NCERT Solutions Class-12-Chapter-4-Determinants
Excercise-4.5
Note: A system of equations \( AX = B \) can be solved using matrices.
1. If \( |A| \neq 0 \), the system is consistent and has a unique solution given by \( X = A^{-1}B \).
2. If \( |A| = 0 \) and \( (\text{adj } A)B \neq O \), the system is inconsistent (no solution).
3. If \( |A| = 0 \) and \( (\text{adj } A)B = O \), the system may be consistent (infinitely many solutions) or inconsistent.
Q1
Examine the consistency of the system of equations:
\( x + 2y = 2 \)
\( 2x + 3y = 3 \)
\( x + 2y = 2 \)
\( 2x + 3y = 3 \)
Solution:
The system can be written as \( AX = B \), where \( A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \).
\( |A| = 1(3) - 2(2) = 3 - 4 = -1 \neq 0 \).
Since \( |A| \neq 0 \), the system is consistent and has a unique solution.
Q2
Examine the consistency of the system of equations:
\( 2x - y = 5 \)
\( x + y = 4 \)
\( 2x - y = 5 \)
\( x + y = 4 \)
Solution:
\( A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \).
\( |A| = 2(1) - (-1)(1) = 2 + 1 = 3 \neq 0 \).
Therefore, the system is consistent.
Q3
Examine the consistency of the system of equations:
\( x + 3y = 5 \)
\( 2x + 6y = 8 \)
\( x + 3y = 5 \)
\( 2x + 6y = 8 \)
Solution:
\( A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \).
\( |A| = 6 - 6 = 0 \). The system may be inconsistent.
Check \( (\text{adj } A)B \). \( \text{adj } A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \).
\( B = \begin{bmatrix} 5 \\ 8 \end{bmatrix} \).
\[ (\text{adj } A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq O \]
Therefore, the system is inconsistent.
Q4
Examine the consistency of the system of equations:
\( x + y + z = 1 \)
\( 2x + 3y + 2z = 2 \)
\( ax + ay + 2az = 4 \)
\( x + y + z = 1 \)
\( 2x + 3y + 2z = 2 \)
\( ax + ay + 2az = 4 \)
Solution:
\( A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix} \).
\[ \begin{aligned} |A| &= 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) \\ &= 4a - 2a - a = a \end{aligned} \]
If \( a \neq 0 \), the system is consistent.
Q5
Examine the consistency of the system:
\( 3x - y - 2z = 2 \)
\( 2y - z = -1 \)
\( 3x - 5y = 3 \)
\( 3x - y - 2z = 2 \)
\( 2y - z = -1 \)
\( 3x - 5y = 3 \)
Solution:
\( A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix} \).
\( |A| = 3(0-5) - (-1)(0-(-3)) + (-2)(0-6) = -15 + 3 + 12 = 0 \).
Calculate \( (\text{adj } A)B \). \( B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix} \).
Adjoint calculation yields non-zero vector product. Thus, inconsistent.
Q6
Examine the consistency of the system:
\( 5x - y + 4z = 5 \)
\( 2x + 3y + 5z = 2 \)
\( 5x - 2y + 6z = -1 \)
\( 5x - y + 4z = 5 \)
\( 2x + 3y + 5z = 2 \)
\( 5x - 2y + 6z = -1 \)
Solution:
\( |A| = 5(18+10) - (-1)(12-25) + 4(-4-15) = 140 - 13 - 76 = 51 \neq 0 \).
The system is consistent.
Q7
Solve system of linear equations using matrix method:
\( 5x + 2y = 4 \)
\( 7x + 3y = 5 \)
\( 5x + 2y = 4 \)
\( 7x + 3y = 5 \)
Solution:
\( A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \).
\( |A| = 15 - 14 = 1 \neq 0 \). So \( A^{-1} \) exists.
\( A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \).
\[ X = A^{-1}B = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \]
So, \( x = 2, y = -3 \).
Q8
Solve system of linear equations using matrix method:
\( 2x - y = -2 \)
\( 3x + 4y = 3 \)
\( 2x - y = -2 \)
\( 3x + 4y = 3 \)
Solution:
\( |A| = 8 - (-3) = 11 \).
\( A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \).
\[ X = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ 3 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -8+3 \\ 6+6 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -5 \\ 12 \end{bmatrix} \]
So, \( x = \frac{-5}{11}, y = \frac{12}{11} \).
Q11
Solve system of linear equations using matrix method:
\( 2x + y + z = 1 \)
\( x - 2y - z = \frac{3}{2} \)
\( 3y - 5z = 9 \)
\( 2x + y + z = 1 \)
\( x - 2y - z = \frac{3}{2} \)
\( 3y - 5z = 9 \)
Solution:
Rewrite second eq: \( 2x - 4y - 2z = 3 \).
\( A = \begin{bmatrix} 2 & 1 & 1 \\ 2 & -4 & -2 \\ 0 & 3 & -5 \end{bmatrix}, B = \begin{bmatrix} 1 \\ 3 \\ 9 \end{bmatrix} \).
\( |A| = 2(20+6) - 1(-10-0) + 1(6-0) = 52 + 10 + 6 = 68 \).
Find \( \text{adj } A \), then \( A^{-1} \), then \( X = A^{-1}B \).
Result: \( x = 1, y = \frac{1}{2}, z = -\frac{3}{2} \).
Q15
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:
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
Solution:
First, find \( |A| \):
\( |A| = 2(0) - (-3)(-2) + 5(1) = -6 + 5 = -1 \).
Calculate Cofactors and Adjoint:
\( \text{adj } A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
\( A^{-1} = \frac{1}{-1} (\text{adj } A) = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \).
The system is \( AX = B \) with \( B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \).
\[ X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]
Thus, \( x = 1, y = 2, z = 3 \).
Q16
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 cost of onion, wheat, rice per kg be \( x, y, z \) respectively.
System of equations:
- \( 4x + 3y + 2z = 60 \)
- \( 2x + 4y + 6z = 90 \Rightarrow x + 2y + 3z = 45 \)
- \( 6x + 2y + 3z = 70 \)
Matrix form \( AX = B \):
\( A = \begin{bmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{bmatrix}, B = \begin{bmatrix} 60 \\ 45 \\ 70 \end{bmatrix} \).
\( |A| = 4(0) - 3(-15) + 2(-10) = 45 - 20 = 25 \).
Find \( X = A^{-1}B \).
Solving yields \( x = 5, y = 8, z = 8 \).
Cost: Onion ₹5/kg, Wheat ₹8/kg, Rice ₹8/kg.