Question

Show that each of the following systems of linear equations is consistent and also find their solutions:
(i) 6x + 4y = 2
9x + 6y = 3

(ii) 2x + 3y = 5
6x + 9y = 15

(iii) 5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5

(iv) x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1

(v) x + y + z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30

(vi) 2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3

Answer 1

$$\begin{gathered} \text{(i) Here,} \\ 6x+4y=2...\left(1\right) \\ 9x+6y=3...\left(2\right) \\ AX=B \\ \mathrm{Here}, \\ A=\left[\begin{array}{cc}6& 4\\ 9& 6\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}2\\ 3\end{array}\right] \\ \left[\begin{array}{cc}6& 4\\ 9& 6\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{cc}6& 4\\ 9& 6\end{array}\right| \\ =36-36 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0. \\ {\text{Let C}}_{ij}{\text{be the co factors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}=6,C_{12}=-9,C_{21}=-4,C_{22}=6 \\ \mathrm{adj}A={\left[\begin{array}{cc}6& -9\\ -4& 6\end{array}\right]}^T \\ =\left[\begin{array}{cc}6& -4\\ -9& 6\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{cc}6& -4\\ -9& 6\end{array}\right]\left[\begin{array}{c}2\\ 3\end{array}\right] \\ =\left[\begin{array}{c}12-12\\ -18+18\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \mathrm{I}\text{f}\left|A\right|\text{=0 and}\left(\text{adj}\mathit{\text{A}}\right)\mathit{\text{B}}\text{=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.} \\ \mathrm{S}\text{ubstituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in the eq. (1), we get} \\ 6x+4k=2 \\ \Rightarrow 6x=2-4k \\ \Rightarrow x=\frac{2-4k}{6} \\ \Rightarrow x=\frac{1-2k}{3} \\ \therefore x=\frac{1-2k}{3}\mathrm{and}y=k \\ \mathrm{These}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=\frac{1-2k}{3}\mathrm{and}y=k\left(\mathrm{where}k\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$


$$\begin{gathered} \text{(ii) Here,} \\ 2x+3y=5...\left(1\right) \\ 6x+9y=15...\left(2\right) \\ \mathrm{or},AX=B \\ \mathrm{where}, \\ 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\\ 15\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2& 3\\ 6& 9\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 15\end{array}\right] \\ \therefore \left|A\right|=\left|\begin{array}{cc}2& 3\\ 6& 9\end{array}\right| \\ =18-18 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0. \\ {\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[a_{ij}\right]\text{. Then,} \\ {C}_{11}=9,C_{12}=-6,C_{21}=-3\mathrm{and}{C}_{22}=2 \\ \therefore \mathrm{adj}A={\left[\begin{array}{cc}9& -6\\ -3& 2\end{array}\right]}^T \\ =\left[\begin{array}{cc}2& -3\\ -6& 9\end{array}\right] \\ \Rightarrow \left(\mathrm{adj}A\right)B=\left[\begin{array}{cc}9& -3\\ -6& 2\end{array}\right]\left[\begin{array}{c}5\\ 15\end{array}\right] \\ =\left[\begin{array}{c}45-45\\ -30+30\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \text{If}\left|A\right|\text{=0 and}\left(\text{adjA}\right)\text{B=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus, AX=B has infinitely many solutions.} \\ \mathrm{S}\text{ubstituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in eq. (1), we get} \\ 2x+3k=5 \\ \Rightarrow 2x=5-3k \\ \Rightarrow x=\frac{5-3k}{2}\mathrm{and}y=k \\ \mathrm{These}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=\frac{5-3k}{2}\mathrm{and}y=k\left(\mathrm{where}k\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$


