Q: Examine the consistency of the system of equations: $$ \begin{aligned} & x+y+z=1 \\ & 2 x+3 y+2 z=2 \\ & a x+a y+2 a z=4 \end{aligned} $$

$$ \begin{aligned} & x+y+z=1 \\ & 2 x+3 y+2 z=2 \end{aligned} $$ The given system of equations is: $a x+a y+2 a z=4$ The given system of equations can be written in the form of $A X=B$, where $$ A=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{array}\right), X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 2 \\ 4 \end{array}\right] $$ Hence, $$ \begin{aligned} |A| & =1(6 a-2 a)-1(4 a-2 a)+1(2 a-3 a) \\ & =4 a-2 a-a \\ & =4 a-3 a \\ & =a \neq 0 \end{aligned} $$ So, $A$ is non-singular. Therefore, $A^{-1}$ exists. Thus, the given system of equations is consistent.

Q: Examine the consistency of the system of equations: $$ \begin{aligned} & 3 x-y-2 z=2 \\ & 2 y-z=-1 \\ & 3 x-5 y=3 \end{aligned} $$

$$ \begin{aligned} & 3 x-y-2 z=2 \\ & 2 y-z=-1 \end{aligned} $$ The given system of equations is: $3 x-5 y=3$ The given system of equations can be written in the form of $A X=B$, where $$ A=\left(\begin{array}{ccc} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{array}\right), X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right] $$ Hence, $$ \begin{aligned} |A| & =3(0-5)-0+3(1+4) \\ & =-15+15 \\ & =0 \end{aligned} $$ So, $A$ is a singular matrix. Now, $$ (\operatorname{adjA})=\left(\begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right) $$ Therefore, $$ \begin{aligned} (\operatorname{adj} A) B & =\left(\begin{array}{ccc} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{array}\right)\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right] \\ & =\left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right] \\ & =\left[\begin{array}{c} -5 \\ -3 \\ -6 \end{array}\right] \\ & \neq 0 \end{aligned} $$ Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

Q: Examine the consistency of the system of equations: $$ \begin{aligned} & 5 x-y+4 z=5 \\ & 2 x+3 y+5 z=2 \\ & 5 x-2 y+6 z=-1 \end{aligned} $$

$$ \begin{aligned} & 5 x-y+4 z=5 \\ & 2 x+3 y+5 z=2 \end{aligned} $$ The given system of equations is: $5 x-2 y+6 z=-1$ The given system of equations can be written in the form of $A X=B$, where $A=\left(\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right), X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}5 \\ 2 \\ -1\end{array}\right]$ Hence, $$ \begin{aligned} |A| & =5(18+10)+1(12-25)+4(-4-15) \\ & =5(28)+1(-13)+4(-19) \\ & =140-13-76 \\ & =51 \neq 0 \end{aligned} $$ So, $A$ is nonsingular. Therefore, $A^{-1}$ exists. Hence, the given system of equations is consistent.

Question 1

Examine the consistency of the system of equations:

$$

\begin{aligned}

& x+y+z=1 \

& 2 x+3 y+2 z=2 \

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

\end{aligned}

$$

Accepted Answer

$$

\begin{aligned}

& x+y+z=1 \

& 2 x+3 y+2 z=2

\end{aligned}

$$

The given system of equations is: $a x+a y+2 a z=4$

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

$$

A=\left(\begin{array}{ccc}

1 & 1 & 1 \

2 & 3 & 2 \

a & a & 2 a

\end{array}ight), X=\left[\begin{array}{c}

x \

y \

z

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

1 \

2 \

4

\end{array}ight]

$$

Hence,

$$

\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 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Thus, the given system of equations is consistent.

Question 2

Examine the consistency of the system of equations:

$x+2 y=2$

$2 x+3 y=3$

Accepted Answer

$$

x+2 y=2

$$

The given system of equations is: $2 x+3 y=3$

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

$$

A=\left(\begin{array}{ll}

1 & 2 \

2 & 3

\end{array}ight), X=\left[\begin{array}{l}

x \

y

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

2 \

3

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =1(3)-2(2) \

& =3-4 \

& =-1 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Thus, the given system of equations is consistent.

