Question

If $x+y+z=4$, $2x+5y-2z=3$, $x+7y-7z=5$, then $z=$

• A. $1$
• B. $2$
• C. $-2$
• D. does not exist

Answer 1

Answer: (D)

The given system of equations can be writtenin matrix form as
$\left[\begin{array}{ccc}1& 1& 1\\ 2& 5& -2\\ 1& 7& -7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 3\\ 5\end{array}\right]$
or $AX=B$, where
$A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 5& -2\\ 1& 7& -7\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]andB=\left[\begin{array}{c}4\\ 3\\ 5\end{array}\right]$

Now, $\left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 2& 5& -2\\ 1& 7& -7\end{array}\right|=-21+12+9=0$

Cofactor matrix of $A=\left[\begin{array}{ccc}-21& 12& 9\\ 14& -8& -6\\ -7& 4& 3\end{array}\right]$

$\therefore$ adj A $={\left[\begin{array}{ccc}-21& 12& 9\\ 14& -8& -6\\ -7& 4& 3\end{array}\right]}^T=\left[\begin{array}{ccc}-21& 14& -7\\ 12& -8& 4\\ 9& -6& 3\end{array}\right]$

Now, $\left(adjA\right)\cdot B=\left[\begin{array}{ccc}-21& 14& -7\\ 12& -8& 4\\ 9& -6& 3\end{array}\right]\left[\begin{array}{c}4\\ 3\\ 5\end{array}\right]$

$\left[\begin{array}{c}-84+42-35\\ 48-24+20\\ 36-18+15\end{array}\right]=\left[\begin{array}{c}-77\\ 44\\ 33\end{array}\right]\ne O$

$\therefore$ The given system of equations has no solution.