Question

If $\left[\begin{array}{ccc}3& 2& -1\\ 4& 9& 2\\ 5& 0& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 7\\ 2\end{array}\right]$, then $(x,y,z)=$

• A. $(1,-1,1)$
• B. $(2,-1,-4)$
• C. $(3,0,6)$
• D. $(2,-1,4)$

Answer 1

Answer: (D)

$(2,-1,4)$

$\left[\begin{array}{ccc}3& 2& -1\\ 4& 9& 2\\ 5& 0& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 7\\ 2\end{array}\right]$

Thus, multiplying

$3x+2y-z=0$ …………..$\left(1\right)$

$4x+9y+2z=7$ ………..$\left(2\right)$

$5x-2z=2$ …………$\left(3\right)$

Thus $z=\frac{5x-2}{2}$

$\therefore 3x+2y=\frac{5x-2}{2}$

$\therefore 6x+4y=5x-2$

$\therefore x+4y=-2$ …………$\left(4\right)$

Also,

$4x+9y+2\left(\frac{5x-2}{2}\right)=7$

$9x+9y=9$

$\therefore x+y=1$ ………..$\left(5\right)$

Solving $\left(4\right)$ and $\left(5\right)$

$3y=-3$

$y=-1$

$\therefore x=1-y$

$=1-(-1)=2$

$\therefore x=2$

$\therefore z=\frac{5x-2}{2}=\frac{5\left(2\right)-2}{2}$

$\therefore z=4$

$=4$

$(x,y,z)=(2,-1,4)$.