RD Sharma Class 12 Exercise 5.4 Determinants Solutions Maths

RD Sharma Class 12 Exercise 5.4 Determinants Solutions Maths - Download PDF Free Online

Updated on Jan 25, 2022 07:11 PM IST

Most class 12 students use the RD Sharma solution books as their companions to solve their doubts. This makes them score good marks in the public examinations due to constant practice in the proper method. Mathematics is a subject where most of the doubt arises while solving a problem. RD Sharma solution Significantly, the Determinants chapter is easy as well as a bit tricky. Even if a student tries to recheck their answers, it takes a lot of time. Therefore, the RD Sharma Class 12th exercise 5.4 books can be used

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RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise

Determinants Excercise: 5.4

Determinants Exercise 5.4 Question 1

Answer:

x = - 6 and y = - 5

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: $\begin{aligned} x - 2 y = 4 \\ - 3 x + 5 y = - 7 \end{aligned}$
Solution:First D: determinant of the coefficient matrix
$D = | \begin{array}{cc} 1 & - 2 \\ - 3 & 5 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})$
$\begin{aligned} = (1) (5) - (- 3) (- 2) \\ = 5 - 6 \\ = - 1 \end{aligned}$
Now,

D \neq 0

If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{1} = | \begin{array}{cc} 4 & - 2 \\ - 7 & 5 \end{array} | \\ = (4) (5) - (- 7) (- 2) \\ = 20 - 14 \end{aligned}

= 6

If we are solving for y, the y column is replaced with constant column i.e.

D_{2} = | \begin{array}{cc} 1 & 4 \\ - 3 & - 7 \end{array} |

Now,

x = \frac{D_{1}}{D} = \frac{6}{- 1} = - 6

y = \frac{D_{2}}{D} = \frac{5}{- 1} = - 5

Hence,

x = - 6 and y = - 5

Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

Determinants Exercise 5.4 Question 2

Answer:

x = - 3 and y = - 7

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:

\begin{aligned} 2 x - y = 1 \\ 7 x - 2 y = - 7 \end{aligned}

Solution: First D: determinant of the coefficient matrix

D = | \begin{array}{ll} 2 & - 1 \\ 7 & - 2 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (2) (- 2) - (- 1) (7) \\ = - 4 + 7 \\ = - 3 \end{aligned}

Now, $D \neq 0$ . If we are solving for x, the x column is replaced with constant column i.e.

$D_{1} = | \begin{array}{cc} 1 & - 1 \\ - 7 & - 2 \end{array} |$

$\begin{aligned} = (1) (- 2) - (- 7) (- 1) \\ = - 2 - 7 \\ = - 9 \end{aligned}$
If we are solving for y, the y column is replaced with constant column i.e
$\begin{aligned} D_{2} = | \begin{array}{cc} 2 & 1 \\ 7 & - 7 \end{array} | \\ = (2) (- 7) - (7) (1) \\ = - 14 - 7 \\ = - 21 \end{aligned}$
Now, $x = \frac{D_{1}}{D} = \frac{- 9}{3} = - 3$
$y = \frac{D_{2}}{D} = \frac{- 21}{3} = - 7$
$H e n c e, x = - 3 a n d y = - 7.$
Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions, that is determinant is zero

determinants Exercise 5.4 Question 3

Answer:

x = 7 and y = - 3

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: $\begin{aligned} 2 x - y = 17 \\ 3 x + 5 y = 6 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{cc} 2 & - 1 \\ 3 & 5 \end{array} | ∵ | \begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (2) (5) - (3) (- 1) \\ = 10 + 3 \\ = 13 \end{aligned}

Now, $D \neq 0$ . If we are solving for x, the x column is replaced with constant column i.e.

$D_{1} = | \begin{array}{cc} 17 & - 1 \\ 6 & 5 \end{array} |$

$\begin{aligned} = (2) (6) - (17) (3) \\ = 12 - 51 \\ = - 39 \\ Now, x = \frac{D_{1}}{D} = \frac{91}{13} = 7 \end{aligned}$

$y = \frac{D_{2}}{D} = \frac{- 39}{13} = - 3$
Hence, x = 7 and y=-3

Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero

determinants Exercise 5.4 Question 4

Answer:

x = 7 and y = - 2

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: $\begin{aligned} 3 x + y = 19 \\ 3 x - y = 23 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{cc} 3 & 1 \\ 3 & - 1 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (- 1) (3) - (3) (1) \\ = - 3 - 3 \\ = - 6 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

