Class 12 ncert solution math exercise 4.6

    EXERCISE 4.6 ( Determinants )

Question 1: Examine the consistency of the system of equations:(Class 12 ncert solution math exercise 4.6)

\begin{aligned} &x+2 y=2 \\ &2 x+3 y=3 \end{aligned}

Solution: 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=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 2 \\ 3 \end{array}\right]

Hence,

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

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:

2 x-y=5

x+y=4

Solution: 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=\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 4 \end{array}\right]

Hence,

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

So, A is non-singular.

Therefore, A^{-1} exists.

Hence, the given system of equations is consistent.

Question 3: Examine the consistency of the system of equations:

x+3 y=5
2 x+6 y=8

Solution: 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=\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 8 \end{array}\right]

Hence,

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

So, A is a singular matrix.

Now,

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

Therefore,

(\operatorname{adj} A) B =\begin{bmatrix} 6 & -5 \\ -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 0

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

Hence, the system of equations is inconsistent.

Question 4: Examine the consistency of the system of equations:

x+y+z=1
2 x+3 y+2 z=2
a x+a y+2 a z=4

Solution: x+y+z=1
2 x+3 y+2 z=2

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=\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2 a \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } \quad B=\left[\begin{array}{l} 1 \\ 2 \\ 4 \end{array}\right]

Hence,

|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

So, A is non-singular.

Therefore, A^{-1} exists.

Thus, the given system of equations is consistent.

Question 5:Examine the consistency of the system of equations:

3 x-y-2 z=2
2 y-z=-1
3 x-5 y=3

Solution:The given system of equations is:

3 x-y-2 z=2
2 y-z=-1
3 x-5 y=3

The given 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=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]

Hence,

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

So, A is a singular matrix.

Now,

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

Therefore,

(\text { adji }) B =\begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}\left[\begin{array}{c} 2 \\ -1 \\ 3 \end{array}\right]

\Rightarrow \left[\begin{array}{c} -10-10+15 \\ -6-6+9 \\ -12-12+18 \end{array}\right] \\ =\left[\begin{array}{l} -5 \\ -3 \\ -6 \end{array}\right] \\ \neq 0

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:

5 x-y+4 z=5

2 x+3 y+5 z=2

5 x-2 y+6 z=-1

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

The given 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=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 5 \\ 2 \\ -1 \end{array}\right]

Hence,

|A| =5(18+10)+1(12-25)+4(-4-15)
=5(28)+1(-13)+4(-19)
=140-13-76
=51 \neq 0

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

Solution: 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=\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{c} 4 \\ 5 \end{array}\right]

Hence,

|A| =15-14
=1 \neq 0

So, A is non-singular.

Therefore, A^{-1} exists.

Now,

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

=\begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}

Then,

\Rightarrow X=A^{-1} B

\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{cc} 3 & -2 \\ -7 & 5 \end{array}\right]\left[\begin{array}{l} 4 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 12-10 \\ -28+25 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 2 \\ -3 \end{array}\right]

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

Solution: 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=\begin{bmatrix}2 & -1 \\ 3 & 4\end{bmatrix}, X=\left[\begin{array}{l}x \\ y\end{array}\right] and B=\left[\begin{array}{c}-2 \\ 3\end{array}\right]

Hence,

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

So, A is non-singular.

Therefore, A^{-1} exists.

Now,

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

=\frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}

Therefore,

\Rightarrow X=A^{-1} B

\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix}=\frac{1}{11}\begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}\left[\begin{array}{c} -2 \\ 3 \end{array}\right]

\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{11}\left[\begin{array}{c} -8+3 \\ 6+6 \end{array}\right]

\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{11}\left[\begin{array}{l} -5 \\ 12 \end{array}\right]

\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} \frac{-5}{11} \\ \frac{12}{11} \end{array}\right]

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

Question 9: Solve system of linear equations, using matrix method.

