Find the value of $ \displaystyle m$ by matrix method for which the simultaneous equations $ \displaystyle 3x + my = 5$ and $ \displaystyle (m + 2) x + 5y = m$ have (i) an infinite number of solutions (ii) no solution.
Solution$ \displaystyle \left. \begin{array}{l}3x+my=5\\(m+2)x+5y=m\end{array} \right\}\ \ ---------(1)$
$ \displaystyle \text{Transforming into matrix form},$
$ \displaystyle \left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right)------(2)$
$ \displaystyle \text{Let }A=\left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right),X=\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ \text{and }B=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right).$
$ \displaystyle \text{Then we have, }AX=B.$
$ \displaystyle \text{If }\det A\ne 0,\text{then }{{A}^{{-1}}}\ \text{exists}\text{.}$
$ \displaystyle {{A}^{{1}}}AX={{A}^{{-1}}}B$
$ \displaystyle IX={{A}^{{-1}}}B$
$ \displaystyle X={{A}^{{-1}}}B$ $ \displaystyle \text{which is a unique solution for }X.$
$ \displaystyle \begin{array}{l}\text{It is given that }X\text{ has}\\\text{(i) an infinite number of solutions and}\\\text{(ii) no solution}\text{.}\end{array}$
$ \displaystyle \text{It means that the system has no unique solution}\text{.}$
$ \displaystyle \text{Thus, }\det A=0.$
$ \displaystyle \therefore 15-2m-{{m}^{2}}=0$
$ \displaystyle \therefore (5+m)(3-m)=0$
$ \displaystyle \therefore m=-5\ \text{or }m=3$
$ \displaystyle \text{When }m=-5,\ \text{System (1) becomes}$
$ \displaystyle 3x-5y=5\text{ and }-3x+5y=-5$
$ \displaystyle \begin{array}{l}\therefore \text{The two equations represents the same }\\\text{straight line and there will be infinite }\\\text{number of solutions}\text{.}\end{array}$
$ \displaystyle \text{When }m=3,\ \text{System (1) becomes}$
$ \displaystyle 3x+3y=5\text{ and }5x+5y=-3$
$ \displaystyle \begin{array}{l}\therefore \text{The two lines are parallel and }\\\text{there will be no solution}\text{.}\\\text{ }\end{array}$