Question 3

Examine the consistency of the system of equations:

$2 x-y=5$

$x+y=4$

Accepted Answer

$$

2 x-y=5

$$

The given system of equations is: $x+y=4$

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

$$

A=\left(\begin{array}{cc}

2 & -1 \

1 & 1

\end{array}ight), X=\left[\begin{array}{l}

x \

y

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

5 \

4

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =2(1)-1(-1) \

& =2+1 \

& =3 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Hence, the given system of equations is consistent.

Question 4

Examine the consistency of the system of equations:

$x+3 y=5$

$2 x+6 y=8$

Accepted Answer

$$

x+3 y=5

$$

The given system of equations is: $2 x+6 y=8$

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

$$

A=\left(\begin{array}{ll}

1 & 3 \

2 & 6

\end{array}ight), X=\left[\begin{array}{l}

x \

y

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

5 \

8

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =1(6)-3(2) \

& =6-6 \

& =0

\end{aligned}

$$

So, $A$ is a singular matrix.

Now,

$$

(\operatorname{adj} A)=\left(\begin{array}{cc}

6 & -3 \

-2 & 1

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

(\operatorname{adj} A) B & =\left(\begin{array}{cc}

6 & -5 \

-2 & 1

\end{array}ight)\left[\begin{array}{l}

5 \

8

\end{array}ight] \

& =\binom{30-24}{-10+8} \

& =\left[\begin{array}{c}

6 \

-2

\end{array}ight] \

& 
eq 0

\end{aligned}

$$

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

Hence, the system of equations is inconsistent.

Question 5

Examine the consistency of the system of equations:

$$

\begin{aligned}

& 3 x-y-2 z=2 \

& 2 y-z=-1 \

& 3 x-5 y=3

\end{aligned}

$$

Accepted Answer

$$

\begin{aligned}

& 3 x-y-2 z=2 \

& 2 y-z=-1

\end{aligned}

$$

The given system of equations is: $3 x-5 y=3$

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

$$

A=\left(\begin{array}{ccc}

3 & -1 & -2 \

0 & 2 & -1 \

3 & -5 & 0

\end{array}ight), X=\left[\begin{array}{l}

x \

y \

z

\end{array}ight] 	ext { and } B=\left[\begin{array}{c}

2 \

-1 \

3

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =3(0-5)-0+3(1+4) \

& =-15+15 \

& =0

\end{aligned}

$$

So, $A$ is a singular matrix.

Now,

$$

(\operatorname{adjA})=\left(\begin{array}{ccc}

-5 & 10 & 5 \

-3 & 6 & 3 \

-6 & 12 & 6

\end{array}ight)

$$

Therefore,

$$

\begin{aligned}

(\operatorname{adj} A) B & =\left(\begin{array}{ccc}

-5 & 10 & 5 \

-3 & 6 & 3 \

-6 & 12 & 6

\end{array}ight)\left[\begin{array}{c}

2 \

-1 \

3

\end{array}ight] \

& =\left[\begin{array}{c}

-10-10+15 \

-6-6+9 \

-12-12+18

\end{array}ight] \

& =\left[\begin{array}{c}

-5 \

-3 \

-6

\end{array}ight] \

& 
eq 0

\end{aligned}

$$

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

Hence, the system of equations is inconsistent.

Question 6

Examine the consistency of the system of equations:

$$

\begin{aligned}

& 5 x-y+4 z=5 \

& 2 x+3 y+5 z=2 \

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

\end{aligned}

$$

Accepted Answer

$$

\begin{aligned}

& 5 x-y+4 z=5 \

& 2 x+3 y+5 z=2

\end{aligned}

$$

The given system of equations is: $5 x-2 y+6 z=-1$

The given system of equations can be written in the form of $A X=B$, where $A=\left(\begin{array}{ccc}5 & -1 & 4 \ 2 & 3 & 5 \ 5 & -2 & 6\end{array}ight), X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$ and $B=\left[\begin{array}{c}5 \ 2 \ -1\end{array}ight]$

Hence,

$$

\begin{aligned}

|A| & =5(18+10)+1(12-25)+4(-4-15) \

& =5(28)+1(-13)+4(-19) \

& =140-13-76 \

& =51 
eq 0

\end{aligned}

$$

So, $A$ is nonsingular.

