Exercise 4.5 Solutions | Class 12 Maths

Examine the consistency of the system of equations in Exercises 1 to 6.

1.
${\begin{cases} x + 2 y = 2 \\ 2 x + 3 y = 3 \end{cases}$

Solution. From the first, $x = 2 - 2 y$ . Substitute: $2 (2 - 2 y) + 3 y = 3 \Rightarrow 4 - y = 3 \Rightarrow y = 1$ . Then $x = 0$ .
Unique solution: $(x, y) = (0, 1)$

2.
${\begin{cases} 2 x - y = 5 \\ x + y = 4 \end{cases}$

Solution. From second $y = 4 - x$ . Substitute: $2 x - (4 - x) = 5 \Rightarrow 3 x = 9 \Rightarrow x = 3$ . Then $y = 1$
Solution: $(3, 1)$

3.
${\begin{cases} x + 3 y = 5 \\ 2 x + 6 y = 8 \end{cases}$

Solution. The second equation is $2 (x + 3 y) = 10$ , but RHS is 8, not 10. The two equations are inconsistent (parallel, no solution).
No solution.

4.
${\begin{cases} x + y + z = 1 \\ 2 x + 3 y + 2 z = 2 \\ a x + a y + 2 a z = 4 \end{cases}$

Solution. From (1) & (2): subtract $2$ (1) from (2) → $y = 0$ . Then $x + z = 1$ .
Third equation is $a (x + y + 2 z) = 4$ . With $y = 0$ this becomes $a (x + 2 z) = 4$ . Using $x = 1 - z$ :

$a ((1 - z) + 2 z) = a (1 + z) = 4 \Rightarrow 1 + z = \frac{4}{a} (if a \neq 0) .$

So for $a \neq 0$ : $z = \frac{4}{a} - 1, x = 2 - \frac{4}{a}, y = 0$ (unique solution).
If $a = 0$ : third eqn becomes $0 = 4$ impossible → inconsistent.

So: consistent (unique) for $a \neq 0$ with $(x, y, z) = (2 - \frac{4}{a}, 0, \frac{4}{a} - 1)$ . Inconsistent when $a = 0$ . depending on $a$ .

5.
${\begin{cases} 3 x - y - 2 z = 2 \\ 2 y - z = - 1 \\ 3 x - 5 y = 3 \end{cases}$

Solution. From second: $z = 2 y + 1$ . From third: $x = (3 + 5 y) / 3$ . Substitute into first:

$3 \cdot \frac{3 + 5 y}{3} - y - 2 (2 y + 1) = 2 \Rightarrow (3 + 5 y) - y - 4 y - 2 = 2 \Rightarrow 1 = 2,$

contradiction. So inconsistent (no solution).

6.
${\begin{cases} 5 x - y + 4 z = 5 \\ 2 x + 3 y + 5 z = 2 \\ 5 x - 2 y + 6 z = - 1 \end{cases}$

Solution. Subtract (1) from (3): $- y + 2 z = - 6 \Rightarrow y - 2 z = 6$ $. Put in (2):$

$2 x + 3 (6 + 2 z) + 5 z = 2 \Rightarrow 2 x + 18 + 11 z = 2 \Rightarrow x = - 8 - \frac{11}{2} z$

Put $x, y$ into (1): solving gives $z = - 2$ , then $y = 2$ , $x = 3$
Unique solution: (x,y,z) = (3,2,-2)

Solve system of linear equations, using matrix method, in Exercises 7 to 14.

Question 7

Solve by matrix method:

${\begin{cases} 5 x + 2 y = 4, \\ 7 x + 3 y = 5. \end{cases}$

Matrix form: $A x = b$ with

$A = [\begin{array}{cc} 5 & 2 \\ 7 & 3 \end{array}], x = [\begin{array}{c} x \\ y \end{array}], b = [\begin{array}{c} 4 \\ 5 \end{array}]$

$\det A = 5 \cdot 3 - 7 \cdot 2 = 15 - 14 = 1$
$A^{- 1} = [\begin{array}{cc} 3 & - 2 \\ - 7 & 5 \end{array}]$ (since $1 / \det A = 1$ ).
Hence $x = A^{- 1} b = [\begin{array}{cc} 3 & - 2 \\ - 7 & 5 \end{array}] [\begin{array}{c} 4 \\ 5 \end{array}] = [\begin{array}{c} 2 \\ - 3 \end{array}]$