D_{1} = | \begin{array}{cc} 19 & 1 \\ 23 & - 1 \end{array} |

\begin{aligned} = (19) (- 1) - (23) (1) \\ = - 19 - 23 \\ = - 42 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{2} & = | \begin{array}{ll} 3 & 19 \\ 3 & 23 \end{array} | \\ = (3) (23) - (19) (3) \\ = 69 - 57 \\ = 12 \end{aligned}

\begin{aligned} Now, x = \frac{D_{1}}{D} = \frac{- 42}{- 6} = 7 \\ y = \frac{D_{2}}{D} = \frac{12}{- 6} = - 2 \end{aligned}

Hence

x = 7 and y = - 2

determinants Exercise 5.4 Question 5

Answer: $x = \frac{- 5}{11} and y = \frac{12}{11}$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: $\begin{aligned} 2 x - y = - 2 \\ 3 x + 4 y = 3 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{cc} 2 & - 1 \\ 3 & 4 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (2) (4) - (3) (- 1) \\ = 8 + 3 \\ = 11 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

D_{1} = | \begin{array}{cc} - 2 & - 1 \\ 3 & 4 \end{array} |

\begin{aligned} = (4) (- 2) - (3) (- 1) \\ = - 8 + 3 \\ = - 5 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

D_{2} = | \begin{array}{cc} 2 & - 2 \\ 3 & 3 \end{array} |

\begin{aligned} = (2) (3) - (3) (- 2) \\ = 6 + 6 \\ = 12 \end{aligned}

Now,

\begin{aligned} x = \frac{D_{1}}{D} = \frac{- 5}{11} \\ y = \frac{D_{2}}{D} = \frac{12}{11} \end{aligned}

Hence, $x = \frac{- 5}{11} and y = \frac{12}{11}$

determinants Exercise 5.4 Question 6

Answer: $x = 2 and y = \frac{- 2}{a}$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:
$\begin{aligned} 3 x + a y = 4 \\ 2 x + a y = 2, a \neq 0 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

\begin{aligned} D & = | \begin{array}{ll} 3 & a \\ 2 & a \end{array} | & ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1}) \\ = 3 a - 2 a \\ = a \end{aligned}

Now, $D \neq 0$ . If we are solving for x, the x column is replaced with constant column i.e.

$\begin{aligned} D_{1} & = | \begin{array}{ll} 4 & a \\ 2 & a \end{array} | \\ = 4 a - 2 a \\ = 2 a \end{aligned}$
If we are solving for y, the y column is replaced with constant column i.e.
$\begin{aligned} D_{2} = | \begin{array}{ll} 3 & 4 \\ 2 & 2 \end{array} | \\ (2) \\ \begin{aligned} = (2) (3) - (4) \\ = 6 - 8 \\ = - 2 \end{aligned} \end{aligned}$
Now,
$\begin{array}{r} x = \frac{D_{1}}{D} = \frac{2 a}{a} = 2 \\ y = \frac{D_{2}}{D} = \frac{- 2}{a} \end{array}$
Hence, $x = 2 and y = \frac{- 2}{a}$
Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero

determinants Exercise 5.4 Question 7

Answer:

x = \frac{16}{3} and y = \frac{- 2}{9}

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:
$\begin{aligned} 2 x + 3 y = 10 \\ x + 6 y = 4 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{ll} 2 & 3 \\ 1 & 6 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (2) (6) - (3) (1) \\ = 12 - 3 \\ = 9 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{1} = | \begin{array}{cc} 10 & 3 \\ 4 & 6 \end{array} | \\ = 60 - 12 \\ = 48 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{2} & = | \begin{array}{ll} 2 & 10 \\ 1 & 4 \end{array} | \\ = 8 - 10 \\ = - 2 \end{aligned}

Now,

\begin{matrix} x = \frac{D_{1}}{D} = \frac{48}{9} = \frac{16}{3} \\ y = \frac{D_{2}}{D} = \frac{- 2}{9} \end{matrix}

Hence,

x = \frac{16}{3} and y = \frac{- 2}{9}

deteminants Exercise 5.4 Question 8

Answer:

x = \frac{9}{2} and y = \frac{- 7}{2}

Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:
$\begin{aligned} 5 x + 7 y = - 2 \\ 4 x + 6 y = - 3 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (5) (6) - (7) (4) \\ = 30 - 28 \\ = 2 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{1} & = | \begin{array}{rr} - 2 & 7 \\ - 3 & 6 \end{array} | \\ = - 12 + 21 \\ = 9 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{2} = | \begin{array}{ll} 5 & - 2 \\ 4 & - 3 \end{array} | \\ = - 15 + 8 \\ = - 7 \\ Now, x = \frac{D_{1}}{D} = \frac{9}{2} \\ y = \frac{D_{2}}{D} = \frac{- 7}{2} \end{aligned}