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

Hence, the given system of equations is consistent.

Question 7

Solve system of linear equations, using matrix method.

$5 x+2 y=4$

$7 x+3 y=5$

Accepted Answer

$$

5 x+2 y=4

$$

The given system of equations is: $7 x+3 y=5$

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

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

Hence,

$$

\begin{aligned}

|A| & =15-14 \

& =1 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{aligned}

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

& =\left(\begin{array}{cc}

3 & -2 \

-7 & 5

\end{array}ight)

\end{aligned}

$$

Then,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left(\begin{array}{cc}

3 & -2 \

-7 & 5

\end{array}ight)\left[\begin{array}{l}

4 \

5

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left[\begin{array}{c}

12-10 \

-28+25

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left[\begin{array}{c}

2 \

-3

\end{array}ight]

\end{aligned}

$$

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

Question 8

Solve system of linear equations, using matrix method.

$2 x-y=-2$

$3 x+4 y=3$

Accepted Answer

$$

2 x-y=-2

$$

The given system of equations is: $3 x+4 y=3$

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

$$

A=\left(\begin{array}{cc}

2 & -1 \

3 & 4

\end{array}ight), X=\left[\begin{array}{l}

x \

y

\end{array}ight] 	ext { and } B=\left[\begin{array}{c}

-2 \

3

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =8+3 \

& =11 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{aligned}

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

& =\frac{1}{11}\left(\begin{array}{cc}

4 & 1 \

-3 & 2

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left(\begin{array}{cc}

4 & 1 \

-3 & 2

\end{array}ight)\left[\begin{array}{c}

-2 \

3

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left[\begin{array}{c}

-8+3 \

6+6

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left[\begin{array}{c}

-5 \

12

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left[\begin{array}{c}

\frac{-5}{11} \

\frac{12}{11}

\end{array}ight]

\end{aligned}

$$

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

Question 9

Solve system of linear equations, using matrix method.

$$

\begin{aligned}

& 4 x-3 y=3 \

& 3 x-5 y=7

\end{aligned}

$$

Accepted Answer

$$

4 x-3 y=3

$$

The given system of equations is: $3 x-5 y=7$

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

$$

A=\left(\begin{array}{ll}

4 & -3 \

3 & -5

\end{array}ight), X=\left[\begin{array}{l}

x \

y

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

3 \

7

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =-20+9 \

& =-11 \

& 
eq 0

\end{aligned}

$$

So, $A$ is nonsingular.

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

Now,

$$

\begin{aligned}

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

& =-\frac{1}{11}\left(\begin{array}{ll}

-5 & 3 \

-3 & 4

\end{array}ight)=\frac{1}{11}\left(\begin{array}{ll}

5 & -3 \

3 & -4

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left(\begin{array}{ll}

5 & -3 \

3 & -4

\end{array}ight)\left[\begin{array}{l}

3 \

7

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left(\begin{array}{ll}

5 & -3 \

3 & -4

\end{array}ight)\left[\begin{array}{l}

3 \

7

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left[\begin{array}{c}

15-21 \

9-28

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{11}\left[\begin{array}{c}

-6 \

-19

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left[\begin{array}{c}

\frac{-6}{11} \

\frac{-19}{11}

\end{array}ight]

\end{aligned}

$$

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

Question 10

Solve system of linear equations, using matrix method.

