Question

The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)0 \\ \text{The given system of equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}2& 1& -1\\ 1& -3& 2\\ 1& 4& -3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}7\\ 1\\ 5\end{array}\right] \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}2& 1& -1\\ 1& -3& 2\\ 1& 4& -3\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}7\\ 1\\ 5\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|\text{=2}\left(9-8\right)-1\left(-3-2\right)-\mathrm{1}\left(4+3\right) \\ =2+5-7 \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A=}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-3& 2\\ 4& -3\end{array}\right|=1,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 2\\ 1& -3\end{array}\right|=5,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& -3\\ 1& 4\end{array}\right|=7 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& -1\\ 4& -3\end{array}\right|=-1,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& -1\\ 1& -3\end{array}\right|=-5,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 1\\ 1& 4\end{array}\right|=-7 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& -1\\ -3& 2\end{array}\right|=-1,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& -1\\ 1& 2\end{array}\right|=-5,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 1\\ 1& -3\end{array}\right|=-7 \\ \mathrm{adj}A={\left[\begin{array}{ccc}1& 5& 7\\ -1& -5& -7\\ -1& -5& -7\end{array}\right]}^T \\ =\left[\begin{array}{ccc}1& -1& -1\\ 5& -5& -5\\ 7& -7& -7\end{array}\right] \\ \Rightarrow \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}1& -1& -1\\ 5& -5& -5\\ 7& -7& -7\end{array}\right]\left[\begin{array}{c}7\\ 1\\ 5\end{array}\right] \\ =\left[\begin{array}{c}7-1-5\\ 35-5-25\\ 49-7-35\end{array}\right] \\ =\left[\begin{array}{c}1\\ 5\\ 7\end{array}\right]\ne 0 \\ \text{So, the given system of equations has no solution.} \end{gathered}$$