Exercise 4.5 Practice
Application of Matrices: Solving Systems of Linear Equations
Q1
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Examine the consistency of the system of equations:
\( x + 2y = 2 \)
\( 2x + 3y = 3 \)
\( x + 2y = 2 \)
\( 2x + 3y = 3 \)
Step 1: Write in Matrix Form
\( AX = B \), where \( A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} \), \( X = \begin{bmatrix} x \\ y
\end{bmatrix} \), \( B = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \).
Step 2: Find Determinant
\( |A| = 1(3) - 2(2) = 3 - 4 = -1 \).
Since \( |A| \ne 0 \), the system is consistent (Unique Solution).
Q2
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Solve the system of equations using matrix method:
\( 5x + 2y = 4 \)
\( 7x + 3y = 5 \)
\( 5x + 2y = 4 \)
\( 7x + 3y = 5 \)
Step 1: Matrix Form
\( A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, B = \begin{bmatrix} 4 \\ 5 \end{bmatrix} \).
Step 2: Find Inverse
\( |A| = 15 - 14 = 1 \).
\( A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \).
\( A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \).
Step 3: Solve X = A⁻¹B
\( X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix}
12-10 \\ -28+25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix} \).
\( x = 2, y = -3 \)
Q3
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Solve system of linear equations using matrix method:
\( 3x - 2y + 3z = 8 \)
\( 2x + y - z = 1 \)
\( 4x - 3y + 2z = 4 \)
\( 3x - 2y + 3z = 8 \)
\( 2x + y - z = 1 \)
\( 4x - 3y + 2z = 4 \)
Step 1: Determinant
\( |A| = 3(2-3) - (-2)(4+4) + 3(-6-4) \)
\( = -3 + 16 - 30 = -17 \ne 0 \). Inverse exists.
\( = -3 + 16 - 30 = -17 \ne 0 \). Inverse exists.
Step 2: Cofactors
\( A_{11}=-1, A_{12}=-8, A_{13}=-10 \)
\( A_{21}=-5, A_{22}=-6, A_{23}=1 \)
\( A_{31}=-1, A_{32}=9, A_{33}=7 \)
\( A_{21}=-5, A_{22}=-6, A_{23}=1 \)
\( A_{31}=-1, A_{32}=9, A_{33}=7 \)
Step 3: Adjoint & Inverse
\( \text{adj } A = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \).
\( A^{-1} = -\frac{1}{17} (\text{adj } A) \).
\( A^{-1} = -\frac{1}{17} (\text{adj } A) \).
Step 4: Solve
Multiply \( A^{-1} B \), where \( B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} \).
\( X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
\( X = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
\( x=1, y=2, z=3 \)
Q4
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Solve the following system of equations by matrix method:
\( x - y + z = 4 \)
\( 2x + y - 3z = 0 \)
\( x + y + z = 2 \)
\( x - y + z = 4 \)
\( 2x + y - 3z = 0 \)
\( x + y + z = 2 \)
Determinant
\( |A| = 1(1+3) - (-1)(2+3) + 1(2-1) \)
\( = 4 + 5 + 1 = 10 \ne 0 \).
\( = 4 + 5 + 1 = 10 \ne 0 \).
Adjoint A
\( \text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \).
Solution
\( X = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0
\\ 2 \end{bmatrix} \)
\( = \frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} \).
\( = \frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} \).
\( x=2, y=-1, z=1 \)
Q5
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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:
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
\( 2x - 3y + 5z = 11 \)
\( 3x + 2y - 4z = -5 \)
\( x + y - 2z = -3 \)
Find Inverse
\( |A| = -1 \).
\( \text{adj } A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
\( A^{-1} = -1 \times (\text{adj } A) \).
\( \text{adj } A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \).
\( A^{-1} = -1 \times (\text{adj } A) \).
Solve
The system coefficients match matrix A.
\( X = A^{-1} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \).
Calculations yield \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
\( X = A^{-1} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \).
Calculations yield \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \).
\( x=1, y=2, z=3 \)
Q6
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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 the cost of each item per kg by matrix
method.
Formulate Equations
Let cost of onion, wheat, rice be \( x, y, z \) respectively.
\( 4x + 3y + 2z = 60 \)
\( 2x + 4y + 6z = 90 \)
\( 6x + 2y + 3z = 70 \)
\( 4x + 3y + 2z = 60 \)
\( 2x + 4y + 6z = 90 \)
\( 6x + 2y + 3z = 70 \)
Matrix Method
\( A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix} \). Find \( A^{-1} \).
Solve
\( |A| = 50 \).
\( X = A^{-1} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} \).
Result: \( x=5, y=8, z=8 \).
\( X = A^{-1} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} \).
Result: \( x=5, y=8, z=8 \).
Onion: ₹5/kg, Wheat: ₹8/kg, Rice: ₹8/kg.