Question

Show that each one of the following systems of linear equation is inconsistent:
(i) 2x + 5y = 7
6x + 15y = 13

(ii) 2x + 3y = 5
6x + 9y = 10

(iii) 4x − 2y = 3
6x − 3y = 5

(iv) 4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1

(v) 3x − y − 2z = 2
2y − z = −1
3x − 5y = 3

(vi) x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4

Answer 1

$$\begin{gathered} \text{(i) The given system of equations can be expressed as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{cc}2& 5\\ 6& 15\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}7\\ 13\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}2& 5\\ 6& 15\end{array}\right| \\ =\left(30-30\right) \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A =}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}=-1^{1+1}\left(15\right)=15,C_{12}=-1^{1+2}\left(6\right)=-6 \\ {C}_{21}=-1^{2+1}\left(5\right)=-5,C_{22}=-1^{2+2}\left(2\right)=2 \\ \mathrm{adj}A={\left[\begin{array}{cc}15& -6\\ -5& 2\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{cc}15& -5\\ -6& 2\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{cc}15& -5\\ -6& 2\end{array}\right]\left[\begin{array}{c}7\\ 13\end{array}\right] \\ =\left[\begin{array}{c}105-65\\ -42+26\end{array}\right] \\ =\left[\begin{array}{c}40\\ -16\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is inconsistent.} \end{gathered}$$


$$\begin{gathered} \text{(ii) The given system of equations can be expressed as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{cc}2& 3\\ 6& 9\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}5\\ 10\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}2& 3\\ 6& 9\end{array}\right| \\ =\left(18-18\right) \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A =}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left(9\right)=9,C_{12}={\left(-1\right)}^{1+2}\left(6\right)=-6 \\ {C}_{21}={\left(-1\right)}^{2+1}\left(3\right)=-3,C_{22}={\left(-1\right)}^{2+2}\left(2\right)=2 \\ \mathrm{adj}A={\left[\begin{array}{cc}9& -6\\ -3& 2\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{cc}9& -3\\ -6& 2\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{cc}9& -3\\ -6& 2\end{array}\right]\left[\begin{array}{c}5\\ 10\end{array}\right] \\ =\left[\begin{array}{c}45-30\\ -30+20\end{array}\right] \\ =\left[\begin{array}{c}15\\ -10\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is inconsistent.} \end{gathered}$$


$$\begin{gathered} \text{(iii) The given system of equations can be expressed as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{cc}4& -2\\ 6& -3\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}3\\ 5\end{array}\right] \\ \left|A\right|=\left|\begin{array}{cc}4& -2\\ 6& -3\end{array}\right| \\ =\left(-12+12\right) \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A =}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}=-{\left(1\right)}^{1+1}\left(-3\right)=-3,C_{12}=-{\left(1\right)}^{1+2}\left(6\right)=-6 \\ {C}_{21}=-{\left(1\right)}^{2+1}\left(-2\right)=2,C_{22}=-{\left(1\right)}^{2+2}\left(4\right)=4 \\ \mathrm{adj}A={\left[\begin{array}{cc}-3& -6\\ 2& 4\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{cc}-3& 2\\ -6& 4\end{array}\right] \\ \left(adjA\right)B=\left[\begin{array}{cc}-3& 2\\ -6& 4\end{array}\right]\left[\begin{array}{c}3\\ 5\end{array}\right] \\ =\left[\begin{array}{c}-9+10\\ -18+20\end{array}\right] \\ =\left[\begin{array}{c}1\\ 2\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is inconsistent.} \end{gathered}$$


