Rbse Solutions Class 12 Maths: Exercise 4.5

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Here are the solutions for Exercise 4.5, covering the consistency of linear equations and solving systems using the matrix method.

Rbse Solutions Class 12 Maths: Exercise 4.5

A system of linear equations $AX = B$ is consistent if it has one or infinitely many solutions.

  1. If $|A| \ne 0$, the system is consistent and has a unique solution.
  2. If $|A| = 0$, calculate $\text{adj}(A)B$.
    • If $\text{adj}(A)B \ne O$ (where $O$ is the zero matrix), the system is inconsistent (no solution).
    • If $\text{adj}(A)B = O$, the system is consistent and has infinitely many solutions.

Examine the Consistency of the System of Equations (1 to 6)

1. $x + 2y = 2$ and $2x + 3y = 3$

$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$

$|A| = 1(3) – 2(2) = 3 – 4 = -1$.

Since $|A| \ne 0$, the system is consistent (Unique Solution).

2. $2x – y = 5$ and $x + y = 4$

$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$

$|A| = 2(1) – (-1)(1) = 2 + 1 = 3$.

Since $|A| \ne 0$, the system is consistent (Unique Solution).

3. $x + 3y = 5$ and $2x + 6y = 8$

$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$

$|A| = 1(6) – 3(2) = 6 – 6 = 0$. (Check $\text{adj}(A)B$)

$\text{adj}(A) = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$

$\text{adj}(A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30 – 24 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix}$

Since $|A| = 0$ and $\text{adj}(A)B \ne O$, the system is inconsistent (No Solution).

4. $x + y + z = 1$, $2x + 3y + 2z = 2$, and $ax + ay + 2az = 4$

$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$

$|A| = 1 \begin{vmatrix} 3 & 2 \\ a & 2a \end{vmatrix} – 1 \begin{vmatrix} 2 & 2 \\ a & 2a \end{vmatrix} + 1 \begin{vmatrix} 2 & 3 \\ a & a \end{vmatrix}$

$|A| = 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)$

$|A| = 4a – 2a – a = a$

  • Case 1: If $a \ne 0$, then $|A| \ne 0$, and the system is consistent (Unique Solution).
  • Case 2: If $a = 0$, then $|A| = 0$. The original equations become:
    • $x + y + z = 1$
    • $2x + 3y + 2z = 2$
    • $0x + 0y + 0z = 4$ (i.e., $0 = 4$)Since $0 = 4$ is a false statement, the system is inconsistent when $a=0$.

5. $3x – y – 2z = 2$, $2y – z = -1$, and $3x – 5y = 3$

$A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$

$|A| = 3 \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} – (-1) \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2) \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}$

$|A| = 3(0 – 5) + 1(0 – (-3)) – 2(0 – 6)$

$|A| = 3(-5) + 3 – 2(-6) = -15 + 3 + 12 = 0$. (Check $\text{adj}(A)B$)

$\text{adj}(A) = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}$, $B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

$\text{adj}(A)B = \begin{bmatrix} -5(2) + 10(-1) + 5(3) \\ -3(2) + 6(-1) + 3(3) \\ -6(2) + 12(-1) + 6(3) \end{bmatrix} = \begin{bmatrix} -10 – 10 + 15 \\ -6 – 6 + 9 \\ -12 – 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix}$

Since $|A| = 0$ and $\text{adj}(A)B \ne O$, the system is inconsistent (No Solution).

6. $5x – y + 4z = 5$, $2x + 3y + 5z = 2$, and $5x – 2y + 6z = -1$

$A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix}$

$|A| = 5 \begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix} – (-1) \begin{vmatrix} 2 & 5 \\ 5 & 6 \end{vmatrix} + 4 \begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix}$

$|A| = 5(18 – (-10)) + 1(12 – 25) + 4(-4 – 15)$

$|A| = 5(28) + 1(-13) + 4(-19) = 140 – 13 – 76 = 51$.

Since $|A| \ne 0$, the system is consistent (Unique Solution).

Solve System of Linear Equations, Using Matrix Method (7 to 14)

The solution is given by $X = A^{-1} B$, where $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

7. $5x + 2y = 4$ and $7x + 3y = 5$

$A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

  1. $|A| = 5(3) – 2(7) = 15 – 14 = 1$
  2. $\text{adj}(A) = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
  3. $A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$
  4. $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}$$$\mathbf{x = 2, \quad y = -3}$$

8. $2x – y = -2$ and $3x + 4y = 3$

$A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, $B = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$

  1. $|A| = 2(4) – (-1)(3) = 8 + 3 = 11$
  2. $\text{adj}(A) = \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$
  3. $A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$
  4. $X = A^{-1} B = \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}$$$\mathbf{x = -5/11, \quad y = 12/11}$$

9. $4x – 3y = 3$ and $3x – 5y = 7$

$A = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

  1. $|A| = 4(-5) – (-3)(3) = -20 + 9 = -11$
  2. $\text{adj}(A) = \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$
  3. $A^{-1} = \frac{1}{-11} \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$
  4. $X = A^{-1} B = \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} = -\frac{1}{11} \begin{bmatrix} 6 \\ 19 \end{bmatrix}$$$\mathbf{x = -6/11, \quad y = -19/11}$$

10. $5x + 2y = 3$ and $3x + 2y = 5$

$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$

  1. $|A| = 5(2) – 2(3) = 10 – 6 = 4$
  2. $\text{adj}(A) = \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$
  3. $A^{-1} = \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$
  4. $X = A^{-1} B = \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} = \frac{1}{4} \begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}$$$\mathbf{x = -1, \quad y = 4}$$

