NCERT Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.6

Jolly September 06, 2022

NCERT Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.6

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-4 (Determinants)Exercise 4.6 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of  NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-4 (Determinants)Exercise 4.6

Exercise 4.6 

Set 1

Examine the consistency of the system of equations in Exercises 1 to 6.

Question 1. x + 2y = 2

                            2x + 3y = 3

Solution:

Matrix form of the given equations is AX = B

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

∴  \begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}2 \\3 \\\end{bmatrix}

Now, |A| = {\begin{vmatrix}1&2\\2&3\end{vmatrix}} = 3-4 = -1 ≠ 0

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 2. 2x – y = 5 

                            x + y = 4

Solution:

Matrix form of the given equations is AX = B

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

∴\begin{bmatrix}2 & -1 \\1 & 1 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}5 \\4 \\\end{bmatrix}

Now, |A| ={\begin{vmatrix}2&-1\\1&1\end{vmatrix}} = 2-(-1) = 3 ≠ 0

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 3. x + 3y = 5 

                            2x + 6y = 8

Solution:

Matrix form of the given equations is AX = B

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

∴ \begin{bmatrix}1 & 3 \\2 & 6 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}5 \\8 \\\end{bmatrix}

Now, |A| ={\begin{vmatrix}1&3\\2&6\end{vmatrix}} = 6-6=0

And, 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} ≠0

∵ Have no common solution.

∴ System of equation is inconsistent.

Question 4. x + y + z = 1 

                            2x + 3y + 2z = 2 

                            ax + ay + 2az = 4

Solution:

Matrix form of the given equations is AX = B

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

∴  \begin{bmatrix}1 & 1 & 1\\2 & 3 & 2\\a & a & 2a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}

Now, |A| = {\begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}} = 1(6a-2a)-1(4a-2a)+1(2a-3a)=4a-2a-a=a≠0

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Question 5. 3x – y – 2z = 2 

                            2y – z = -1 

                            3x – 5y = 3 

Solution:

Matrix form of the given equations is AX = B

where, A ={\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}}, B=\begin{bmatrix}2\\-1\\3\end{bmatrix}and, X =\begin{bmatrix}x\\y\\z\end{bmatrix}

∴{\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}

Now, |A| ={\begin{vmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{vmatrix}} = 3(0-5)-(-1)(0+3)+(-2)(0-6)=3(-5)+3+12=-15+15=0

And, adj. A =\begin{bmatrix}-5 & 10 & 5\\-3 & 6 & 3\\-6 & 12 & 6\end{bmatrix}

∴ (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} ≠0

∴ System of equation is inconsistent.

Question 6. 5x – y + 4z = 5

                            2x + 3y + 5z = 2 

                             5x – 2y + 6z = –1

Solution:

Matrix form of the given equations is AX = B

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

∴\begin{bmatrix}5 & -1 & 4\\2 & 3 & 5\\5 & -2 & 6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\2\\-1\end{bmatrix}

Now, |A| =\begin{vmatrix}5 & -1 & 4\\2 & 3 & 5\\5 & -2 & 6\end{vmatrix}=5(8+10)-(-1)(12-25)+4(-4-15)=140-13-76=140-89=51≠0

∵ Inverse of matrix exists, unique solution.

∴ System of equation is consistent.

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7. 5x + 2y = 4 

                            7x + 3y = 5

Solution:

Matrix form of the given equations is AX = B

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

∴\begin{bmatrix}5 &2 \\7 & 3 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}4 \\5 \\\end{bmatrix}

Now, |A|=\begin{vmatrix}5 &2 \\7 & 3 \\\end{vmatrix}=15-14=1 ≠0

∴Unique solution

Now, X = A-1B =\frac{1}{|A|}(adj.A)B

\begin{bmatrix}x\\y\\\end{bmatrix}=\frac{1}{1}\begin{bmatrix}3&-2\\-7&5\\\end{bmatrix}\begin{bmatrix}4\\5\\\end{bmatrix}=\begin{bmatrix}12-10 \\-28+25 \\\end{bmatrix}=\begin{bmatrix}2 \\- 3 \\\end{bmatrix}

Therefore, x=2 and y=-3

Question 8. 2x – y = -2 
                            3x + 4y = 3

Solution:

