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Q14:
Solve the system of linear equations, using matrix method: $x – y + 2z = 7$, $3x + 4y – 5z = – 5$, $2x – y + 3z = 12$
Solution :
Step 1: Matrix Representation of the System
Given the system of linear equations:
- $x - y + 2z = 7$
- $3x + 4y - 5z = -5$
- $2x - y + 3z = 12$
The system can be represented as $AX = B$, where:
$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$Step 2: Verification of Non-Singularity
We calculate the determinant $|A|$ to determine invertibility [Per the Invertibility Criterion for Square Matrices]:
$|A| = 1(4 \cdot 3 - (-5)(-1)) - (-1)(3 \cdot 3 - (-5)(2)) + 2(3(-1) - 4 \cdot 2)$ $|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)$ $|A| = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4$Since $|A| = 4 \neq 0$, $A^{-1}$ exists, and the system has a unique solution.
Step 3: Determination of the Adjoint Matrix
To find $A^{-1}$, we compute the cofactor matrix $C_{ij}$.
$C_{11} = +(12-5) = 7, \quad C_{12} = -(9+10) = -19, \quad C_{13} = +(-3-8) = -11$ $C_{21} = -(-3+2) = 1, \quad C_{22} = +(3-4) = -1, \quad C_{23} = -(-1+2) = -1$ $C_{31} = +(5-8) = -3, \quad C_{32} = -(-5-6) = 11, \quad C_{33} = +(4+3) = 7$The Adjoint matrix $Adj(A)$ is the transpose of the cofactor matrix:
$Adj(A) = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$Step 4: Calculation of Inverse and Solution Matrix
Per $A^{-1} = \frac{1}{|A|} Adj(A)$:
$A^{-1} = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$The solution is $X = A^{-1}B$:
$X = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$ $x = \frac{1}{4} [7(7) + 1(-5) + (-3)(12)] = \frac{1}{4} [49 - 5 - 36] = \frac{8}{4} = 2$ $y = \frac{1}{4} [-19(7) + (-1)(-5) + 11(12)] = \frac{1}{4} [-133 + 5 + 132] = \frac{4}{4} = 1$ $z = \frac{1}{4} [-11(7) + (-1)(-5) + 7(12)] = \frac{1}{4} [-77 + 5 + 84] = \frac{12}{4} = 3$Visual Representation of System Geometry
Final Solution: x = 2, y = 1, z = 3
More Questions from Class 12 Mathematics Determinants EXERCISE 4.5
- Q1: Examine the consistency of the system of equations: $x + 2y = 2$, $2x + 3y = 3$
- Q2: Examine the consistency of the system of equations: $2x – y = 5$, $x + y = 4$
- Q3: Examine the consistency of the system of equations: $x + 3y = 5$, $2x + 6y = 8$
- Q4: Examine the consistency of the system of equations: $x + y + z = 1$, $2x + 3y + 2z = 2$, $ax + ay + 2az = 4$
- Q5: Examine the consistency of the system of equations: $3x–y – 2z = 2$, $2y – z = –1$, $3x – 5y = 3$
- Q6: Examine the consistency of the system of equations: $5x – y + 4z = 5$, $2x + 3y + 5z = 2$, $5x – 2y + 6z = –1$
- Q7: Solve the system of linear equations, using matrix method: $5x + 2y = 4$, $7x + 3y = 5$
- Q8: Solve the system of linear equations, using matrix method: $2x – y = –2$, $3x + 4y = 3$
- Q9: Solve the system of linear equations, using matrix method: $4x – 3y = 3$, $3x – 5y = 7$
- Q10: Solve the system of linear equations, using matrix method: $5x + 2y = 3$, $3x + 2y = 5$
- Q11: Solve the system of linear equations, using matrix method: $2x + y + z = 1$, $x – 2y – z = \frac{3}{2}$, $3y – 5z = 9$
- Q12: Solve the system of linear equations, using matrix method: $x – y + z = 4$, $2x + y – 3z = 0$, $x + y + z = 2$
- Q13: Solve the system of linear equations, using matrix method: $2x + 3y +3 z = 5$, $x – 2y + z = – 4$, $3x – y – 2z = 3$
- 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 $2x – 3y + 5z = 11$, $3x + 2y – 4z = – 5$, $x + y – 2z = – 3$
- 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.
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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