$$\begin{gathered} \text{(iii) Here,} \\ 5x+3y+7z=4...\left(1\right) \\ 3x+26y+2z=9...\left(2\right) \\ 7x+2y+10z=5...\left(3\right) \\ \mathrm{or},AX=B \\ \mathrm{where}, \\ A=\left[\begin{array}{ccc}5& 3& 7\\ 3& 26& 2\\ 7& 2& 10\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}4\\ 9\\ 5\end{array}\right] \\ \left[\begin{array}{ccc}5& 3& 7\\ 3& 26& 2\\ 7& 2& 10\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 9\\ 5\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}5& 3& 7\\ 3& 26& 2\\ 7& 2& 10\end{array}\right| \\ =5\left(260-4\right)-3\left(30-14\right)+7(6-182) \\ =1280-48-1232 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)B=0. \\ {\text{Let C}}_{ij}{\text{be the co-factors 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}26& 2\\ 2& 10\end{array}\right|=256,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}3& 2\\ 7& 10\end{array}\right|=-16,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}3& 26\\ 7& 2\end{array}\right|=-176 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}3& 7\\ 2& 10\end{array}\right|=-16,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}5& 7\\ 7& 10\end{array}\right|=1,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}5& 3\\ 7& 2\end{array}\right|=11 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}3& 7\\ 26& 2\end{array}\right|=-176,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}5& 7\\ 3& 2\end{array}\right|=11,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}5& 3\\ 3& 26\end{array}\right|=121 \\ \mathrm{adj}A={\left[\begin{array}{ccc}256& -16& -176\\ -16& 1& 11\\ -176& 11& 121\end{array}\right]}^T \\ =\left[\begin{array}{ccc}256& -16& -176\\ -16& 1& 11\\ -176& 11& 121\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}256& -16& -176\\ -16& 1& 11\\ -176& 11& 121\end{array}\right]\left[\begin{array}{c}4\\ 9\\ 5\end{array}\right] \\ =\left[\begin{array}{c}1024-144-880\\ -64+9+55\\ -704+99+605\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \text{if}\left|A\right|\text{=0 and}\left(\text{adjA}\right)\text{B=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.} \\ \mathrm{Substituting}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get} \\ 5x+3y=4-7k\mathrm{and}3x+26y=9-2k \\ \left[\begin{array}{cc}5& 3\\ 3& 26\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}4-7k\\ 9-2k\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}5& 3\\ 3& 26\end{array}\right| \\ =130-9 \\ =121\ne 0 \\ \mathrm{adj}A=\left|\begin{array}{cc}26& -3\\ -3& 5\end{array}\right| \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{121}\left[\begin{array}{cc}26& -3\\ -3& 5\end{array}\right] \\ \therefore X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{121}\left[\begin{array}{cc}26& -3\\ -3& 5\end{array}\right]\left[\begin{array}{c}4-7k\\ 9-2k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{121}\left[\begin{array}{c}104-182k-27+6k\\ -12+21k+45-10k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{77-176k}{121}\\ \frac{33+11k}{121}\end{array}\right] \\ \Rightarrow x=\frac{11\left(7-16k\right)}{121},y=\frac{11\left(3+k\right)}{121}\mathrm{and}z=k \\ \therefore x=\frac{7-16k}{11},y=\frac{3+k}{11}\mathrm{and}z=k \\ \mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=\frac{7-16k}{11},y=\frac{3+k}{11}\mathrm{and}z=k\left(\mathrm{where}k\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$