Hence,

x = \frac{9}{2} and y = \frac{- 7}{2}

determinants Exercise 5.4 Question 9

Answer: $x = \frac{- 10}{37} and y = \frac{92}{37}$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:
$\begin{aligned} 9 x + 5 y = 10 \\ 3 y - 2 x = 8 \Rightarrow - 2 x + 3 y = 8 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{cc} 9 & 5 \\ - 2 & 3 \end{array} |

∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (9) (3) - (5) (- 2) \\ = 27 + 10 \\ = 37 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{1} & = | \begin{array}{cc} 10 & 5 \\ 8 & 3 \end{array} | \\ = 30 - 40 \\ = - 10 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{2} = | \begin{array}{cc} 9 & 10 \\ - 2 & 8 \end{array} | \\ = 72 + 20 \\ = 92 \\ Now, x = \frac{D_{1}}{D} = \frac{- 10}{37} \\ y = \frac{D_{2}}{D} = \frac{92}{37} \end{aligned}

Hence

y = 92 / 37 and x = - 10 / 37

determinants Exercise 5.4 Question 10

Answer: $x = \frac{7}{5} and y = \frac{- 1}{5}$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given:
$\begin{aligned} x + 2 y = 1 \\ 3 x + y = 4 \end{aligned}$
Solution:
First D: determinant of the coefficient matrix

D = | \begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = (1) (1) - (3) (2) \\ = 1 - 6 \\ = - 5 \end{aligned}

Now,

D \neq 0

. If we are solving for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{1} & = | \begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array} | \\ = 1 - 8 \\ = - 7 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{2} & = | \begin{array}{ll} 1 & 1 \\ 3 & 4 \end{array} | \\ = 4 - 3 \\ = 1 \end{aligned}

Now,

\begin{aligned} x = \frac{D_{1}}{D} = \frac{- 7}{- 5} = \frac{7}{5} \\ y = \frac{D_{2}}{D} = \frac{1}{- 5} = \frac{- 1}{5} \end{aligned}

Hence, $x = \frac{7}{5} and y = \frac{- 1}{5}$
Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

determinants Exercise 5.4 Question 11

Answer:

x = - 1, y = 2 and z = 3

Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 3 x + y + z = 2 \\ 2 x - 4 y + 3 z = - 1 \end{aligned}$
$4 x + y - 3 z = - 11$
Solution:
First take coefficient of variables x, y and z.

| A | = | \begin{array}{ccc} 3 & 1 & 1 \\ 2 & - 4 & 3 \\ 4 & 1 & - 3 \end{array} |

∴

(Taking first row for solving determinant)

\begin{aligned} = 3 (12 - 3) - 1 (- 6 - 12) + 1 (2 + 16) \\ = 3 (9) - 1 (- 18) + (1) (18) \\ = 27 + 18 + 18 \\ = 63 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

D_{x} = | \begin{array}{ccc} 2 & 1 & 1 \\ - 1 & - 4 & 3 \\ - 11 & 1 & - 3 \end{array} |

\begin{aligned} = 2 (12 - 3) - 1 (3 + 33) + 1 (- 1 - 44) \\ = 2 (9) - 1 (36) + 1 (- 45) \\ = 18 - 36 - 45 \\ = - 63 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 3 & 2 & 1 \\ 2 & - 1 & 3 \\ 4 & - 11 & - 3 \end{array} | \\ = 3 (3 + 33) - 2 (- 6 - 12) + 1 (- 22 + 4) \\ = 3 (36) - 2 (- 18) + 1 (- 18) \\ = 108 + 36 - 18 \\ = 126 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} & = | \begin{array}{ccc} 3 & 1 & 2 \\ 2 & - 4 & - 1 \\ 4 & 1 & - 11 \end{array} | \\ = 3 (44 + 1) - 1 (- 22 + 4) + 2 (2 + 16) \\ = 3 (45) - 1 (- 18) + 2 (18) \\ = 135 + 18 + 36 \\ = 189 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 63}{63} = - 1 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{126}{63} = 2 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{189}{63} = 3 \end{aligned}

Concept: Determinant solving of 3 x 3 matrix (Cramer’s rule).
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

determinants Exercise 5.4 Question 12

Answer: $x = - 1, y = - 5 and z = 8$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} x - 4 y - z = 11 \\ 2 x - 5 y + 2 z = 39 \\ - 3 x + 2 y + z = 1 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

| A | = | \begin{array}{ccc} 1 & - 4 & - 1 \\ 2 & - 5 & 2 \\ - 3 & 2 & 1 \end{array} |