$$\begin{gathered} \text{(iv) The given system of equations can be written as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}4& -5& -2\\ 5& -4& 2\\ 2& 2& 8\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}2\\ -2\\ -1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}4& -5& -2\\ 5& -4& 2\\ 2& 2& 8\end{array}\right| \\ =4\left(-32-4\right)+5\left(40-4\right)-2(10+8) \\ =-144+180-36 \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-4& 2\\ 2& 8\end{array}\right|=28,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}5& 2\\ 2& 8\end{array}\right|=-36,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}5& -4\\ 2& 2\end{array}\right|=18 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-5& -2\\ 2& 8\end{array}\right|=36,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}4& -2\\ 2& 8\end{array}\right|=36,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}4& -5\\ 2& 2\end{array}\right|=-18 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-5& -2\\ -4& 2\end{array}\right|=-18,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}4& -2\\ 5& 2\end{array}\right|=-18,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}4& -5\\ 5& -4\end{array}\right|=9 \\ \mathrm{adj}A={\left[\begin{array}{ccc}28& -36& 18\\ 36& 36& -18\\ -18& -18& 9\end{array}\right]}^{\mathrm{T}} \\ =\left[\begin{array}{ccc}28& 36& -18\\ -36& 36& -18\\ 18& -18& 9\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}28& 36& -18\\ -36& 36& -18\\ 18& -18& 9\end{array}\right]\left[\begin{array}{c}2\\ -2\\ -1\end{array}\right] \\ =\left[\begin{array}{c}56-72+18\\ -72-72+18\\ 36+36-9\end{array}\right] \\ =\left[\begin{array}{c}2\\ -126\\ 63\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is consistent.} \end{gathered}$$


$$\begin{gathered} \text{(v) The given system of equations can be written as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right| \\ =3\left(0-5\right)+1\left(0+3\right)-2(0-6) \\ =-15+3+12 \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}2& -1\\ -5& 0\end{array}\right|=-5,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}0& -1\\ 3& 0\end{array}\right|=-3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}0& 2\\ 3& -5\end{array}\right|=-6 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-1& -2\\ -5& 0\end{array}\right|=10,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}3& -2\\ 3& 0\end{array}\right|=6,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}3& -1\\ 3& -5\end{array}\right|=12 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-1& -2\\ 2& -1\end{array}\right|=5,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}3& -2\\ 0& -1\end{array}\right|=3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}3& -1\\ 0& 2\end{array}\right|=6 \\ \mathrm{adj}A={\left[\begin{array}{ccc}-5& -3& -6\\ 10& 6& 12\\ 5& 3& 6\end{array}\right]}^T \\ =\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right] \\ =\left[\begin{array}{c}-10-10+15\\ -6-6+9\\ -12-12+18\end{array}\right] \\ =\left[\begin{array}{c}-5\\ -3\\ -6\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is consistent.} \end{gathered}$$


$$\begin{gathered} \text{(vi) The given system of equations can be written as follows:} \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{ccc}1& 1& -2\\ 1& -2& 1\\ -2& 1& 1\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}5\\ -2\\ 4\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& -2\\ 1& -2& 1\\ -2& 1& 1\end{array}\right| \\ =1\left(-2-1\right)-1\left(1+2\right)-2(1-4) \\ =-3-3+6 \\ =0 \\ {\text{Let C}}_{ij}{\text{be the cofactors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}={\left(-1\right)}^{1+1}\left|\begin{array}{cc}-2& 1\\ 1& 1\end{array}\right|=-3,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 1\\ -2& 1\end{array}\right|=-3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& -2\\ -2& 1\end{array}\right|=-3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& -2\\ 1& 1\end{array}\right|=-3,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& -2\\ -2& 1\end{array}\right|=-3,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ -2& 1\end{array}\right|=-3 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& -2\\ -2& 1\end{array}\right|=-3,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& -2\\ 1& 1\end{array}\right|=-3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 1& -2\end{array}\right|=-3 \\ \mathrm{adj}A={\left[\begin{array}{ccc}-3& -3& -3\\ -3& -3& -3\\ -3& -3& -3\end{array}\right]}^T \\ =\left[\begin{array}{ccc}-3& -3& -3\\ -3& -3& -3\\ -3& -3& -3\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}-3& -3& -3\\ -3& -3& -3\\ -3& -3& -3\end{array}\right]\left[\begin{array}{c}5\\ -2\\ 4\end{array}\right] \\ =\left[\begin{array}{c}-15+6-12\\ -15+6-12\\ -15+6-12\end{array}\right] \\ =\left[\begin{array}{c}-21\\ -21\\ -21\end{array}\right]\ne 0 \\ \text{Hence, the given system of equations is consistent.} \end{gathered}$$