$$\begin{gathered} \text{(iv) Here,} \\ x-y+z=3...\left(1\right) \\ 2x+y-z=2...\left(2\right) \\ -x-2y+2z=1...\left(3\right) \\ \mathrm{or},AX=B \\ \mathrm{where}, \\ A=\left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ -1& -2& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right] \\ \left[\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ -1& -2& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ -1& -2& 2\end{array}\right| \\ =1\left(2-2\right)+1\left(4-1\right)+1(-4+1) \\ =0+3-3 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)B=0. \\ {\text{Let C}}_{ij}{\text{be the co-factors 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}1& -1\\ -2& 2\end{array}\right|=0,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}2& -1\\ -1& 2\end{array}\right|=-3,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}2& 1\\ -1& -2\end{array}\right|=-3 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}-1& 1\\ -2& 2\end{array}\right|=0,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ -1& 2\end{array}\right|=3,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& -1\\ -1& -2\end{array}\right|=3 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right|=0,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 2& -1\end{array}\right|=3,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|=3 \\ \mathrm{adj}A={\left[\begin{array}{ccc}0& -3& -3\\ 0& 3& 3\\ 0& 3& 3\end{array}\right]}^T \\ =\left[\begin{array}{ccc}0& 0& 0\\ -3& 3& 3\\ -3& 3& 3\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}0& 0& 0\\ -3& 3& 3\\ -3& 3& 3\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right] \\ =\left[\begin{array}{c}0\\ -9+6+3\\ -9+6+3\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \text{If}\left|A\right|\text{=0 and}\left(\text{adj}\mathit{\text{A}}\right)\mathit{\text{B}}\text{=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus,}\mathit{\text{AX=B}}\text{has infinitely many solutions.} \\ \mathrm{Substituting}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq.}\left(\text{1}\right)\text{and eq.}\left(\text{2}\right)\text{, we get} \\ x-y=3-k\mathrm{and}2x+y=2+k \\ \left[\begin{array}{cc}1& -1\\ 2& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}3-k\\ 2+k\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right| \\ =1+2=3\ne 0 \\ \mathrm{adj}A=\left|\begin{array}{cc}1& 2\\ -1& 1\end{array}\right| \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{3}\left[\begin{array}{cc}1& 1\\ -2& 1\end{array}\right] \\ \therefore X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}1& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}3-k\\ 2+k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}3-k+2+k\\ -6+2k+2+k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{5}{3}\\ \frac{3k-4}{3}\end{array}\right] \\ \therefore x=\frac{5}{3},y=\frac{3k-4}{3}\mathrm{and}z=k \\ \mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=\frac{5}{3},y=\frac{3k-4}{3}\mathrm{and}z=k\left(\mathrm{where}k\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$


$$\begin{gathered} \text{(v) Here,} \\ x+y+z=6...\left(1\right) \\ x+2y+3z=14\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right) \\ x+4y+7z=30...\left(3\right) \\ \mathrm{or},AX=B \\ \mathrm{where}, \\ A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}6\\ 14\\ 30\end{array}\right] \\ \left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}6\\ 14\\ 30\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right| \\ =1\left(14-12\right)-1\left(7-3\right)+1(4-2) \\ =2-4+2 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0. \\ {\text{Let C}}_{ij}{\text{be the co-factors 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& 3\\ 4& 7\end{array}\right|=2,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}1& 3\\ 1& 7\end{array}\right|=-4,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}1& 2\\ 1& 4\end{array}\right|=2 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}1& 1\\ 4& 7\end{array}\right|=-3,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}1& 1\\ 1& 7\end{array}\right|=6,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& 1\\ 1& 4\end{array}\right|=-3 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=1,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|=-2,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|=1 \\ \mathrm{adj}A={\left[\begin{array}{ccc}2& -4& 2\\ -3& 6& -3\\ 1& -2& 1\end{array}\right]}^T \\ =\left[\begin{array}{ccc}2& -3& 1\\ -4& 6& -2\\ 2& -3& 1\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}2& -3& 1\\ -4& 6& -2\\ 2& -3& 1\end{array}\right]\left[\begin{array}{c}6\\ 14\\ 30\end{array}\right] \\ =\left[\begin{array}{c}12-42+30\\ -24+84-60\\ 12-42+30\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \text{If}\left|A\right|\text{=0 and}\left(\text{ad}\mathit{\text{jA}}\right)\mathit{\text{B}}\text{=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.} \\ \mathrm{Subs}\text{tituting}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get} \\ x+y=6-k\mathrm{and}x+2y=14-3k \\ \left[\begin{array}{cc}1& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6-k\\ 14+3k\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right| \\ =2-1=1\ne 0 \\ \mathrm{adj}A=\left|\begin{array}{cc}2& -1\\ -1& 1\end{array}\right| \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{1}\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right] \\ \therefore X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{1}\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\left[\begin{array}{c}6-k\\ 14-3k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{1}\left[\begin{array}{c}12-2k-14+3k\\ -6+k+14-3k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{k-2}{1}\\ \frac{8-2k}{1}\end{array}\right] \\ \therefore x=k-2,y=8-2k\mathrm{and}z=k \\ \mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=k-2,y=8-2k\mathrm{and}z=k\left(\mathrm{where}k\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$