11. $2x + y + z = 1$, $x – 2y – z = 3/2$, and $3y – 5z = 9$

$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 3/2 \\ 9 \end{bmatrix}$

  1. $|A|$: (Expand along $\mathbf{C_1}$ for the zero)$|A| = 2 \begin{vmatrix} -2 & -1 \\ 3 & -5 \end{vmatrix} – 1 \begin{vmatrix} 1 & 1 \\ 3 & -5 \end{vmatrix} + 0$$|A| = 2(10 – (-3)) – 1(-5 – 3) = 2(13) – 1(-8) = 26 + 8 = 34$
  2. $\text{adj}(A)$: (Cofactors used in the Adjoint matrix)$$\text{adj}(A) = \begin{bmatrix} 13 & 8 & 3 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$$
  3. $A^{-1}$: $A^{-1} = \frac{1}{34} \text{adj}(A)$
  4. $X = A^{-1} B$:$$X = \frac{1}{34} \begin{bmatrix} 13 & 8 & 3 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\ 3/2 \\ 9 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13(1) + 8(3/2) + 3(9) \\ 5(1) – 10(3/2) + 3(9) \\ 3(1) – 6(3/2) – 5(9) \end{bmatrix}$$$$X = \frac{1}{34} \begin{bmatrix} 13 + 12 + 27 \\ 5 – 15 + 27 \\ 3 – 9 – 45 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 52 \\ 17 \\ -51 \end{bmatrix}$$$$\mathbf{x = 52/34 = 26/17, \quad y = 17/34 = 1/2, \quad z = -51/34 = -3/2}$$

12. $x – y + z = 4$, $2x + y – 3z = 0$, and $x + y + z = 2$

$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

  1. $|A|$: (Expand along $\mathbf{R_1}$)$|A| = 1(1 – (-3)) – (-1)(2 – (-3)) + 1(2 – 1)$$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$
  2. $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$$
  3. $A^{-1}$: $A^{-1} = \frac{1}{10} \text{adj}(A)$
  4. $X = A^{-1} B$:$$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} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$$$\mathbf{x = 2, \quad y = -1, \quad z = 1}$$

13. $2x + 3y + 3z = 5$, $x – 2y + z = -4$, and $3x – y – 2z = 3$

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

  1. $|A|$: (Expand along $\mathbf{R_1}$)$|A| = 2(4 – (-1)) – 3(-2 – 3) + 3(-1 – (-6))$$|A| = 2(5) – 3(-5) + 3(5) = 10 + 15 + 15 = 40$
  2. $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$$
  3. $A^{-1}$: $A^{-1} = \frac{1}{40} \text{adj}(A)$
  4. $X = A^{-1} B$:$$X = \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}{40} \begin{bmatrix} 25 – 12 + 27 \\ 25 + 52 + 3 \\ 25 – 44 – 21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$$$\mathbf{x = 1, \quad y = 2, \quad z = -1}$$

14. $x – y + 2z = 7$, $3x + 4y – 5z = -5$, and $2x – y + 3z = 12$

$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$

  1. $|A|$: (Expand along $\mathbf{R_1}$)$|A| = 1(12 – 5) – (-1)(9 – (-10)) + 2(-3 – 8)$$|A| = 1(7) + 1(19) + 2(-11) = 7 + 19 – 22 = 4$
  2. $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$$
  3. $A^{-1}$: $A^{-1} = \frac{1}{4} \text{adj}(A)$
  4. $X = A^{-1} B$:$$X = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} = \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}$$$$\mathbf{x = 2, \quad y = 1, \quad z = 3}$$

15. Find $A^{-1}$ and use it to solve the system of equations.

Given $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$ and the system:

$2x – 3y + 5z = 11$

$3x + 2y – 4z = -5$

$x + y – 2z = -3$

The coefficient matrix of the system is exactly $A$. $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$.

  1. $|A|$: (Expand along $\mathbf{R_1}$)$|A| = 2(-4 – (-4)) – (-3)(-6 – (-4)) + 5(3 – 2)$$|A| = 2(0) + 3(-2) + 5(1) = 0 – 6 + 5 = -1$
  2. $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$$
  3. $A^{-1}$:$$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}$$
  4. $X = A^{-1} B$:$$X = \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}$$$$\mathbf{x = 1, \quad y = 2, \quad z = 3}$$

16. Cost of Onion, Wheat, and Rice per kg (Matrix Method)

Let $x, y, z$ be the cost per kg of onion, wheat, and rice, respectively.

The system of equations is:

$4x + 3y + 2z = 60$

$2x + 4y + 6z = 90$

$6x + 2y + 3z = 70$

$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$

  1. $|A|$: (Expand along $\mathbf{R_1}$)$|A| = 4(12 – 12) – 3(6 – 36) + 2(4 – 24)$$|A| = 4(0) – 3(-30) + 2(-20) = 0 + 90 – 40 = 50$
  2. $\text{adj}(A)$:$$\text{adj}(A) = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$$
  3. $A^{-1}$:$$A^{-1} = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$$
  4. $X = A^{-1} B$:$$X = \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} 0 – 450 + 700 \\ 1800 + 0 – 1400 \\ -1200 + 900 + 700 \end{bmatrix}$$$$X = \frac{1}{50} \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$$

$$\mathbf{\text{Cost per kg: Onion = ₹ 5, Wheat = ₹ 8, Rice = ₹ 8}}$$

Ex.4.1
Ex.4.2
Ex.4.3
Ex.4.4
Ex.4.5
Ex.4.misce.

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