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

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Solution

## $\text{(i) Here,}\phantom{\rule{0ex}{0ex}}6x+4y=2...\left(1\right)\phantom{\rule{0ex}{0ex}}9x+6y=3...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}AX=B\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}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]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{cc}6& 4\\ 9& 6\end{array}\right|\phantom{\rule{0ex}{0ex}}=36-36\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{C}_{11}=6,{C}_{12}=-9,{C}_{21}=-4,{C}_{22}=6\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{cc}6& -9\\ -4& 6\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}6& -4\\ -9& 6\end{array}\right]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}12-12\\ -18+18\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{S}\text{ubstituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in the eq. (1), we get}\phantom{\rule{0ex}{0ex}}6x+4k=2\phantom{\rule{0ex}{0ex}}⇒6x=2-4k\phantom{\rule{0ex}{0ex}}⇒x=\frac{2-4k}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1-2k}{3}\phantom{\rule{0ex}{0ex}}\therefore x=\frac{1-2k}{3}\mathrm{and}y=k\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$ $\text{(ii) Here,}\phantom{\rule{0ex}{0ex}}2x+3y=5...\left(1\right)\phantom{\rule{0ex}{0ex}}6x+9y=15...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or},AX=B\phantom{\rule{0ex}{0ex}}\mathrm{where},\phantom{\rule{0ex}{0ex}}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]\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \left|A\right|=\left|\begin{array}{cc}2& 3\\ 6& 9\end{array}\right|\phantom{\rule{0ex}{0ex}}=18-18\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{C}_{11}=9,{C}_{12}=-6,{C}_{21}=-3\mathrm{and}{C}_{22}=2\phantom{\rule{0ex}{0ex}}\therefore \mathrm{adj}A={\left[\begin{array}{cc}9& -6\\ -3& 2\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& -3\\ -6& 9\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}45-45\\ -30+30\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus, AX=B has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{S}\text{ubstituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in eq. (1), we get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2x+3k=5\phantom{\rule{0ex}{0ex}}⇒2x=5-3k\phantom{\rule{0ex}{0ex}}⇒x=\frac{5-3k}{2}\mathrm{and}y=k\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$ $\text{(iii) Here,}\phantom{\rule{0ex}{0ex}}5x+3y+7z=4...\left(1\right)\phantom{\rule{0ex}{0ex}}3x+26y+2z=9...\left(2\right)\phantom{\rule{0ex}{0ex}}7x+2y+10z=5...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or},AX=B\phantom{\rule{0ex}{0ex}}\mathrm{where},\phantom{\rule{0ex}{0ex}}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]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{ccc}5& 3& 7\\ 3& 26& 2\\ 7& 2& 10\end{array}\right|\phantom{\rule{0ex}{0ex}}=5\left(260-4\right)-3\left(30-14\right)+7\left(6-182\right)\phantom{\rule{0ex}{0ex}}=1280-48-1232\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\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.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{ccc}256& -16& -176\\ -16& 1& 11\\ -176& 11& 121\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}256& -16& -176\\ -16& 1& 11\\ -176& 11& 121\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}1024-144-880\\ -64+9+55\\ -704+99+605\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\text{}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get}\phantom{\rule{0ex}{0ex}}5x+3y=4-7k\mathrm{and}3x+26y=9-2k\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{cc}5& 3\\ 3& 26\end{array}\right|\phantom{\rule{0ex}{0ex}}=130-9\phantom{\rule{0ex}{0ex}}=121\ne 0\phantom{\rule{0ex}{0ex}}\mathrm{adj}A=\left|\begin{array}{cc}26& -3\\ -3& 5\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A\phantom{\rule{0ex}{0ex}}=\frac{1}{121}\left[\begin{array}{cc}26& -3\\ -3& 5\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore X={A}^{-1}B\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{77-176k}{121}\\ \frac{33+11k}{121}\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{11\left(7-16k\right)}{121},y=\frac{11\left(3+k\right)}{121}\mathrm{and}z=k\phantom{\rule{0ex}{0ex}}\therefore x=\frac{7-16k}{11},y=\frac{3+k}{11}\mathrm{and}z=k\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$ $\text{(iv) Here,}\phantom{\rule{0ex}{0ex}}x-y+z=3...\left(1\right)\phantom{\rule{0ex}{0ex}}2x+y-z=2...\left(2\right)\phantom{\rule{0ex}{0ex}}-x-2y+2z=1...\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{or},AX=B\phantom{\rule{0ex}{0ex}}\mathrm{where},\phantom{\rule{0ex}{0ex}}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]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ -1& -2& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}=1\left(2-2\right)+1\left(4-1\right)+1\left(-4+1\right)\phantom{\rule{0ex}{0ex}}=0+3-3\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\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.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{ccc}0& -3& -3\\ 0& 3& 3\\ 0& 3& 3\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}0& 0& 0\\ -3& 3& 3\\ -3& 3& 3\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ -9+6+3\\ -9+6+3\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus,}\mathit{\text{AX=B}}\text{has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\text{}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq.