad \ldots \text{(i)}\)
\(x-2 y-2 z\quad\ldots\text{(ii)}\)
\(x+3 y+z=4\quad \ldots\text{(iii)}\)
Solving we get \(x=1, y=2\) and \(z=-3\)
Now, \(2 x-y+z=2 \times 1-2+(-3)=-3\)
\(x+y+z=0 \quad \ldots \text{(i)}\)
\(x-2 y-2 z\quad\ldots\text{(ii)}\)
\(x+3 y+z=4\quad \ldots\text{(iii)}\)
Solving we get \(x=1, y=2\) and \(z=-3\)
Now, \(2 x-y+z=2 \times 1-2+(-3)=-3\)
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