∵

(Taking first row for solving determinant)

\begin{aligned} = 1 (- 5 - 4) + 4 (2 + 6) - 1 (4 - 15) \\ = 1 (- 9) + 4 (8) - 1 (- 11) \\ = - 9 + 32 + 11 \\ = 34 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 11 & - 4 & - 1 \\ 39 & - 5 & 2 \\ 1 & 2 & 1 \end{array} | \\ = 11 (- 5 - 4) + 4 (39 - 2) - 1 (78 + 5) \\ = 11 (- 9) + 4 (37) - 1 (83) \\ = - 99 + 148 - 83 \\ = 34 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 1 & 11 & - 1 \\ 2 & 39 & 2 \\ - 3 & 1 & 1 \end{array} | \\ = 1 (39 - 2) - 11 (2 + 6) - 1 (2 + 117) \\ = 1 (37) - 11 (8) - 1 (119) \\ = 37 - 88 - 119 \\ = - 170 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} & = | \begin{array}{ccc} 1 & - 4 & 11 \\ 2 & - 5 & 39 \\ - 3 & 2 & 1 \end{array} | \\ = 1 (- 5 - 78) + 4 (2 + 117) + 11 (4 - 15) \end{aligned}

\begin{aligned} = - 83 + 4 (119) + 11 (- 11) \\ = 272 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 34}{34} = - 1 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 170}{34} = - 5 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{272}{34} = 8 \end{aligned}

Concept: Determinant solving of 3 x 3 matrix (Cramer’s rule)
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero

deteminants Exercise 5.4 Question 13

Answer:

x = 1, y = 2 and z = 1

Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 6 x + y - 3 z = 5 \\ x + 3 y - 2 z = 5 \\ 2 x + y + 4 z = 8 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

| A | = | \begin{array}{ccc} 6 & 1 & - 3 \\ 1 & 3 & - 2 \\ 2 & 1 & 4 \end{array} |

∵

(Taking first row for solving determinant)

\begin{aligned} = 6 (12 + 2) - 1 (4 + 4) - 3 (1 - 6) \\ = 6 (14) - 1 (8) - 3 (- 5) \\ = 84 - 8 + 15 \\ = 91 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

D_{\bar{x}} = | \begin{array}{ccc} 5 & 1 & - 3 \\ 5 & 3 & - 2 \\ 8 & 1 & 4 \end{array} |

\begin{aligned} = 5 (12 + 2) - 1 (20 + 16) - 3 (5 - 24) \\ = 5 (14) - 1 (36) - 3 (- 19) \\ = 70 - 36 + 57 \\ = 91 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

D_{x} = | \begin{array}{ccc} 6 & 5 & - 3 \\ 1 & 5 & - 2 \\ 2 & 8 & 4 \end{array} |

\begin{aligned} = 6 (20 + 16) - 5 (4 + 4) - 3 (8 + 10) \\ = 6 (36) + 3 (- 2) - 5 (8) \\ = 216 - 6 - 40 \\ = 182 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} = | \begin{array}{lll} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{array} | \\ = 6 (24 - 5) - 1 (8 - 10) + 5 (1 - 6) \\ = 6 (19) - 1 (- 2) + 5 (- 5) \\ = 114 + 2 - 25 \\ = 91 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{91}{91} = 1 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{182}{91} = 2 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{91}{91} = 1 \end{aligned}

Concept: Determinant solving of 3 x 3 matrix (Cramer’s rule)
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

deteminants Exercise 5.4 Question 14

Answer: $x = 3, y = 2 and z = 1$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} x + y = 5 \\ y + z = 3 \\ x + z = 4 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

\begin{aligned} | A | & = | \begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array} | \\ = 1 (1) - 1 (- 1) + 0 (- 1) \\ = 1 + 1 \end{aligned}

∵

(Taking first row for solving determinant)