$5 x+2 y=3$

$3 x+2 y=5$

Accepted Answer

$$

5 x+2 y=3

$$

The given system of equations is: $3 x+2 y=5$

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

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

Hence,

$$

\begin{aligned}

|A| & =10-6 \

& =4 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{aligned}

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

& =\frac{1}{4}\left(\begin{array}{cc}

2 & -2 \

-3 & 5

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{4}\left(\begin{array}{cc}

2 & -2 \

-3 & 5

\end{array}ight)\left[\begin{array}{l}

3 \

5

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{4}\left(\begin{array}{cc}

2 & -2 \

-3 & 5

\end{array}ight)\left[\begin{array}{l}

3 \

5

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{4}\left[\begin{array}{c}

6-10 \

-9+25

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\frac{1}{4}\left[\begin{array}{c}

-4 \

16

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y

\end{array}ight]=\left[\begin{array}{c}

-1 \

4

\end{array}ight]

\end{aligned}

$$

Hence, $x=-1$ and $y=4$

Question 11

Solve system of linear equations, using matrix method.

$$

\begin{aligned}

& 2 x+y+z=1 \

& x-2 y-z=\frac{3}{2} \

& 3 y-5 z=9

\end{aligned}

$$

Accepted Answer

$$

\begin{aligned}

& 2 x+y+z=1 \

& x-2 y-z=\frac{3}{2}

\end{aligned}

$$

The given system of equations is: $3 y-5 z=9$

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

$$

A=\left(\begin{array}{ccc}

2 & 1 & 1 \

1 & -2 & -1 \

0 & 3 & -5

\end{array}ight), X=\left[\begin{array}{c}

x \

y \

z

\end{array}ight] 	ext { and } \quad B=\left[\begin{array}{c}

1 \

\frac{3}{2} \

9

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =2(10+3)-1(-5-3)+0 \

& =2(13)-1(-8) \

& =26+8 \

& =34 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{array}{lll}

A_{11}=13 & A_{12}=5 & A_{13}=3 \

A_{21}=8 & A_{22}=-10 & A_{23}=-6 \

a_{31}=1 & A_{32}=3 & A_{33}=-5

\end{array}

$$

Hence,

$$

\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}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{34}\left(\begin{array}{ccc}

13 & 8 & 1 \

5 & -10 & 3 \

3 & -6 & -5

\end{array}ight)\left[\begin{array}{l}

1 \

\frac{3}{2} \

9

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{34}\left(\begin{array}{ccc}

13 & 8 & 1 \

5 & -10 & 3 \

3 & -6 & -5

\end{array}ight)\left[\begin{array}{l}

1 \

\frac{3}{2} \

9

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{34}\left[\begin{array}{c}

13+12+9 \

5-15+27 \

3-9-45

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{34}\left[\begin{array}{c}

34 \

17 \

-51

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{c}

1 \

\frac{1}{2} \

-\frac{3}{2}

\end{array}ight]

\end{aligned}

$$

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

Question 12

Solve system of linear equations, using matrix method.

$x-y+z=4$

$2 x+y-3 z=0$

$x+y+z=2$

Accepted Answer

$$

\begin{aligned}

& x-y+z=4 \

& 2 x+y-3 z=0

\end{aligned}

$$

The given system of equations is: $x+y+z=2$

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

$$

A=\left(\begin{array}{ccc}

1 & -1 & 1 \

2 & 1 & -3 \

1 & 1 & 1

\end{array}ight), X=\left[\begin{array}{l}

x \

y \

z

\end{array}ight] 	ext { and } B=\left[\begin{array}{l}

4 \

0 \

2

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =1(1+3)+1(2+3)+1(2-1) \

& =4+5+1 \

& =10 \

& 
eq 0

\end{aligned}

$$

So, $A$ is nonsingular.

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

Now,

$$

\begin{array}{lll}

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

\end{array}

$$

Hence,

$$

\begin{aligned}

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

& =\frac{1}{10}\left(\begin{array}{ccc}

4 & 2 & 2 \

-5 & 0 & 5 \

1 & -2 & 3

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{10}\left(\begin{array}{ccc}

4 & 2 & 2 \

-5 & 0 & 5 \

1 & -2 & 3

\end{array}ight)\left[\begin{array}{l}

4 \

0 \

2

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{10}\left(\begin{array}{ccc}

4 & 2 & 2 \

-5 & 0 & 5 \

1 & -2 & 3

\end{array}ight)\left[\begin{array}{l}

4 \

0 \

2

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{10}\left[\begin{array}{c}

16+0+4 \

-20+0+10 \

4+0+6

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{10}\left[\begin{array}{c}

20 \

-10 \

10

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{c}

2 \

-1 \

1

\end{array}ight]

\end{aligned}

$$

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

Question 13

Solve system of linear equations, using matrix method.

