Exercise 4.5, Question 1
\(x + 2y = 2,\ 2x + 3y = 3\)
Show solution
- \[\begin{aligned}|A| &= \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} \\ &= -1 \ne 0\end{aligned}\]
Answer: Consistent (unique solution)
NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5
Exercise 4.5: Solving systems of linear equations. Write the system as \(AX = B\). If \(|A| \ne 0\) there is exactly one solution, \(X = A^{-1}B\) (consistent). If \(|A| = 0\), work out \((\operatorname{adj} A)B\): non-zero means no solution (inconsistent); zero means either infinitely many solutions or none, which must then be checked directly.
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Exercise 4.5 questions and solutions
Exercise 4.5, Question 2
\(2x - y = 5,\ x + y = 4\)
Show solution
- \(|A| = 2 + 1 = 3 \ne 0\).
Answer: Consistent (unique solution)
Exercise 4.5, Question 3
\(x + 3y = 5,\ 2x + 6y = 8\)
Show solution
- \(|A| = 6 - 6 = 0\).
- \[\begin{aligned}(\operatorname{adj} A)B &= \begin{bmatrix}6 & -3 \\ -2 & 1\end{bmatrix}\begin{bmatrix}5 \\ 8\end{bmatrix} \\ &= \begin{bmatrix}6 \\ -2\end{bmatrix} \ne O\end{aligned}\]: no solution. (Indeed \(2x + 6y\) would have to be both 10 and 8.)
Answer: Inconsistent
Exercise 4.5, Question 4
\(x + y + z = 1,\ 2x + 3y + 2z = 2,\ ax + ay + 2az = 4\)
Show solution
- \[\begin{aligned}|A| &= \begin{vmatrix}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a\end{vmatrix} \\ &= 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) \\ &= a\end{aligned}\]
- \(a \ne 0\): \(|A| \ne 0\), a unique solution. The textbook's intended answer is 'consistent' (it assumes \(a \ne 0\)).
Answer: Consistent for \(a \ne 0\) (unique solution); for \(a = 0\) the third equation reads \(0 = 4\), so the system is inconsistent
Exercise 4.5, Question 5
\(3x - y - 2z = 2,\ 2y - z = -1,\ 3x - 5y = 3\)
Show solution
- \[\begin{aligned}|A| &= 3(0 - 5) + 1(0 + 3) - 2(0 - 6) \\ &= -15 + 3 + 12 \\ &= 0\end{aligned}\]
- \[\begin{aligned}(\operatorname{adj} A)B &= \begin{bmatrix}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 3\end{bmatrix} \\ &= \begin{bmatrix}-5 \\ -3 \\ -6\end{bmatrix} \ne O\end{aligned}\]
Answer: Inconsistent
Exercise 4.5, Question 6
\(5x - y + 4z = 5,\ 2x + 3y + 5z = 2,\ 5x - 2y + 6z = -1\)
Show solution
- \[\begin{aligned}|A| &= 5(18 + 10) + 1(12 - 25) + 4(-4 - 15) \\ &= 140 - 13 - 76 \\ &= 51 \ne 0\end{aligned}\]
Answer: Consistent (unique solution)
Exercise 4.5, Question 7
\(5x + 2y = 4,\ 7x + 3y = 5\)
Show solution
- \[A = \begin{bmatrix}5 & 2 \\ 7 & 3\end{bmatrix}\], \(|A| = 1\), \[A^{-1} = \begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}\]
- \[\begin{aligned}X &= A^{-1}B \\ &= \begin{bmatrix}3 & -2 \\ -7 & 5\end{bmatrix}\begin{bmatrix}4 \\ 5\end{bmatrix} \\ &= \begin{bmatrix}2 \\ -3\end{bmatrix}\end{aligned}\]
Answer: \(x = 2,\ y = -3\)
Exercise 4.5, Question 8
\(2x - y = -2,\ 3x + 4y = 3\)
Show solution
- \(|A| = 8 + 3 = 11\), \[A^{-1} = \tfrac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac{1}{11}\begin{bmatrix}4 & 1 \\ -3 & 2\end{bmatrix}\begin{bmatrix}-2 \\ 3\end{bmatrix} \\ &= \tfrac{1}{11}\begin{bmatrix}-5 \\ 12\end{bmatrix}\end{aligned}\]
Answer: \(x = -\tfrac{5}{11},\ y = \tfrac{12}{11}\)
Exercise 4.5, Question 9
\(4x - 3y = 3,\ 3x - 5y = 7\)
Show solution
- \(|A| = -20 + 9 = -11\), \[A^{-1} = -\tfrac{1}{11}\begin{bmatrix}-5 & 3 \\ -3 & 4\end{bmatrix}\]
- \[\begin{aligned}X &= -\tfrac{1}{11}\begin{bmatrix}-5 & 3 \\ -3 & 4\end{bmatrix}\begin{bmatrix}3 \\ 7\end{bmatrix} \\ &= -\tfrac{1}{11}\begin{bmatrix}6 \\ 19\end{bmatrix}\end{aligned}\]
Answer: \(x = -\tfrac{6}{11},\ y = -\tfrac{19}{11}\)
Exercise 4.5, Question 10
\(5x + 2y = 3,\ 3x + 2y = 5\)
Show solution
- \(|A| = 10 - 6 = 4\), \[A^{-1} = \tfrac14\begin{bmatrix}2 & -2 \\ -3 & 5\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac14\begin{bmatrix}2 & -2 \\ -3 & 5\end{bmatrix}\begin{bmatrix}3 \\ 5\end{bmatrix} \\ &= \tfrac14\begin{bmatrix}-4 \\ 16\end{bmatrix}\end{aligned}\]