= 2

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{\overset{―}{y}} = | \begin{array}{lll} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{array} | \\ = 5 (1) - 1 (3 - 4) + 0 (0 - 4) \\ = 5 + 1 \\ = 6 \end{aligned}

\begin{aligned} D_{y} & = | \begin{array}{lll} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{array} | \\ = 1 (3 - 4) - 5 (0 - 1) + 0 (0 - 3) \\ = - 1 + 5 \\ = 4 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} = | \begin{array}{lll} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{array} | \\ = 1 (4 - 0) - 1 (0 - 3) + 5 (0 - 1) \\ = 4 + 3 - 5 \\ = 2 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{6}{2} = 3 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{4}{2} = 2 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{2}{2} = 1 \end{aligned}

deteminants Exercise 5.4 Question 15

Answer:

x = 5, y = - 3 and z = - 2

Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 2 y - 3 z = 0 \\ x + 3 y = - 4 \\ 3 x + 4 y = 3 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

\begin{aligned} | A | & = | \begin{array}{ccc} 0 & 2 & - 3 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \end{array} | \\ = 0 (0) - 2 (0 - 0) - 3 (4 - 9) \\ = - 3 (- 5) \\ = 15 \end{aligned}

∵

(Taking first row for solving determinant)
Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 0 & 2 & - 3 \\ - 4 & 3 & 0 \\ 3 & 4 & 0 \end{array} | \\ = 0 (0) - 2 (0) - 3 (- 16 - 9) \\ = - 3 (- 25) \\ = 75 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{y} & = | \begin{array}{ccc} 0 & 0 & - 3 \\ 1 & - 4 & 0 \\ 3 & 3 & 0 \end{array} | \\ = 0 (0) - 0 (0) - 3 (3 + 12) \\ = - 3 (15) \\ = - 45 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} & = | \begin{array}{ccc} 0 & 2 & 0 \\ 1 & 3 & - 4 \\ 3 & 4 & 3 \end{array} | \\ = 0 (9 + 16) - 2 (3 + 12) + 0 (4 - 9) \\ = 0 - 2 (15) + 0 \\ = - 30 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{75}{15} = 5 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 45}{15} = - 3 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 30}{15} = - 2 \end{aligned}

determinants Exercise 5.4 Question 16

Answer: $x = 1, y = - 1 and z = - 1$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 5 x - 7 y + z = 11 \\ 6 x - 8 y - z = 15 \\ 3 x + 2 y - 6 z = 7 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

| A | = | \begin{array}{ccc} 5 & - 7 & 1 \\ 6 & - 8 & - 1 \\ 3 & 2 & - 6 \end{array} |

∵

(Taking first row for solving determinant)

\begin{aligned} = 5 (48 + 2) + 7 (- 36 + 3) + 1 (12 + 24) \\ = 5 (50) + 7 (- 33) + 1 (36) \\ = 250 - 231 + 36 \\ = 55 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 11 & - 7 & 1 \\ 15 & - 8 & - 1 \\ 7 & 2 & - 6 \end{array} | \\ = 11 (48 + 2) + 7 (- 90 + 7) + 1 (30 + 56) \\ = 11 (50) + 7 (- 83) + 86 \\ = 550 - 581 + 86 \\ = 55 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{y} & = | \begin{array}{ccc} 5 & 11 & 1 \\ 6 & 15 & - 1 \\ 3 & 7 & - 6 \end{array} | \\ = 5 (- 90 + 7) - 11 (- 36 + 3) + 1 (42 - 45) \\ = 5 (- 83) - 11 (- 33) + 1 (- 3) \end{aligned}

\begin{aligned} = - 415 + 363 - 3 \\ = - 55 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} & = | \begin{array}{ccc} 5 & - 7 & 11 \\ 6 & - 8 & 15 \\ 3 & 2 & 7 \end{array} | \\ = 5 (- 56 - 30) + 7 (42 - 45) + 11 (12 + 24) \\ = 5 (- 86) + 7 (- 3) + 11 (36) \\ = - 430 - 21 + 396 \\ = - 55 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{55}{55} = 1 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 55}{55} = - 1 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 55}{55} = - 1 \end{aligned}

deteminants Exercise 5.4 Question 17

Answer:

x = 2, y = - 3 and z = - 4

Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 2 x - 3 y - 4 z = 29 \\ - 2 x + 5 y - z = - 15 \\ 3 x - y + 5 z = - 11 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

D = | \begin{array}{ccc} 2 & - 3 & - 4 \\ - 2 & 5 & - 1 \\ 3 & - 1 & 5 \end{array} |