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

$x-2 y+z=-4$

$3 x-y-2 z=3$

Accepted Answer

$$

\begin{aligned}

& 2 x+3 y+3 z=5 \

& x-2 y+z=-4

\end{aligned}

$$

The given system of equations is: $3 x-y-2 z=3$

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

$$

A=\left(\begin{array}{ccc}

2 & 3 & 3 \

1 & -2 & 1 \

3 & -1 & -2

\end{array}ight), X=\left[\begin{array}{l}

x \

y \

z

\end{array}ight] 	ext { and } B=\left[\begin{array}{c}

5 \

-4 \

3

\end{array}ight]

$$

Hence,

$$

\begin{aligned}

|A| & =2(4+1)-3(-2-3)+3(-1+6) \

& =10+15+15 \

& =40 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{array}{lll}

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

\end{array}

$$

Hence,

$$

\begin{aligned}

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

& =\frac{1}{40}\left(\begin{array}{ccc}

5 & 3 & 9 \

5 & -13 & 1 \

5 & 11 & -7

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{40}\left(\begin{array}{ccc}

5 & 3 & 9 \

5 & -13 & 1 \

5 & 11 & -7

\end{array}ight)\left[\begin{array}{c}

5 \

-4 \

3

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{40}\left(\begin{array}{ccc}

5 & 3 & 9 \

5 & -13 & 1 \

5 & 11 & -7

\end{array}ight)\left[\begin{array}{c}

5 \

-4 \

3

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{40}\left[\begin{array}{c}

25-12+27 \

25+52+3 \

25-44-21

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{40}\left[\begin{array}{c}

40 \

80 \

-40

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{c}

1 \

2 \

-1

\end{array}ight]

\end{aligned}

$$

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

Question 14

Solve system of linear equations, using matrix method.

$x-y+2 z=7$

$3 x+4 y-5 z=-5$

$2 x-y+3 z=12$

Accepted Answer

$$

\begin{aligned}

& x-y+2 z=7 \

& 3 x+4 y-5 z=-5

\end{aligned}

$$

The given system of equations is: $2 x-y+3 z=12$

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

$A=\left(\begin{array}{ccc}1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3\end{array}ight), X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$ and $B=\left[\begin{array}{c}7 \ -5 \ 12\end{array}ight]$

Hence,

$$

\begin{aligned}

|A| & =1(12-5)+1(9+10)+2(-3-8) \

& =7+19-22 \

& =4 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{array}{lll}

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

\end{array}

$$

Hence,

$$

\begin{aligned}

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

& =\frac{1}{4}\left(\begin{array}{ccc}

7 & 1 & -3 \

-19 & -1 & 11 \

-11 & -1 & 7

\end{array}ight)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{4}\left(\begin{array}{ccc}

7 & 1 & -3 \

-19 & -1 & 11 \

-11 & -1 & 7

\end{array}ight)\left[\begin{array}{c}

7 \

-5 \

12

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{4}\left(\begin{array}{ccc}

7 & 1 & -3 \

-19 & -1 & 11 \

-11 & -1 & 7

\end{array}ight)\left[\begin{array}{c}

7 \

-5 \

12

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{4}\left[\begin{array}{c}

49-5-36 \

-133+5+132 \

-77+5+84

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{4}\left[\begin{array}{c}

49-5-36 \

-133+5+132 \

-77+5+84

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{4}\left[\begin{array}{c}

8 \

4 \

12

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{l}

2 \

1 \

3

\end{array}ight]

\end{aligned}

$$

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

Question 15

$A=\left(\begin{array}{ccc}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}ight)$, find $A^{-1}$. Using $A^{-1}$ solve the system of equations

