TS EAMCET · Maths · Matrices
If \(\left[\begin{array}{ccc}2 & 1 & 1 \\ 0 & 3 & -1 \\ 1 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]\), then \(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\)
- A \(\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]+K\left[\begin{array}{c}3 \\ 1 \\ -2\end{array}\right], K \in \mathrm{R}\)
- B \(\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]+K\left[\begin{array}{c}1 \\ -2 \\ 3\end{array}\right], K \in \mathrm{R}\)
- C \(\left[\begin{array}{l}1 \\ 1 \\ 0\end{array}\right]+K\left[\begin{array}{c}-2 \\ 1 \\ 3\end{array}\right], K \in \mathrm{R}\)
- D \(\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]+K\left[\begin{array}{c}-2 \\ 1 \\ 3\end{array}\right], K \in \mathrm{R}\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]+K\left[\begin{array}{c}-2 \\ 1 \\ 3\end{array}\right], K \in \mathrm{R}\)
Step-by-step Solution
Detailed explanation
It is given that, \(\left[\begin{array}{ccc}2 & 1 & 1 \\0 & 3 & -1 \\1 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}1 \\1 \\0\end{array}\right]\) \(\Rightarrow \quad 2 x+y+z=1\) \(\ldots(\mathrm{i})\) \(3 y-z=1\)…
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