# Solve the following system of equations by matrix method:

Question:

Solve the following system of equations by matrix method:

(i) $5 x+7 y+2=0$

$4 x+6 y+3=0$

(ii) $5 x+2 y=3$

$3 x+2 y=5$

(iii) $3 x+4 y-5=0$

$x-y+3=0$

(iv) $3 x+y=19$

$3 x-y=23$

(v) $3 x+7 y=4$

$x+2 y=-1$

(vi) $3 x+y=23$

$5 x+3 y=12$

Solution:

(i) The given system of equations can be written in matrix form as folllows:

$\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-2 \\ -3\end{array}\right]$

$\mathrm{AX}=\mathrm{B}$

Here,

$A=\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}-2 \\ -3\end{array}\right]$

Now,

$|A|=\left[\begin{array}{ll}5 & 7 \\ 4 & 6\end{array}\right]$

$=30-28$

$=2 \neq 0$

The given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $A=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(6)=6, C_{12}=(-1)^{1+2}(4)=-4$

$C_{21}=(-1)^{2+1}(7)=-7, C_{22}=(-1)^{2+2}(5)$

$=5$

$\operatorname{adj} A=\left[\begin{array}{cc}6 & -4 \\ -7 & 5\end{array}\right]^{T}$

$=\left[\begin{array}{cc}6 & -7 \\ -4 & 5\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$\Rightarrow A^{-1}=\frac{1}{2}\left[\begin{array}{cc}6 & -7 \\ -4 & 5\end{array}\right]$

$X=A^{-1} B$

$=\frac{1}{2}\left[\begin{array}{cc}6 & -7 \\ -4 & 5\end{array}\right]\left[\begin{array}{l}-2 \\ -3\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{c}-12+21 \\ 8-15\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{9}{2} \\ \frac{-7}{2}\end{array}\right]$

$\therefore x=\frac{9}{2}$ and $y=\frac{-7}{2}$

(ii) The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}3 \\ 5\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 5\end{array}\right]$

Now,

$|A|=\left[\begin{array}{ll}5 & 2 \\ 3 & 2\end{array}\right]$

$=10-6$

$=4 \neq 0$

The given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(2)=2, C_{12}=(-1)^{1+2}(3)=-3$

$C_{21}=(-1)^{2+1}(2)=-2, C_{22}=(-1)^{2+2}(5)=5$

$\therefore \operatorname{adj} A=\left[\begin{array}{cc}2 & -3 \\ -2 & 5\end{array}\right]^{T}$

$=\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]$

$X=A^{-1} B$

$=\frac{1}{4}\left[\begin{array}{cc}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{l}3 \\ 5\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{c}6-10 \\ -9+25\end{array}\right]$

$\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{-4}{4} \\ \frac{16}{4}\end{array}\right]$

$\therefore x=-1$ and $y=4$

(iii) The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{cc}3 & 4 \\ 1 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}5 \\ -3\end{array}\right]$

Now,

$|A|=\left[\begin{array}{cc}3 & 4 \\ 1 & -1\end{array}\right]$

$=-3-4$

$=-7 \neq 0$

So, the given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(-1)=-1, C_{12}=(-1)^{1+2}(1)=-1$

$C_{21}=(-1)^{2+1}(4)=-4, C_{22}=(-1)^{2+2}(3)=3$

$\operatorname{adj} A=\left[\begin{array}{cc}-1 & -1 \\ -4 & 3\end{array}\right]^{T}=\left[\begin{array}{cc}-1 & -4 \\ -1 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-7}\left[\begin{array}{cc}-1 & -4 \\ -1 & 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{7}{-7} \\ \frac{-14}{-7}\end{array}\right]$

$\therefore x=-1$ and $y=2$

(iv) The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{cc}3 & 1 \\ 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}19 \\ 23\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{cc}3 & 1 \\ 3 & -1\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}19 \\ 23\end{array}\right]$

Now,

$|A|=\left[\begin{array}{cc}3 & 1 \\ 3 & -1\end{array}\right]$

$=-3-3$

$=-6 \neq 0$

So, the given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(-1)=-1, C_{12}=(-1)^{1+2}(3)=-3$