Answer: $x = 2, y = - 3.$

Question 8

${\begin{cases} 2 x - y = - 2, \\ 3 x + 4 y = 3. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 2 & - 1 \\ 3 & 4 \end{array}], b = [\begin{array}{c} - 2 \\ 3 \end{array}]$
$\det A = 2 \cdot 4 - 3 \cdot (- 1) = 8 + 3 = 11$

$A^{- 1} = \frac{1}{11} [\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}]$
$x = A^{- 1} b = \frac{1}{11} [\begin{array}{cc} 4 & 1 \\ - 3 & 2 \end{array}] [\begin{array}{c} - 2 \\ 3 \end{array}] = [\begin{array}{c} - \frac{5}{11} \\ \frac{12}{11} \end{array}]$

Answer: $x = - \frac{5}{11}, y = \frac{12}{11} .$

Question 9

${\begin{cases} 4 x - 3 y = 3, \\ 3 x - 5 y = 7. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 4 & - 3 \\ 3 & - 5 \end{array}], b = [\begin{array}{c} 3 \\ 7 \end{array}]$
$\det A = 4 (- 5) - 3 (- 3) = - 20 + 9 = - 11$

$A^{- 1} = \frac{1}{- 11} [\begin{array}{cc} - 5 & 3 \\ - 3 & 4 \end{array}] = \frac{1}{11} [\begin{array}{cc} 5 & - 3 \\ 3 & - 4 \end{array}]$ $x = A^{- 1} b = \frac{1}{11} [\begin{array}{cc} 5 & - 3 \\ 3 & - 4 \end{array}] [\begin{array}{c} 3 \\ 7 \end{array}] = [\begin{array}{c} - \frac{6}{11} \\ - \frac{19}{11} \end{array}]$

Answer: $x = - \frac{6}{11}, y = - \frac{19}{11} .$

Question 10

${\begin{cases} 5 x + 2 y = 3, \\ 3 x + 2 y = 5. \end{cases}$

Matrix form: $A = [\begin{array}{cc} 5 & 2 \\ 3 & 2 \end{array}], b = [\begin{array}{c} 3 \\ 5 \end{array}]$
Subtract second from first: $2 x = - 2 \Rightarrow x = - 1$ Then $5 (- 1) + 2 y = 3 \Rightarrow - 5 + 2 y = 3 \Rightarrow 2 y = 8 \Rightarrow y = 4.$

(Or use inverse: $\det A = 5 \cdot 2 - 3 \cdot 2 = 10 - 6 = 4$ , $A^{- 1} = \frac{1}{4} [\begin{array}{cc} 2 & - 2 \\ - 3 & 5 \end{array}]$ giving same result.)

Answer: $x = - 1, y = 4.$

Question 11

Solve the system

${\begin{cases} 2 x + y + z = 1, \\ x - 2 y - z = \frac{3}{2}, \\ 2 x + y - 3 z = 0. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 1 & - 2 & - 1 & \frac{3}{2} \\ 2 & 1 & - 3 & 0 \end{array}]$

To avoid fractions, multiply row2 by 2 first:

$R_{2} \leftarrow 2 R_{2} : [\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 2 & - 4 & - 2 & 3 \\ 2 & 1 & - 3 & 0 \end{array}]$

Now eliminate using $R_{1}$ :

$R_{2} \leftarrow R_{2} - R_{1}$
$[0, - 5, - 3 ∣ 2]$
$R_{3} \leftarrow R_{3} - R_{1}$
$[0, 0, - 4 ∣ - 1]$

Matrix becomes

$[\begin{array}{cccc} 2 & 1 & 1 & 1 \\ 0 & - 5 & - 3 & 2 \\ 0 & 0 & - 4 & - 1 \end{array}]$

Back-substitution:

From row3: $- 4 z = - 1 \Rightarrow z = \frac{1}{4}$

Row2: $- 5 y - 3 z = 2 \Rightarrow - 5 y - 3 \cdot \frac{1}{4} = 2 \Rightarrow - 5 y = \frac{11}{4} \Rightarrow y = - \frac{11}{20}$

Row1: $2 x + y + z = 1 \Rightarrow 2 x + (- \frac{11}{20}) + \frac{1}{4} = 1$
Compute $y + z = - \frac{11}{20} + \frac{5}{20} = - \frac{6}{20} = - \frac{3}{10}$
So $2 x - \frac{3}{10} = 1 \Rightarrow 2 x = \frac{13}{10} \Rightarrow x = \frac{13}{20}$

Answer: $x = \frac{13}{20}, y = - \frac{11}{20}, z = \frac{1}{4}$

Question 12

Solve the system

${\begin{cases} x - y + z = 4, \\ 2 x + y - 3 z = 0, \\ 3 y - 5 z = 9. \end{cases}$

Solution (matrix / substitution).