Answer: \(x = -1,\ y = 4\)
Exercise 4.5, Question 11
\(2x + y + z = 1,\ x - 2y - z = \tfrac32,\ 3y - 5z = 9\)
Show solution
- \[A = \begin{bmatrix}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{bmatrix}\], \[\begin{aligned}|A| &= 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) \\ &= 34\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac{1}{34}\begin{bmatrix}13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5\end{bmatrix}\begin{bmatrix}1 \\ \frac{3}{2} \\ 9\end{bmatrix} \\ &= \tfrac{1}{34}\begin{bmatrix}34 \\ 17 \\ -51\end{bmatrix}\end{aligned}\]
Answer: \(x = 1,\ y = \tfrac12,\ z = -\tfrac32\)
Exercise 4.5, Question 12
\(x - y + z = 4,\ 2x + y - 3z = 0,\ x + y + z = 2\)
Show solution
- \[\begin{aligned}|A| &= 1(1 + 3) + 1(2 + 3) + 1(2 - 1) \\ &= 10\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac{1}{10}\begin{bmatrix}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{bmatrix}\begin{bmatrix}4 \\ 0 \\ 2\end{bmatrix} \\ &= \tfrac{1}{10}\begin{bmatrix}20 \\ -10 \\ 10\end{bmatrix}\end{aligned}\]
Answer: \(x = 2,\ y = -1,\ z = 1\)
Exercise 4.5, Question 13
\(2x + 3y + 3z = 5,\ x - 2y + z = -4,\ 3x - y - 2z = 3\)
Show solution
- \[\begin{aligned}|A| &= 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) \\ &= 10 + 15 + 15 \\ &= 40\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac{1}{40}\begin{bmatrix}5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7\end{bmatrix}\begin{bmatrix}5 \\ -4 \\ 3\end{bmatrix} \\ &= \tfrac{1}{40}\begin{bmatrix}40 \\ 80 \\ -40\end{bmatrix}\end{aligned}\]
Answer: \(x = 1,\ y = 2,\ z = -1\)
Exercise 4.5, Question 14
\(x - y + 2z = 7,\ 3x + 4y - 5z = -5,\ 2x - y + 3z = 12\)
Show solution
- \[\begin{aligned}|A| &= 1(12 - 5) + 1(9 + 10) + 2(-3 - 8) \\ &= 7 + 19 - 22 \\ &= 4\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac14\begin{bmatrix}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{bmatrix}\begin{bmatrix}7 \\ -5 \\ 12\end{bmatrix} \\ &= \tfrac14\begin{bmatrix}8 \\ 4 \\ 12\end{bmatrix}\end{aligned}\]
Answer: \(x = 2,\ y = 1,\ z = 3\)
Exercise 4.5, Question 15
\(A = \begin{bmatrix}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{bmatrix}\). Find \(A^{-1}\) and use it to solve \(2x - 3y + 5z = 11,\ 3x + 2y - 4z = -5,\ x + y - 2z = -3\).
Show solution
- \[\begin{aligned}|A| &= 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) \\ &= 0 - 6 + 5 \\ &= -1\end{aligned}\]
- \[\operatorname{adj} A = \begin{bmatrix}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{bmatrix}\], so \(A^{-1} = -\operatorname{adj} A\).
- \[\begin{aligned}X &= A^{-1}\begin{bmatrix}11 \\ -5 \\ -3\end{bmatrix} \\ &= -\begin{bmatrix}-1 \\ -2 \\ -3\end{bmatrix}\end{aligned}\]
Answer: \[A^{-1} = \begin{bmatrix}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{bmatrix}\]; \(x = 1,\ y = 2,\ z = 3\)
Exercise 4.5, Question 16
4 kg onion, 3 kg wheat and 2 kg rice cost Rs 60; 2 kg onion, 4 kg wheat and 6 kg rice cost Rs 90; 6 kg onion, 2 kg wheat and 3 kg rice cost Rs 70. Find the price per kg of each by the matrix method.
Show solution
- Prices x, y, z (Rs/kg): \(4x + 3y + 2z = 60\), \(2x + 4y + 6z = 90\), \(6x + 2y + 3z = 70\).
- \[\begin{aligned}|A| &= 4(12 - 12) - 3(6 - 36) + 2(4 - 24) \\ &= 0 + 90 - 40 \\ &= 50\end{aligned}\]; \[\operatorname{adj} A = \begin{bmatrix}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{bmatrix}\]
- \[\begin{aligned}X &= \tfrac{1}{50}\begin{bmatrix}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{bmatrix}\begin{bmatrix}60 \\ 90 \\ 70\end{bmatrix} \\ &= \tfrac{1}{50}\begin{bmatrix}250 \\ 400 \\ 400\end{bmatrix}\end{aligned}\]
Answer: Onion Rs 5/kg, wheat Rs 8/kg, rice Rs 8/kg
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