∵

(Taking first row for solving determinant)

\begin{aligned} = 2 (25 - 1) + 3 (- 10 + 3) - 4 (2 - 15) \\ = 2 (24) + 3 (- 7) - 4 (- 13) \\ = 48 - 21 + 52 \\ = 79 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 29 & - 3 & - 4 \\ - 15 & 5 & - 1 \\ - 11 & - 1 & 5 \end{array} | \\ = 29 (25 - 1) + 3 (- 75 - 11) - 4 (15 + 55) \\ = 29 (24) + 3 (- 86) - 4 (70) \\ = 696 - 258 - 280 \\ = 158 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 2 & 29 & - 4 \\ - 2 & - 15 & - 1 \\ 3 & - 11 & 5 \end{array} | \\ = 2 (- 75 - 11) - 29 (- 10 + 3) - 4 (22 + 45) \\ = 2 (- 86) - 29 (- 7) - 4 (67) \\ = - 172 + 203 - 268 \\ = - 237 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} = | \begin{array}{ccc} 2 & - 3 & 29 \\ - 2 & 5 & - 15 \\ 3 & - 1 & - 11 \end{array} | \\ = 2 (- 55 - 15) + 3 (22 + 45) + 29 (2 - 15) \\ = 2 (- 70) + 3 (67) + 29 (- 13) \\ = - 140 + 201 - 377 \\ = - 316 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{158}{79} = 2 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 237}{79} = - 3 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 316}{79} = - 4 \end{aligned}

determinants Exercise 5.4 Question 18

Answer: $x = - 2, y = 3 and z = - 4$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} x + y = 1 \\ x + z = - 6 \\ x - y - 2 z = 3 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

| A | = | \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & - 1 & - 2 \end{array} |

∵

(Taking first row for solving determinant)

\begin{aligned} = 1 (0 + 1) - 1 (- 2 - 1) + 0 (- 1 - 0) \\ = 1 + 3 \\ = 4 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\begin{aligned} D_{x} & = | \begin{array}{ccc} 1 & 1 & 0 \\ - 6 & 0 & 1 \\ 3 & - 1 & - 2 \end{array} | \\ = 1 (0 + 1) - 1 (12 - 3) + 0 (6 - 0) \\ = 1 - 9 \\ = - 8 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\begin{aligned} D_{y} & = | \begin{array}{ccc} 1 & 1 & 0 \\ 1 & - 6 & 1 \\ 1 & 3 & - 2 \end{array} | \\ = 1 (12 - 3) - 1 (- 2 - 1) + 0 (3 + 6) \\ = 9 + 3 \\ = 12 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} D_{z} & = | \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & - 6 \\ 1 & - 1 & 3 \end{array} | \\ = 1 (0 - 6) - 1 (3 + 6) + 1 (- 1 + 0) \\ = - 6 - 9 - 1 \\ = - 16 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 8}{4} = - 2 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{12}{4} = 3 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 16}{4} = - 4 \end{aligned}

Concept: Determinant solving of 3 x 3 matrix (Cramer’s rule)
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.

determinants Exercise 5.4 Question 19

Answer: $x = \frac{- (c - d) (d - b)}{(a - b) (c - a)}, y = \frac{- (a - d) (d - c)}{(a - b) (b - c)} and z = \frac{- (b - d) (d - a)}{(b - c) (c - a)}$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} x + y + z + 1 = 0 \\ a x + b y + c z + d = 0 \\ a^{2} x + b^{2} y + c^{2} z + d^{2} = 0 \end{aligned}$
Solution:
First take coefficient of variables x, y and z.

\begin{aligned} | A | = | \begin{array}{lll} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array} | \\ C_{2} \to C_{2} - C_{1} \\ C_{3} \to C_{3} - C_{1} \end{aligned}

= | \begin{array}{ccc} 1 & 0 & 0 \\ a & b - a & c - a \\ a^{2} & b^{2} - a^{2} & c^{2} - a^{2} \end{array} |

Now taking (b-a) and (c-a) from

C_{2}

and

C_{3}

respectively,

= (b - a) (c - a) | \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b + a & c + a \end{array} |

Expanding along

R_{1}

\begin{aligned} = (b - a) (c - a) [1 (c + a - b - a)] \\ = (b - a) (c - a) (c - b) \\ = (a - b) (b - c) (c - a) \end{aligned}

Now for x, the x column is replaced with constant column i.e.

