Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.5

Last Updated : 03 Apr, 2024

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

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

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= $\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}$ , 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.

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