Question

What values of x,y,z satisfy the following system of linear equations $\left[\begin{array}{ccc}1& 2& 3\\ 1& 3& 4\\ 2& 2& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 8\\ 12\end{array}\right]$

• A. x = 6, y = 3, z = 2
• B. x = 12, y = 3, z = -4
• C. x = 6, y = 6, z = -4
• D. x = 12, y = -3, z = 4

Answer 1

The correct option is C x = 6, y = 6, z = -4

[A : B] = $\left[\begin{array}{ccccc}1& 2& 3& :& 6\\ 1& 3& 4& :& 8\\ 2& 2& 3& :& 12\end{array}\right]$

$R_2\to R_2-R_1:R_3\to R_3-2R_1$

$\sim \left[\begin{array}{ccccc}1& 2& 3& :& 6\\ 0& 1& 1& :& 2\\ 0& -2& -3& :& 0\end{array}\right]$

$R_3\to R_3+2R_2$

$\sim \left[\begin{array}{ccccc}1& 2& 3& :& 6\\ 0& 1& 1& :& 2\\ 0& 0& -1& :& 4\end{array}\right]$ ....(1)

$\rho \left(A\right)=3,\rho \left(A|B\right)=3$
Number of unknowns = 3

So unique solution exist.

Now given system is equivalent to (1) so Ax = B

$\Rightarrow \left[\begin{array}{ccc}1& 2& 3\\ 1& 3& 4\\ 2& 2& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 8\\ 12\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ 0& 0& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 2\\ 4\end{array}\right]$

On comparision
z = -4,
y + z = 2
x + 2y + 3z = 6
Hence (z = - 4, y = 6, x = 6)