4 x-3 y=3

3 x-5 y=7

Solution: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=\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}, X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 3 \\ 7 \end{array}\right]

Hence,

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

So, A is nonsingular. Therefore, A^{-1} exists.

Now,

A^{-1} =\frac{1}{|A|}(\operatorname{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,

\Rightarrow X=A^{-1} B

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

=\frac{1}{11}\begin{bmatrix}5 & -3 \\3 & -4\end{bmatrix}\left[\begin{array}{l}3 \\7\end{array}\right]

\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\frac{1}{11}\left[\begin{array}{rr}5 & -3 \\3 & -4\end{array}\right]\left[\begin{array}{l}3 \\7\end{array}\right]

\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}15-21 \\9-28\end{array}\right]

\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\frac{1}{11}\left[\begin{array}{c}-6 \\-19\end{array}\right]

\Rightarrow\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{c}\frac{-6}{11} \\\frac{-19}{11}\end{array}\right]

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

Solution: 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}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]

Hence,

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

So, A is non-singular.

Therefore, A^{-1} exists.

Now,

A^{-1} =\frac{1}{|A|}(\text { adjA })

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

Therefore,

\Rightarrow X=A^{-1} B

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

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

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

\Rightarrow\left[\begin{array}{l} x \\ y \end{array}\right]=\frac{1}{4}\left[\begin{array}{l} -4 \\ 16 \end{array}\right]

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

Hence, x=-1 and y=4

Question 11: Solve system of linear equations, using matrix method.

2 x+y+z=1
x-2 y-z=\frac{3}{2}
3 y-5 z=9

Solution:The given system of equations is:
2 x+y+z=1
x-2 y-z=\frac{3}{2}
3 y-5 z=9

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} \text { and } B=\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}

Hence,

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

So, A is non-singular.

Therefore, A^{-1} exists.

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

Hence,

Therefore,

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

=\frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}

\Rightarrow X=A^{-1} B

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}\begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 13+12+9 \\ 5-15+27 \\ 3-9-45 \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{34}\begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} \frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}

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

Solution:The given system of equations is:
x-y+z=4
2 x+y-3 z=0
x+y+z=2

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} \text { and } B=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}

Hence,

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

So, A is nonsingular.

Therefore, A^{-1} exists.

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

Hence,

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

=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}

Therefore,

\Rightarrow X=A^{-1} B

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix}{l} 4 \\ 0 \\ 2 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\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}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{10}\begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}

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

Solution:The given system of equations is:

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

The given system of equations can be written in the form of 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} \text { and } B=\begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}

Hence,

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

So, A is non-singular.

Therefore, A^{-1} 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

Hence,

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

=\frac{1}{40}\begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}

Therefore,

\Rightarrow X=A^{-1} B

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\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} 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}.

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{40}\begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}

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

Solution:The given system of equations is:
x-y+2 z=7
3 x+4 y-5 z=-5
2 x-y+3 z=12

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}

Hence,

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

So, A is non-singular.

Therefore, A^{-1} exists

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

Hence,

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

=\frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}

Therefore,

\Rightarrow X=A^{-1} B
\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\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} 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}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}

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

Question 15: A=\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4 \\1 & 1 & -2\end{bmatrix} \text {, find } A^{-1} \text {. Using } A^{-1} \text { solve the system of equations }
2 x-3 y+5 z=11
3 x+2 y-4 z=-5
x+y-2 z=-3

Solution: It is given that 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

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

Hence,

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

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}

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

Therefore,

\Rightarrow X=A^{-1} B

\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}

\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}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}

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.

Solution: Let the cost of onions, wheat, and rice per \mathrm{kg} in ₹ be x, y and 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

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

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}

Therefore,

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

So, A is non-singular.

Therefore, A^{-1} exists.

Now,

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

Therefore,

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

=\frac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}

Hence,

\Rightarrow X=A^{-1} B

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\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 & -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}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\frac{1}{50}\begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix}

\Rightarrow\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}

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.

 

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