$C_{21}=(-1)^{2+1}(1)=-1, C_{22}=(-1)^{2+2}(3)=3$

$\operatorname{adj} A=\left[\begin{array}{cc}-1 & -3 \\ -1 & 3\end{array}\right]^{T}$

$=\left[\begin{array}{cc}-1 & -1 \\ -3 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-6}\left[\begin{array}{cc}-1 & -1 \\ -3 & 3\end{array}\right]$

$X=A^{-1} B$

$=\frac{1}{-6}\left[\begin{array}{cc}-1 & -1 \\ -3 & 3\end{array}\right]\left[\begin{array}{l}19 \\ 23\end{array}\right]$

$=\frac{1}{-6}\left[\begin{array}{l}-19-23 \\ -57+69\end{array}\right]$

$=\left[\begin{array}{l}x \\ y\end{array}\right]$

$=\left[\begin{array}{c}\frac{-42}{-6} \\ \frac{12}{-6}\end{array}\right]$

$\therefore x=7$ and $y=-2$

(v) The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{ll}3 & 7 \\ 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}4 \\ -1\end{array}\right]$

$A X=B$

Here,

$A=\left[\begin{array}{ll}3 & 7 \\ 1 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}4 \\ -1\end{array}\right]$

Now,

$|A|=\left[\begin{array}{ll}3 & 7 \\ 1 & 2\end{array}\right]$

$=6-7$

$=-1 \neq 0$

So, the given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(2)=2, C_{12}=(-1)^{1+2}(1)=-1$

$C_{21}=(-1)^{2+1}(7)=-7, C_{22}=(-1)^{2+2}(3)$

$=3$

$\operatorname{adj} A=\left[\begin{array}{cc}2 & -1 \\ -7 & 3\end{array}\right]^{T}$

$=\left[\begin{array}{cc}2 & -7 \\ -1 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{-1}\left[\begin{array}{cc}2 & -7 \\ -1 & 3\end{array}\right]$

$X=A^{-1} B$

$=\frac{1}{-1}\left[\begin{array}{cc}2 & -7 \\ -1 & 3\end{array}\right]\left[\begin{array}{c}4 \\ -1\end{array}\right]$

$=\frac{1}{-1}\left[\begin{array}{c}8+7 \\ -4-3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}\frac{15}{-1} \\ \frac{-7}{-1}\end{array}\right]$

$\therefore x=-15$ and $y=7$

(vi) The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{ll}3 & 1 \\ 5 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}7 \\ 12\end{array}\right]$

$A X=\mathrm{B}$

Here,

$A=\left[\begin{array}{ll}3 & 1 \\ 5 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{c}7 \\ 12\end{array}\right]$

Now,

$|A|=\left[\begin{array}{ll}3 & 1 \\ 5 & 3\end{array}\right]$

$=9-5$

$=4 \neq 0$

So, the given system has a unique solution given by $X=A^{-1} B$.

Let $\mathrm{C}_{i j}$ be the cofactors of the elements $\mathrm{a}_{i j}$ in $A=\left[a_{i j}\right]$. Then,

$C_{11}=(-1)^{1+1}(3)=3, C_{12}=(-1)^{1+2}(5)=-5$

$C_{21}=(-1)^{2+1}(1)=-1, C_{22}=(-1)^{2+2}(3)=3$

$\operatorname{adj} A=\left[\begin{array}{cc}3 & -5 \\ -1 & 3\end{array}\right]^{T}$

$=\left[\begin{array}{cc}3 & -1 \\ -5 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$=\frac{1}{4}\left[\begin{array}{cc}3 & -1 \\ -5 & 3\end{array}\right]$

$X=A^{-1} B$

$=\frac{1}{4}\left[\begin{array}{cc}3 & -1 \\ -5 & 3\end{array}\right]\left[\begin{array}{c}7 \\ 12\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{c}21-12 \\ -35+36\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}\frac{9}{4} \\ \frac{1}{4}\end{array}\right]$

$\therefore x=\frac{9}{4}$ and $y=\frac{1}{4}$