Matrix form of the given equations is AX = B

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

∴\begin{bmatrix}2 &-1 \\3 & 4 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}-2 \\3 \\\end{bmatrix}

Now, |A|=\begin{vmatrix}2 &-1 \\3 & 4 \\\end{vmatrix}=8-(-3)=8+3=11≠0

∴Unique solution

Now, X = A-1B \frac{1}{|A|}(adj.A)B

\begin{bmatrix}x\\y\\\end{bmatrix}=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\\\end{bmatrix}\begin{bmatrix}-2\\3\\\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-8+3 \\6+6 \\\end{bmatrix}=\begin{bmatrix}-5/11 \\12/11 \\\end{bmatrix}

Therefore, x=-5/11 and y=12/11

Question 9. 4x – 3y = 3

                            3x – 5y = 7

Solution:

Matrix form of the given equations is AX = B

where, A=\begin{bmatrix}4 &-3 \\3 & -5 \\\end{bmatrix}, B=\begin{bmatrix}3 \\7 \\\end{bmatrix}, X=\begin{bmatrix}x\\y\\\end{bmatrix}

∴\begin{bmatrix}4 &-3 \\3 & -5 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}3 \\7 \\\end{bmatrix}

Now, |A|=\begin{vmatrix}4 &-3 \\3 & -5 \\\end{vmatrix}=-20-(-9)=-20+9=-11≠0

∴Unique solution\begin{vmatrix}4 &-3 \\3 & -5 \\\end{vmatrix}=-20-(-9)=-20+9=-11≠0n

Now, X =A-1B \frac{1}{|A|}A(adj.A)B

\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}=\begin{bmatrix}-6/11 \\-19/11 \\\end{bmatrix}

Therefore, x= -6/11 and y= -19/11

Question 10. 5x + 2y = 3 
                               3x + 2y = 5

Solution:

Matrix form of the given equations is AX = B

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

∴\begin{bmatrix}5 &2 \\3 & 2 \\\end{bmatrix}\begin{bmatrix}x\\y\\\end{bmatrix}=\begin{bmatrix}3 \\5 \\\end{bmatrix}

Now, |A|=\begin{vmatrix}5 &2 \\3 & 2 \\\end{vmatrix}=10-6=4≠0

∴Unique solution

Now, X = A-1B\frac{1}{|A|}A(adj.A)B

\begin{bmatrix}x\\y\\\end{bmatrix}=\frac{1}{4}\begin{bmatrix}2&-2\\-3&5\\\end{bmatrix}\begin{bmatrix}3\\5\\\end{bmatrix}=\frac{1}{4}\begin{bmatrix}6-10 \\-9+25 \\\end{bmatrix}=\begin{bmatrix}-1 \\4 \\\end{bmatrix}

Therefore, x= -1 and y= 4

Exercise 4.6 

Set 2

Question 11. 2x + y + z = 1 

                               x – 2y – z = 3/2

                               3y – 5z = 9 

Solution:

Matrix form of the given equation is AX = B

i.e.\begin{bmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\3/2\\9\end{bmatrix}

∴ |A| =\begin{vmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{vmatrix}=2(10+3)-1(-5-0)+1(3-0)=26+5+3≠0

∴ Solution is unique.

Now, X = A-1B = \frac{1}{|A|}(adj.A)B

\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\\3/2\\ 9\end{bmatrix} =\frac{1}{34}\begin{bmatrix}13+12+9\\5-15+27\\3-9-45\end{bmatrix}=\frac{1}{34}\begin{bmatrix}34\\17\\-51\end{bmatrix}=\begin{bmatrix}1\\1/2\\-3/2\end{bmatrix}

Therefore, x=1, y=1/2, z=3/2

Question 12.  x – y + z = 4

                                2x + y – 3z = 0

                                x + y + z = 2

Solution:

Matrix form of the given equation is AX = B

i.e\begin{bmatrix}1 & -1 & 1\\2 & 1 & -3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}

∴ |A| =\begin{vmatrix}1 & -1 & 1\\2 & 1 & -3\\1 & 1 & 1\end{vmatrix}=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10≠0

∴ Solution is unique.