D_{x} = | \begin{array}{ccc} 1 & 1 & 1 \\ d & b & c \\ d^{2} & b^{2} & c^{2} \end{array} |

\begin{aligned} C_{2} \to C_{2} - C_{1} \\ C_{3} \to C_{3} - C_{1} \end{aligned}

D_{x}

= - | \begin{array}{ccc} 1 & 0 & 0 \\ d & b - d & c - d \\ d^{2} & b^{2} - d^{2} & c^{2} - d^{2} \end{array} |

Now taking (b-d) and (c-d) from

C_{2}

and

C_{3}

respectively,

= - (b - d) (c - d) | \begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & b + d & c + d \end{array} |

Expanding along

R_{1}

\begin{aligned} = - (b - d) (c - d) [1 (c + d - b - d)] \\ = - (b - d) (c - d) (c - b) \\ = - (b - c) (c - d) (d - b) \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

D_{x} = | \begin{array}{lll} 1 & 1 & 1 \\ a & d & c \\ a^{2} & d^{2} & c^{2} \end{array} |

\begin{aligned} C_{2} \to C_{2} - C_{1} \\ C_{3} \to C_{3} - C_{1} \end{aligned}

D_{y} = | \begin{array}{ccc} 1 & 0 & 0 \\ a & d - a & c - a \\ a^{2} & d^{2} - a^{2} & c^{2} - a^{2} \end{array} |

Now taking (d-a) and (c-a) from

C_{2}

and

C_{3}

respectively,

= - (d - a) (c - a) | \begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & d + a & c + a \end{array} |

Expanding along

R_{1}

\begin{aligned} = - (d - a) (c - a) [1 (c + a - d - a)] \\ = - (d - a) (c - a) (c - d) \\ = - (a - d) (d - c) (c - a) \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

D_{z} = | \begin{array}{ccc} 1 & 1 & 1 \\ a & b & d \\ a^{2} & b^{2} & d^{2} \end{array} |

\begin{aligned} C_{2} \to C_{2} - C_{1} \\ C_{3} \to C_{3} - C_{1} \end{aligned}

\begin{aligned} D_{z} \\ = | \begin{array}{ccc} 1 & 0 & 0 \\ a & b - a & d - a \\ a^{2} & b^{2} - a^{2} & d^{2} - a^{2} \end{array} | \end{aligned}

Now taking (b-a) and (d-a) from

C_{2}

and

C_{3}

respectively,

= - (b - a) (d - a) | \begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b + a & d + a \end{array} |

Expanding along $R_{1},$

$\begin{aligned} = - (b - a) (d - a) [1 (d + a - b - a)] \\ = - (b - a) (d - a) (d - b) \\ = - (a - b) (b - d) (d - a) \end{aligned}$
By Cramer’s rule,
$\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- (b - c) (c - d) (d - b)}{(a - b) (b - c) (c - a)} = \frac{- (c - d) (d - b)}{(a - b) (c - a)} \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- (a - d) (d - c) (c - a)}{(a - b) (b - c) (c - a)} = \frac{- (a - d) (d - c)}{(a - b) (b - c)} \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- (a - b) (b - d) (d - a)}{(a - b) (b - c) (c - a)} = \frac{- (b - d) (d - a)}{(b - c) (c - a)} \end{aligned}$
Concept: Solving matrix of order 3x3 (Elementary row and column operations)

deteminants Exercise 5.4 Question 20

Answer: $x = - 2, y = 3, z = \frac{3}{2} and w = \frac{- 1}{2}$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} x + y + z + w = 2 \\ x - 2 y + 2 z + 2 w = - 6 \\ 2 x + y - 2 z + 2 w = - 5 \\ 3 x - y + 3 z - 3 w = - 3 \end{aligned}$
Solution:
Solving determinant,

| A | = | \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & - 2 & 2 & 2 \\ 2 & 1 & - 2 & 2 \\ 3 & - 1 & 3 & - 3 \end{array} |

\begin{aligned} C_{2} & \to C_{2} - C_{1} \\ C_{3} & \to C_{3} - C_{1} \\ C_{4} & \to C_{4} - C_{1} \end{aligned}

= | \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & - 3 & 1 & 1 \\ 2 & - 1 & - 4 & 0 \\ 3 & - 4 & 0 & - 6 \end{array} |

Expanding along

R_{1}

= 1 | \begin{array}{ccc} - 3 & 1 & 1 \\ - 1 & - 4 & 0 \\ - 4 & 0 & - 6 \end{array} |

\begin{aligned} C_{2} \to C_{2} - C_{3} \\ C_{1} \to C_{1} + 3 C_{3} \end{aligned}

\begin{aligned} = | \begin{array}{ccc} 0 & 0 & 1 \\ - 1 & - 4 & 0 \\ - 22 & 6 & - 6 \end{array} | \\ = 1 (- 6 - 88) \end{aligned}

= - 94

D_{x} = | \begin{array}{cccc} 2 & 1 & 1 & 1 \\ - 6 & - 2 & 2 & 2 \\ - 5 & 1 & - 2 & 2 \\ - 3 & - 1 & 3 & - 3 \end{array} | = 188

