Solution

The given system of equations is:

$x+2 y=2$

$2 x+3 y=3$

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 1 & 2 \\ 2 & 3\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} $ and $B= \begin{bmatrix} 2 \\ 3 \end{bmatrix} $.

Now,

$|A|=1(3)-2(2)=3-4=-1 \neq 0$

$\therefore A$ is non-singular.

Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Solution

The given system of equations is:

$2 x-y=5$

$x+y=4$

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 2 & -1 \\ 1 & 1\end{bmatrix} , X= \begin{bmatrix} x \\ z\end{bmatrix} $ and $B= \begin{bmatrix} 5 \\ 4 \end{bmatrix} $.

Now,

$|A|=2(1)-(-1)(1)=2+1=3 \neq 0$

$\therefore A$ is non-singular.

Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Solution

Answer

The given system of equations is:

$x+3 y=5$

$2 x+6 y=8$

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 1 & 3 \\ 2 & 6\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} $ and $B= \begin{bmatrix} 5 \\ 8 \end{bmatrixs} $.

Now,

$|A|=1(6)-3(2)=6-6=0$

$\therefore A$ is a singular matrix.

Now,

$(adj A)= \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} $

$(adj A) B= \begin{bmatrix} 6 & -3 \\ -2 & 1\end{bmatrix} \begin{bmatrix} 5 \\ 8\end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8\end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Solution

The given system of equations is:

$x+y+z=1$

$2 x+3 y+2 z=2$

$a x+a y+2 a z=4$

This system of equations can be written in the form $A X=B$, where

$ A= \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B= \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} \text{. } $

Now,

$ \begin{aligned} |A| & =1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a) \\ & =4 a-2 a-a=4 a-3 a=a \neq 0 \end{aligned} $

$\therefore A$ is non-singular.

Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Solution

The given system of equations is:

$3 x-y-2 z=2$

$2 y-z=-1$

$3 x-5 y=3$

This system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ and $B= \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} $.

Now,

$|A|=3(0-5)-0+3(1+4)=-15+15=0$

$\therefore A$ is a singular matrix.

Now,

$(adj A)= \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix} $

$\therefore(adj A) B= \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3\end{bmatrix} = \begin{bmatrix} -10-10+15 \\ -6-6+9 \\ -12-12+18\end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix} \neq O$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Solution

The given system of equations is:

$5 x-y+4 z=5$

$2 x+3 y+5 z=2$

$5 x-2 y+6 z=-1$

This system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ and $B= \begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix} $.

Now,

$$ \begin{aligned} |A| & =5(18+10)+1(12-25)+4(-4-15) \\ & =5(28)+1(-13)+4(-19) \\ & =140-13-76 \\ & =51 \neq 0 \end{aligned} $$

$\therefore A$ is non-singular.

Therefore, $A^{-1}$ exists.

Hence, the given system of equations is consistent.

Solution

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 5 & 2 \\ 7 & 3\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} $ and $B= \begin{bmatrix} 4 \\ 5 \end{bmatrix} $.

Now, $|A|=15-14=1 \neq 0$.

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

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

$\therefore A^{-1}= \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} $

$\therefore X=A^{-1} B= \begin{bmatrix} 3 & -2 \\ -7 & 5\end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 12-10 \\ -28+25\end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} $

Hence, $x=2$ and $y=-3$.

Solution

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 2 & -1 \\ 3 & 4\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} $ and $B= \begin{bmatrix} -2 \\ 3 \end{bmatrix} $.

Now,

$|A|=8+3=11 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

$A^{-1}=\frac{1}{|A|} adj A=\frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} $

$\therefore X=A^{-1} B=\frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2\end{bmatrix} \begin{bmatrix} -2 \\ 3 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} x \\ y\end{bmatrix} =\frac{1}{11} \begin{bmatrix} -8+3 \\ 6+6\end{bmatrix} =\frac{1}{11} \begin{bmatrix} -5 \\ 12\end{bmatrix} = \begin{bmatrix} -\frac{5}{11} \\ \frac{12}{11} \end{bmatrix} $

Hence, $x=\frac{-5}{11}$ and $y=\frac{12}{11}$.

