Matrices and Determinants
8.0 Cramer's rule
8.0 Cramer's rule
Let us consider a system of equations ${a_1}x + {b_1}y + {c_1}z = {d_1}$
${a_2}x + {b_2}y + {c_2}z = {d_2}$
${a_3}x + {b_3}y + {c_3}z = {d_3}$
Here $\Delta = \left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,$ ${\Delta _1} = \left| {\begin{array}{c} {{d_1}}&{{b_1}}&{{c_1}} \\ {{d_2}}&{{b_2}}&{{c_2}} \\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|$
${\Delta _2} = \left| {\begin{array}{c} {{a_1}}&{{d_1}}&{{c_1}} \\ {{a_2}}&{{d_2}}&{{c_2}} \\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right|,$ ${\Delta _3} = \left| {\begin{array}{c} {{a_1}}&{{b_1}}&{{d_1}} \\ {{a_2}}&{{b_2}}&{{d_2}} \\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|$
By Cramer's rule , we have $x = \frac{{{\Delta _1}}}{\Delta },y = \frac{{{\Delta _2}}}{\Delta }\& z = \frac{{{\Delta _3}}}{\Delta }$
Question 14.
Solve the following system of linear equations using cramer's method
$x + 2y = 0$
$ - x + y + z = 1$
$x + 2y + 3z = 0$
Solution:
\[\Delta = \left| {\begin{array}{c} 1&2&0 \\ { - 1}&1&1 \\ 1&2&3 \end{array}} \right| = \left| {\begin{array}{c} 1&1 \\ 2&3 \end{array}} \right| - 2\left| {\begin{array}{c} { - 1}&1 \\ 1&3 \end{array}} \right| = 9\]
\[{\Delta _1} = \left| {\begin{array}{c} 0&2&0 \\ 1&1&1 \\ 0&2&3 \end{array}} \right| = - 2\left| {\begin{array}{c} 1&1 \\ 0&3 \end{array}} \right| = - 6\]
\[{\Delta _2} = \left| {\begin{array}{c} 1&0&0 \\ { - 1}&1&1 \\ 1&0&3 \end{array}} \right| = 1\left| {\begin{array}{c} 1&1 \\ 0&3 \end{array}} \right| = 3\]
\[{\Delta _3} = \left| {\begin{array}{c} 1&2&0 \\ { - 1}&1&1 \\ 1&2&0 \end{array}} \right| = 0\]
$(\because {R_1} = {R_3})$
$x = \frac{{{\Delta _1}}}{{{\Delta _{}}}} = - \frac{6}{9} = \frac{{ - 2}}{3}$ , $y = \frac{{{\Delta _2}}}{{{\Delta _{}}}} = \frac{3}{9} = \frac{1}{3}$ , $z = 0$
Question 15.
Solve the following equations by cramer's rule
$\begin{equation} \begin{aligned} x + y + z = 9 \\ 2x + 5y + 7z = 52 \\ 2x + y - z = 0 \\\end{aligned} \end{equation} $
Solution:
Here \[\Delta = \left| {\begin{array}{c} 1&1&1 \\ 2&5&7 \\ 2&1&{ - 1} \end{array}} \right|\;applying\left[ {{C_2} \to {C_2} - {C_{1,}}{C_3} \to {C_3} - {C_1}} \right]\]
\[\therefore \Delta = \left| {\begin{array}{c} 1&0&0 \\ 2&3&5 \\ 2&{ - 1}&{ - 3} \end{array}} \right| = 1.\left( { - 9 + 5} \right) = - 4\]
\[{\Delta _1} = \left| {\begin{array}{c} 9&1&1 \\ {52}&5&7 \\ 0&1&{ - 1} \end{array}} \right|\ applying\left[ {{C_2} \to {C_2} + {C_3}} \right]\]
\[\therefore {\Delta _1} = \left| {\begin{array}{c} 9&2&1 \\ {52}&{12}&7 \\ 0&0&{ - 1} \end{array}} \right| = - 1\left( {108 - 104} \right) = - 4\]
\[{\Delta _2} = \left| {\begin{array}{c} 1&9&1 \\ 2&{52}&7 \\ 2&0&{ - 1} \end{array}} \right|\ applying\left[ {{C_1} \to {C_1} + 2{C_3}} \right]\]
\[\therefore {\Delta _2} = \left| {\begin{array}{c} 3&9&1 \\ {16}&{52}&7 \\ 0&0&{ - 1} \end{array}} \right| = - 1\left( {156 - 144} \right) = - 12\]
\[{\Delta _3} = \left| {\begin{array}{c} 1&1&9 \\ 2&5&{52} \\ 2&1&0 \end{array}} \right|\ applying\left[ {{C_1} \to {C_1} - 2{C_2}} \right]\]
\[\therefore {\Delta _3} = \left| {\begin{array}{c} { - 1}&1&9 \\ { - 8}&5&{52} \\ 0&1&0 \end{array}} \right| = - 1\left( { - 52 + 72} \right) = - 20\]
Therefore,by Cramer's rule
$x = \frac{{{\Delta _1}}}{\Delta } = \frac{{ - 4}}{{ - 4}} = 1$
$y = \frac{{{\Delta _2}}}{\Delta } = \frac{{ - 12}}{{ - 4}} = 3$
and $z = \frac{{{\Delta _3}}}{\Delta } = \frac{{ - 20}}{{ - 4}} = 5$
$\therefore x = 1,y = 3,z = 5$
Remarks:-
- If $\Delta \ne 0$,then system will have unique finite solution,and so equations are consistent.
- If $\Delta = 0$,and atleast one of ${\Delta _1},{\Delta _2},{\Delta _3}$ be non zero ,then the system has no solution that is equations are inconsistent .
- If $\Delta = {\Delta _1} = {\Delta _2} = {\Delta _3} = 0$,then equations will have infinite number of solutions,and atleast one cofactor of $\Delta $is non zero that is equations are consistent.