$$\begin{gathered} \text{(vi) Here,} \\ 2x+2y-2z=1...\left(1\right) \\ 4x+4y-z=2...\left(2\right) \\ 6x+6y+2z=3...\left(3\right) \\ \mathrm{or},AX=B \\ \mathrm{where},A=\left[\begin{array}{ccc}2& 2& -2\\ 4& 4& -1\\ 6& 6& 2\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\mathrm{and}B=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \\ \left[\begin{array}{ccc}2& 2& -2\\ 4& 4& -1\\ 6& 6& 2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \\ \left|A\right|=\left|\begin{array}{ccc}2& 2& -2\\ 4& 4& -1\\ 6& 6& 2\end{array}\right| \\ =2\left(8+6\right)-2\left(8+6\right)-2(24-24) \\ =28-28-0 \\ =0 \\ \mathrm{So},A\mathrm{is}\mathrm{singular}.\mathrm{Thus},\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{either}\mathrm{inconsistent}\mathrm{or}\mathrm{it}\mathrm{is}\mathrm{consistent}\mathrm{with} \\ \mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)B=0. \\ {\text{Let C}}_{ij}{\text{be the co-factors 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& -1\\ 6& 2\end{array}\right|=14,C_{12}={\left(-1\right)}^{1+2}\left|\begin{array}{cc}4& -1\\ 6& 2\end{array}\right|=-14,C_{13}={\left(-1\right)}^{1+3}\left|\begin{array}{cc}4& 4\\ 6& 6\end{array}\right|=0 \\ {C}_{21}={\left(-1\right)}^{2+1}\left|\begin{array}{cc}2& -2\\ 6& 2\end{array}\right|=-16,C_{22}={\left(-1\right)}^{2+2}\left|\begin{array}{cc}2& -2\\ 6& 2\end{array}\right|=16,C_{23}={\left(-1\right)}^{2+3}\left|\begin{array}{cc}2& 2\\ 6& 6\end{array}\right|=0 \\ {C}_{31}={\left(-1\right)}^{3+1}\left|\begin{array}{cc}2& -2\\ 4& -1\end{array}\right|=6,C_{32}={\left(-1\right)}^{3+2}\left|\begin{array}{cc}2& -2\\ 4& -1\end{array}\right|=-6,C_{33}={\left(-1\right)}^{3+3}\left|\begin{array}{cc}2& 2\\ 4& 4\end{array}\right|=0 \\ \mathrm{adj}A={\left[\begin{array}{ccc}14& -14& 0\\ -16& 16& 0\\ 6& -6& 0\end{array}\right]}^T \\ =\left[\begin{array}{ccc}14& -16& 6\\ -14& 16& -6\\ 0& 0& 0\end{array}\right] \\ \left(\mathrm{adj}A\right)B=\left[\begin{array}{ccc}14& -16& 6\\ -14& 16& -6\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \\ =\left[\begin{array}{c}14-32+18\\ -14+32-18\\ 0\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \text{If}\left|A\right|\text{=0 and}\left(\text{adj}\mathit{\text{A}}\right)\mathit{\text{B}}\text{=0, then the system is consistent and has infinitely many solutions.} \\ \text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.} \\ \mathrm{Substituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get} \\ 2x-2z=1-2k\mathrm{and}4x-z=2-4k \\ \left[\begin{array}{cc}2& -2\\ 4& -1\end{array}\right]\left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{c}1-2k\\ 2-4k\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{cc}2& -2\\ 4& -1\end{array}\right| \\ =-2+8=6\ne 0 \\ \mathrm{adj}A=\left|\begin{array}{cc}-1& 2\\ -4& 2\end{array}\right| \\ \Rightarrow A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ =\frac{1}{6}\left[\begin{array}{cc}-1& 2\\ -4& 2\end{array}\right] \\ \therefore X=A^{-1}B \\ \Rightarrow \left[\begin{array}{c}x\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{cc}-1& 2\\ -4& 2\end{array}\right]\left[\begin{array}{c}1-2k\\ 2-4k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}-1+2k+4-8k\\ -4+8k+4-8k\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{c}\frac{3-6k}{6}\\ 0\end{array}\right] \\ \therefore x=\frac{1-2k}{2},y=k\hspace{0.17em}\mathrm{and}z=0 \\ \mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}. \\ \mathrm{Thus},x=\frac{1-2k}{2},y=k\mathrm{and}z=0\left(\mathrm{where}k\mathrm{is}\mathrm{real}\mathrm{number}\right)\mathrm{satisfy}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}. \end{gathered}$$