Solution

The given system of equations can be written in the form of $A X=B$, where

$ A= \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix} , X= \begin{bmatrix} x \\ y \end{bmatrix} \text{ and } B= \begin{bmatrix} 3 \\ 7 \end{bmatrix} \text{. } $

Now,

$|A|=-20+9=-11 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now,

$A^{-1}=\frac{1}{|A|}(adj A)=-\frac{1}{11} \begin{bmatrix} -5 & 3 \\ -3 & 4\end{bmatrix} =\frac{1}{11} \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix} $

$\therefore X=A^{-1} B=\frac{1}{11} \begin{bmatrix} 5 & -3 \\ 3 & -4\end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} x \\ y\end{bmatrix} =\frac{1}{11} \begin{bmatrix} 5 & -3 \\ 3 & -4\end{bmatrix} \begin{bmatrix} 3 \\ 7\end{bmatrix} =\frac{1}{11} \begin{bmatrix} 15-21 \\ 9-28\end{bmatrix} =\frac{1}{11} \begin{bmatrix} -6 \\ -19\end{bmatrix} = \begin{bmatrix} -\frac{6}{11} \\ -\frac{19}{11} \end{bmatrix} $

Hence, $x=\frac{-6}{11}$ and $y=\frac{-19}{11}$.

Solution

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 5 & 2 \\ 3 & 2\end{bmatrix} , X= \begin{bmatrix} x \\ y\end{bmatrix} $ and $B= \begin{bmatrix} 3 \\ 5 \end{bmatrix} $.

Now,

$|A|=10-6=4 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Solution

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} $ and $B= \begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix} $.

Now,

$|A|=2(10+3)-1(-5-3)+0=2(13)-1(-8)=26+8=34 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

$\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]^{\top} \\ & =\left[\begin{array}{ccc}M_{11} & -M_{21} & M_{31} \\ -M_{12} & M_{22} & -M_{32} \\ M_{13} & -M_{23} & M_{33}\end{array}\right]=\left[\begin{array}{ccc}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \\ \end{array}\right]\end{aligned}$

$\begin{aligned} & A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ & =\frac{1}{34}\left[\begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array}\right] \end{aligned}$

Solve for the values of $x, y$ and $z$

$A X=B \Rightarrow X=A^{-1} B$

$\begin{aligned} & \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array}\right]\left[\begin{array}{l} 1 \\ \frac{3}{2} \\ 9 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 13(1)+8\left(\frac{3}{2}\right)+1(9) \\ 5(1)+(-10)\left(\frac{3}{2}\right)+3(9) \\ 3(1)+(-6)\left(\frac{3}{2}\right)+(-5)(9) \end{array}\right] \end{aligned}$

$\begin{aligned} & \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 13+12+9 \\ 5-15+27 \\ 3-9-45 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 34 \\ 17 \\ -51 \end{array}\right] \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 1 \\ \frac{1}{2} \\ \frac{-3}{2} \end{array}\right] \end{aligned}$

Hence, $x=1, y=\frac{1}{2}$, and $z=-\frac{3}{2}$.

Solution

The given system of equations can be written in the form of $A X=B$, where

$A= \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ and $B= \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} $.

Now,

$|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

$ \begin{aligned} & \text{ Now, } A _{11}=4, A _{12}=-5, A _{13}=1 \\ & A _{21}=2, A _{22}=0, A _{23}=-2 \\ & A _{31}=2, A _{32}=5, A _{33}=3 \\ & \therefore A^{-1}=\frac{1}{|A|}(adj A)=\frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \\ & \therefore X=A^{-1} B=\frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} \\ & \quad=\frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} \\ & \quad= \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix} \end{aligned} $

Hence, $x=2, y=-1$, and $z=1$.

Solution

The given system of equations can be written in the form $A X=B$, where

$A= \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2\end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z\end{bmatrix} $ and $B= \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} $.