$2 x-3 y+5 z=11$

$3 x+2 y-4 z=-5$

$x+y-2 z=-3$

Accepted Answer

It is given that $A=\left(\begin{array}{ccc}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}ight)$

Therefore,

$$

\begin{aligned}

|A| & =2(-4+4)+3(-6+4)+5(3-2) \

& =0-6+5 \

& =-1 \

& 
eq 0

\end{aligned}

$$

Now,

$$

\begin{array}{lll}

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

\end{array}

$$

Hence,

$$

\begin{aligned}

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

& =-\left(\begin{array}{llc}

0 & -1 & 2 \

2 & -9 & 23 \

1 & -5 & 13

\end{array}ight)=\left(\begin{array}{ccc}

0 & 1 & -2 \

-2 & 9 & -23 \

-1 & 5 & -13

\end{array}ight)

\end{aligned}

$$

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

$A=\left(\begin{array}{ccc}2 & -3 & 5 \ 3 & 2 & -4 \ 1 & 1 & -2\end{array}ight), X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$ and $B=\left[\begin{array}{c}11 \ -5 \ -3\end{array}ight]$

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

Therefore,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left(\begin{array}{ccc}

0 & 1 & -2 \

-2 & 9 & -23 \

-1 & 5 & -13

\end{array}ight)\left[\begin{array}{l}

11 \

-5 \

-3

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{ccc}

0 & 1 & -2 \

-2 & 9 & -23 \

-1 & 5 & -13

\end{array}ight]\left[\begin{array}{l}

11 \

-5 \

-3

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{c}

0-5+6 \

-22-45+69 \

-11-25+39

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{l}

1 \

2 \

3

\end{array}ight]

\end{aligned}

$$

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

Question 16

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹ 60 . The cost of 2 kg onion, 4 kg wheat and 6 kg rice is $₹ 90$. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹ 70 . Find cost of each item per kg by matrix method.

Accepted Answer

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

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

$$

\begin{aligned}

& 4 x+3 y+2 z=60 \

& 2 x+4 y+6 z=90 \

& 6 x+2 y+3 z=70

\end{aligned}

$$

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

$A=\left(\begin{array}{lll}4 & 3 & 2 \ 2 & 4 & 6 \ 6 & 2 & 3\end{array}ight), X=\left[\begin{array}{l}x \ y \ z\end{array}ight]$ and $B=\left[\begin{array}{c}60 \ 90 \ 70\end{array}ight]$

Therefore,

$$

\begin{aligned}

|A| & =4(12-12)-3(6-36)+2(4-24) \

& =0+90-40 \

& =50 \

& 
eq 0

\end{aligned}

$$

So, $A$ is non-singular.

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

Now,

$$

\begin{array}{lll}

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{array}

$$

Therefore,

$$

\begin{aligned}

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

& =\frac{1}{50}\left(\begin{array}{ccc}

0 & -5 & 10 \

30 & 0 & -20 \

-20 & 10 & 10

\end{array}ight)

\end{aligned}

$$

Hence,

$$

\begin{aligned}

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

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{50}\left(\begin{array}{ccc}

0 & -5 & 10 \

30 & 0 & -20 \

-20 & 10 & 10

\end{array}ight)\left[\begin{array}{l}

60 \

90 \

70

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{50}\left(\begin{array}{ccc}

0 & -5 & 10 \

30 & 0 & -20 \

-20 & 10 & 10

\end{array}ight)\left[\begin{array}{l}

60 \

90 \

70

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{50}\left[\begin{array}{c}

0-450+700 \

1800+0-1400 \

-1200+900+700

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\frac{1}{50}\left[\begin{array}{l}

250 \

400 \

400

\end{array}ight] \

& \Rightarrow\left[\begin{array}{l}

x \

y \

z

\end{array}ight]=\left[\begin{array}{l}

5 \

8 \

8

\end{array}ight]

\end{aligned}

$$

Thus, $x=5, y=8$ and $z=8$

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

\section*{MISCELLANEOUS EXERCISE}