Now, X = A-1B =\frac{1}{|A|} (adj.A)B

\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} =\frac{1}{10}\begin{bmatrix}16+0+4\\-20+0+10\\4-0+6\end{bmatrix}=\frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix}=\begin{bmatrix}2\\-1\\ 1\end{bmatrix}

Therefore, x = 2, y = -1, z = 1

Question 13. 2x + 3y +3 z = 5 

                               x – 2y + z = – 4 

                               3x – y – 2z = 3

Solution:

Matrix form of given equation is AX = B

i.e.\begin{bmatrix}2 & 3 & 3\\1 & -2 & -1\\3 & -1 & -2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-4\\3\end{bmatrix}

∴ |A| =\begin{vmatrix}2 & 3 & 3\\1 & -2 & -1\\3 & -1 & -2\end{vmatrix}=2(4+1)-3(-2-3)+3(-1+6)=10+15+15≠0

∴ Solution is unique.

Now, X = A-1B = \frac{1}{|A|}(adj.A)B

\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} =\frac{1}{10}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix}=\frac{1}{10}\begin{bmatrix}40\\80\\-40\end{bmatrix}=\begin{bmatrix}1\\2\\ -1\end{bmatrix}

Therefore, x = 1, y = 2, z = -1

Question 14. x – y + 2z = 7

                               3x + 4y – 5z = – 5

                               2x – y + 3z = 12

Solution:

Matrix form of given equation is AX = B

i.e.\begin{bmatrix}1 & -1 & 2\\3 & 4 & -5\\2 & -1 & 3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-5\\12\end{bmatrix}

∴ |A| =\begin{vmatrix}1 & -1 & 2\\3 & 4 & -5\\2 & -1 & 3\end{vmatrix}=1(12-5)-(-1)(9+10)+2(-3-8)=7+9-22=4≠0

∴ Solution is unique.

Now, X = A-1B =\frac{1}{|A|} (adj.A)B

\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} =\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\ 3\end{bmatrix}

Therefore, x = 2, y = 1, z = 3

Question 15. If A=Solutions Class 12 maths Chapter-4 (Determinants), find A–1. Using A–1 solve the system of equations 

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

Solution:

Given: A=\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}

Now, |A|= \begin{vmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{vmatrix}

∴ |A|= 2(-4+4)-(-3)(-6+4)+5(3-2)=0-6+5= -1≠0

Means, A-1 exists.

And A-1 =\frac{1}{|A|}(adj.A)……(1)

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,

∴ adj. A =\begin{bmatrix}0 & -1 & 2\\2 & -9 & 23\\1 & -5 & 13\end{bmatrix}

From eq. (1),

A-1=\frac{1}{-1}\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, Matrix form of given equation is AX = B

i.e.\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\3\end{bmatrix}

∵ Solution is unique.

∴ X=A-1B

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

Therefore, x = 1, y = 2, z = 3

Question 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is 60 rupees. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is 90 rupees. The cost of 6 kg onion 2 kg wheat and 3 kg rice is 70 rupees. Find cost of each item per kg by matrix method.

Solution:

Let Rs x, Rs y, Rs z per kg be the prices of onion, wheat and rice respectively.

A.T.Q.

4x+3y+2z=60

2x+4y+6z=90

6x+2y+3z=70

Matrix form of equation is AX = B

where, A=\begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix},B=\begin{bmatrix}60\\90\\70\end{bmatrix}and X=\begin{bmatrix}x\\y\\z\end{bmatrix}

=> \begin{bmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}

Now, |A|=\begin{vmatrix}4 & 3 & 2\\2 & 4 & 6\\6 & 2 & 3\end{vmatrix}=4(12-12)-3(6-36)+2(4-24)=0+90-40=50 ≠0

∴ Solution is unique

Now, X=A-1B=\frac{1}{|A|}(adj. A)B……(1)

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,

∴ (adj.A)= \begin{bmatrix}0 & -5 & 10\\30 & 0 & -20\\-20 & 10 & 10\end{bmatrix}

From eqn.(1)

\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} =\frac{1}{50}\begin{bmatrix}-450+700\\1800-1400\\-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, z = 8

Hence, the cost of onion, wheat and rice are Rs. 5, Rs 8 and Rs 8 per kg.

Chapter-4 (Determinants)

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