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The solution of the equation
$\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ is $(x, y, z)=$
Options:
$\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$ is $(x, y, z)=$
Solution:
2744 Upvotes
Verified Answer
The correct answer is:
$(-1,0,2)$
$$
\begin{aligned}
\left[\begin{array}{rrr}
1 & 0 & 1 \\
-1 & 1 & 0 \\
0 & -1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
\Rightarrow\left[\begin{array}{r}
x+0 y+z \\
-x+y+0 z \\
0 x-y+z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
\Rightarrow \quad x+z=1, \\
-x+y=1
\end{aligned}
$$
and $\quad-y+z=2$
On solving these equations, we get
$$
x=-1, y=0, z=2
$$
\begin{aligned}
\left[\begin{array}{rrr}
1 & 0 & 1 \\
-1 & 1 & 0 \\
0 & -1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
\Rightarrow\left[\begin{array}{r}
x+0 y+z \\
-x+y+0 z \\
0 x-y+z
\end{array}\right]=\left[\begin{array}{l}
1 \\
1 \\
2
\end{array}\right] \\
\Rightarrow \quad x+z=1, \\
-x+y=1
\end{aligned}
$$
and $\quad-y+z=2$
On solving these equations, we get
$$
x=-1, y=0, z=2
$$
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