Now,

$|A|=2(4+1)-3(-2-3)+3(-1+6)=2(5)-3(-5)+3(5)=10+15+15=40 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

Now, $A _{11}=5, A _{12}=5, A _{13}=5$

$A _{21}=3, A _{22}=-13, A _{23}=11$

$A _{31}=9, A _{32}=1, A _{33}=-7$

$\therefore A^{-1}=\frac{1}{|A|}(adj A)=\frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} $

$\therefore X=A^{-1} B=\frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} $

$\Rightarrow \begin{bmatrix} x \\ y \\ z\end{bmatrix} =\frac{1}{40} \begin{bmatrix} 25-12+27 \\ 25+52+3 \\ 25-44-21 \end{bmatrix} $

$=\frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} $

$= \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} $

Hence, $x=1, y=2$, and $z=-1$.

Solution

The given system of equations can be written in the form of $A X=B$, where

$ A= \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B= \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \text{. } $

Now,

$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$

Thus, $A$ is non-singular. Therefore, its inverse exists.

$ \begin{aligned} & \text{ Now, } A _{11}=7, A _{12}=-19, A _{13}=-11 \\ & A _{21}=1, A _{22}=-1, A _{23}=-1 \\ & A _{31}=-3, A _{32}=11, A _{33}=7 \\ & \therefore A^{-1}=\frac{1}{|A|}(\text{ adj } A)=\frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \\ & \therefore X=A^{-1} B=\frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\frac{1}{4} \begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix} \\ & \quad=\frac{1}{4} \begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \end{aligned} $

Hence, $x=2, y=1$, and $z=3$.

Solution

$ \begin{aligned} & A= \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} \\ & \therefore|A|=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0 \end{aligned} $

Now, $A _{11}=0, A _{12}=2, A _{13}=1$

$A _{21}=-1, A _{22}=-9, A _{23}=-5$

$A _{31}=2, A _{32}=23, A _{33}=13$

$\therefore A^{-1}=\frac{1}{|A|}(adj A)=- \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} $

Now, the given system of equations can be written in the form of $A X=B$, where

$ A= \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B= \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \text{. } $

The solution of the system of equations is given by $X=A^{-1} B$.

$ \begin{aligned} & X=A^{-1} B \end{aligned} $

$\Rightarrow \begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix}11 \\ -5 \\ -3 \end{bmatrix}$

$ \begin{aligned} & = \begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix} \\ & = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned} $

Hence, $x=1, y=2$, and $z=3$.

Solution

Let the cost of onions, wheat, and rice per $kg$ be $Rs x, Rs y$, and $Rs z$ respectively.

Then, the given situation can be represented by a system of equations as:

$4 x+3 y+2 z=60$

$2 x+4 y+6 z=90$

$6 x+2 y+3 z=70$

This system of equations can be written in the form of $A X=B$, where

$ \begin{aligned} & A= \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix} , X= \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{ and } B= \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} \text{. } \\ & |A|=4(12-12)-3(6-36)+2(4-24)=0+90 \\ & \text{ Now, } \quad \begin{matrix} A _{11}=0, A _{12}=30, A _{13}=-20 \\ A _{21}=-5, A _{22}=0, A _{23}=10 \\ A _{31}=10, A _{32}=-20, A _{33}=10 \end{matrix} \end{aligned} $

$ |A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 \neq 0 $

$ \begin{aligned} & \therefore adj A= \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \\ & \therefore A^{-1}=\frac{1}{|A|} \text{ adj } A=\frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \end{aligned} $

Now,

$ \begin{aligned} & X=A^{-1} B \\ & \Rightarrow X=\frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} \\ & \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 0-450+700 \\ 1800+0-1400 \\ -1200+900+700 \end{bmatrix} \\ & =\frac{1}{50} \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} \\ & = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix} \\ & \therefore x=5, y=8 \text{ and } z=8 . \end{aligned} $

Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.