From the third equation:

$3 y - 5 z = 9 \Rightarrow y = \frac{9 + 5 z}{3}$

Substitute $y$ into the first equation:

$x - \frac{9 + 5 z}{3} + z = 4$

Multiply by 3:

$3 x - (9 + 5 z) + 3 z = 12 \Rightarrow 3 x - 9 - 2 z = 12 \Rightarrow 3 x - 2 z = 21.$

So $\begin{matrix} x = 7 + \frac{2}{3} z . & (*) \end{matrix}$

Now substitute $x$ and $y$ into the second equation $2 x + y - 3 z = 0$

$2 (7 + \frac{2}{3} z) + \frac{9 + 5 z}{3} - 3 z = 0.$

Compute left side:

$14 + \frac{4}{3} z + \frac{9 + 5 z}{3} - 3 z = 14 + \frac{4 z + 9 + 5 z}{3} - 3 z$

$= 14 + \frac{9 + 9 z}{3} - 3 z = 14 + 3 + 3 z - 3 z = 17.$

That gives $17 = 0$ , a contradiction.

Conclusion: The system is inconsistent; no solution.

Answer: $No solution (inconsistent system).$

Question 13

Solve the system

${\begin{cases} 2 x + 3 y + 3 z = 5, \\ x - 2 y + z = - 4, \\ 3 x - y - 2 z = 3. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 2 & 3 & 3 & 5 \\ 1 & - 2 & 1 & - 4 \\ 3 & - 1 & - 2 & 3 \end{array}]$

Use row2 as pivot to eliminate others:

$R_{1} \leftarrow R_{1} - 2 R_{2}$
$[0, 7, 1 ∣ 13]$
$R_{3} \leftarrow R_{3} - 3 R_{2}$
$[0, 5, - 5 ∣ 15]$

So we have

$[\begin{array}{cccc} 1 & - 2 & 1 & - 4 \\ 0 & 7 & 1 & 13 \\ 0 & 5 & - 5 & 15 \end{array}]$

(reordered so row2 is the original $R_{2}$ ).

Eliminate $y$ from row3 using row2:

$R_{3} \leftarrow 7 R_{3} - 5 R_{2}$

$7 R_{3} = [0, 35, - 35 ∣ 105], 5 R_{2} = [0, 35, 5 ∣ 65]$

So $R_{3} \leftarrow [0, 0, - 40 ∣ 40]$ ⇒ $- 40 z = 40 \Rightarrow z = - 1$

Back-substitute:

Row2: $7 y + z = 13 \Rightarrow 7 y - 1 = 13 \Rightarrow 7 y = 14 \Rightarrow y = 2$

Row1 (original row2): $x - 2 y + z = - 4 \Rightarrow x - 4 - 1 = - 4 \Rightarrow x = 1$

Answer: $x = 1, y = 2, z = - 1$

Question 14 (Plain text)

Solve the system

${\begin{cases} x - y + 2 z = 7, \\ 3 x + 4 y - 5 z = - 5, \\ 2 x - y + 3 z = 12. \end{cases}$

Solution (matrix / elimination).

Augmented matrix:

$[\begin{array}{cccc} 1 & - 1 & 2 & 7 \\ 3 & 4 & - 5 & - 5 \\ 2 & - 1 & 3 & 12 \end{array}]$

Eliminate below first pivot $R_{1}$ :

$R_{2} \leftarrow R_{2} - 3 R_{1} : [0, 7, - 11 ∣ - 26]$
$R_{3} \leftarrow R_{3} - 2 R_{1} : [0, 1, - 1 ∣ - 2]$

Swap $R_{2}$ and $R_{3}$ to use the simpler pivot:

$[\begin{array}{cccc} 1 & - 1 & 2 & 7 \\ 0 & 1 & - 1 & - 2 \\ 0 & 7 & - 11 & - 26 \end{array}]$

Eliminate $y$ from row3:

$R_{3} \leftarrow R_{3} - 7 R_{2} : [0, 0, - 4 ∣ - 12]$

So $- 4 z = - 12 \Rightarrow z = 3$

Back-substitute:

Row2: $y - z = - 2 \Rightarrow y - 3 = - 2 \Rightarrow y = 1$

Row1: $x - y + 2 z = 7 \Rightarrow x - 1 + 6 = 7 \Rightarrow x = 2$

Answer: $x = 2, y = 1, z = 3$

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14 (Plain text)

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