Q107 / 180 · COMEDK 2012 · Mathematics
If $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4\end{array}\right], B=\left[\begin{array}{c}7 \\ 16 \\ 22\end{array}\right], X=\left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $A X=B$, then $z$ is equal to
- A1
- B$-1$
- C$-3$
- Correct answerD3
Solution
We have, $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 4\end{array}\right], B=\left[\begin{array}{c}7 \\ 16 \\ 22\end{array}\right]$
$$
X=\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]
$$
Now, $A X=B$
$$
\begin{aligned}
&x+y+z=7 \quad \text{...(i)} \\
&x+2 y+3 z=16 \quad \text{...(ii)} \\
&x+3 y+4 z=22 \quad \text{...(iii)}
\end{aligned}
$$
Subtracting Eq. (i) from Eq. (ii), we get
$$
y+2 z=9 \quad \text{...(iv)}
$$
Subtracting Eq. (iii) from Eq. (ii), we get
$$
y+z=6 \quad \text{...(v)}
$$
Subtracting Eq. (v) from Eq. (iv), we get
$z=3$
Hence, the correct answer is option: 'D'.