}\left(\text{1}\right)\text{and eq.}\left(\text{2}\right)\text{, we get}\phantom{\rule{0ex}{0ex}}x-y=3-k\mathrm{and}2x+y=2+k\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}=1+2=3\ne 0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{adj}A=\left|\begin{array}{cc}1& 2\\ -1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[\begin{array}{cc}1& 1\\ -2& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore X={A}^{-1}B\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{5}{3}\\ \frac{3k-4}{3}\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore x=\frac{5}{3},y=\frac{3k-4}{3}\mathrm{and}z=k\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$ $\text{(v) Here,}\phantom{\rule{0ex}{0ex}}x+y+z=6...\left(1\right)\phantom{\rule{0ex}{0ex}}x+2y+3z=14\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)\phantom{\rule{0ex}{0ex}}x+4y+7z=30...\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{or},AX=B\phantom{\rule{0ex}{0ex}}\mathrm{where},\phantom{\rule{0ex}{0ex}}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]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 4& 7\end{array}\right|\phantom{\rule{0ex}{0ex}}=1\left(14-12\right)-1\left(7-3\right)+1\left(4-2\right)\phantom{\rule{0ex}{0ex}}=2-4+2\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}\mathrm{because}\left(\mathrm{adj}A\right)B\ne 0\mathrm{or}\left(\mathrm{adj}A\right)=0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{ccc}2& -4& 2\\ -3& 6& -3\\ 1& -2& 1\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}2& -3& 1\\ -4& 6& -2\\ 2& -3& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}12-42+30\\ -24+84-60\\ 12-42+30\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{Subs}\text{tituting}\mathit{\text{z}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get}\phantom{\rule{0ex}{0ex}}x+y=6-k\mathrm{and}x+2y=14-3k\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{cc}1& 1\\ 1& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}=2-1=1\ne 0\phantom{\rule{0ex}{0ex}}\mathrm{adj}A=\left|\begin{array}{cc}2& -1\\ -1& 1\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A\phantom{\rule{0ex}{0ex}}=\frac{1}{1}\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore X={A}^{-1}B\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\frac{k-2}{1}\\ \frac{8-2k}{1}\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore x=k-2,y=8-2k\mathrm{and}z=k\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{also}\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$ $\text{(vi) Here,}\phantom{\rule{0ex}{0ex}}2x+2y-2z=1...\left(1\right)\phantom{\rule{0ex}{0ex}}4x+4y-z=2...\left(2\right)\phantom{\rule{0ex}{0ex}}6x+6y+2z=3...\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{or},AX=B\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{ccc}2& 2& -2\\ 4& 4& -1\\ 6& 6& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}=2\left(8+6\right)-2\left(8+6\right)-2\left(24-24\right)\phantom{\rule{0ex}{0ex}}=28-28-0\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\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.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\text{Let C}}_{ij}{\text{be the co-factors of the elements a}}_{ij}\text{in A}\left[{a}_{ij}\right]\text{. Then,}\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}{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\phantom{\rule{0ex}{0ex}}\mathrm{adj}A={\left[\begin{array}{ccc}14& -14& 0\\ -16& 16& 0\\ 6& -6& 0\end{array}\right]}^{T}\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{ccc}14& -16& 6\\ -14& 16& -6\\ 0& 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}14-32+18\\ -14+32-18\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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.}\phantom{\rule{0ex}{0ex}}\text{Thus,}\mathit{\text{AX}}\text{=}\mathit{\text{B}}\text{has infinitely many solutions.}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathit{\text{y}}\text{=}\mathit{\text{k}}\text{in eq. (1) and eq. (2), we get}\phantom{\rule{0ex}{0ex}}2x-2z=1-2k\mathrm{and}4x-z=2-4k\phantom{\rule{0ex}{0ex}}\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]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\left|A\right|=\left|\begin{array}{cc}2& -2\\ 4& -1\end{array}\right|\phantom{\rule{0ex}{0ex}}=-2+8=6\ne 0\phantom{\rule{0ex}{0ex}}\mathrm{adj}A=\left|\begin{array}{cc}-1& 2\\ -4& 2\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{A}^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\left[\begin{array}{cc}-1& 2\\ -4& 2\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore X={A}^{-1}B\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\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]\phantom{\rule{0ex}{0ex}}⇒\left[\begin{array}{c}x\\ z\end{array}\right]=\left[\begin{array}{c}\frac{3-6k}{6}\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\therefore x=\frac{1-2k}{2},y=k \mathrm{and}z=0\phantom{\rule{0ex}{0ex}}\mathrm{These}\mathrm{values}\mathrm{of}x,y\mathrm{and}z\mathrm{satisfy}\mathrm{the}\mathrm{third}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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}.$

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