D_{y} = | \begin{array}{cccc} 1 & 2 & 1 & 1 \\ 1 & - 6 & 2 & 2 \\ 2 & - 5 & - 2 & 2 \\ 3 & - 3 & 3 & - 3 \end{array} | = - 282

D_{z} = | \begin{array}{cccc} 1 & 1 & 2 & 1 \\ 1 & - 2 & - 6 & 2 \\ 2 & 1 & - 5 & 2 \\ 3 & - 1 & - 3 & - 3 \end{array} | = - 141

D_{w} = | \begin{array}{cccc} 1 & 1 & 1 & 2 \\ 1 & - 2 & 2 & - 6 \\ 2 & 1 & - 2 & - 5 \\ 3 & - 1 & 3 & - 3 \end{array} | = 47

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{188}{- 94} = - 2 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 282}{- 94} = 3 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 141}{- 94} = \frac{3}{2} \\ \Rightarrow w = \frac{D_{w}}{D} = \frac{47}{- 94} = \frac{- 1}{2} \end{aligned}

Concept: Solving matrix of order 4x4 (Elementary row and column operations)

determinants Exercise 5.4 Question 21

Answer: $x = 1, y = \frac{2}{7}, z = \frac{2}{7} and w = \frac{- 1}{7}$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
$\begin{aligned} 2 x - 3 z + w = 1 \\ x - y + 2 w = 1 \\ - 3 y + z + w = 1 \\ x + y + z = 1 \end{aligned}$
Solution:
Solving determinant,

| A | = | \begin{array}{cccc} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} |

\begin{aligned} C_{2} & \to C_{2} - C_{1} \\ C_{3} & \to C_{3} - C_{1} \end{aligned}

= | \begin{array}{cccc} 2 & - 2 & - 5 & 1 \\ 1 & - 2 & - 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{array} |

Expanding along

R_{1}

= - 1 | \begin{array}{ccc} - 2 & - 5 & 1 \\ - 2 & - 1 & 2 \\ - 3 & 1 & 1 \end{array} |

\begin{aligned} C_{2} \to C_{2} - C_{3} \\ C_{1} \to C_{1} + 3 C_{3} \end{aligned}

\begin{aligned} = - 1 | \begin{array}{ccc} 1 & - 6 & 1 \\ 4 & - 3 & 2 \\ 0 & 0 & 1 \end{array} | \\ = - 1 (- 3 + 24) \\ = - 21 \end{aligned}

\begin{aligned} D_{x} = | \begin{array}{cccc} 1 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 2 \\ 1 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} | = - 21 \\ D_{y} = | \begin{array}{cccc} 2 & 1 & 1 & 0 \end{array} | = - 6 \\ D_{z} = | \begin{array}{cccc} 2 & 1 & - 3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & - 1 & 1 & 2 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} | = - 6 \\ D_{w} = | \begin{array}{cccc} 2 & 0 & - 3 & 1 \\ 1 & - 1 & 0 & 1 \\ 0 & - 3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array} | = 3 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 21}{- 21} = 1 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 6}{- 21} = \frac{2}{7} \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 6}{- 21} = \frac{2}{7} \\ \Rightarrow w = \frac{D_{w}}{D} = \frac{3}{- 21} = \frac{- 1}{7} \end{aligned}

Concept: Solving matrix of order 4x4 (Elementary row and column operations)

determinants Exercise 5.4 Question 22

Answer: $x = \infty and y = \infty$
Hint: Solving determinant gives zero.
Given:
$\begin{aligned} 2 x - y = 5 \\ 4 x - 2 y = 7 \end{aligned}$
Solution:
$2 x - y = 5$ .....(1)
$\begin{aligned} \Rightarrow 4 x - 2 y = 7 & \Rightarrow 2 (2 x - y) = 7 \\ \Rightarrow 2 x - y = \frac{7}{2} \end{aligned}$ .....(2)
Now, different value of 2x – y is not possible. So, the linear equations are inconsistent.
Solving determinant,

| A | = | \begin{array}{ll} 2 & - 1 \\ 4 & - 2 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = - 4 + 4 \\ D = 0 \end{aligned}

\begin{aligned} D_{x} = | \begin{array}{ll} 5 & - 1 \\ 7 & - 2 \end{array} | = - 10 + 7 = - 3 \neq 0 \\ D_{y} = | \begin{array}{ll} 2 & 5 \\ 4 & 7 \end{array} | = 14 - 20 = - 6 \neq 0 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 3}{0} = \infty \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 6}{0} = \infty \\ Since, D = 0 and D_{x} and D_{y} \neq 0 \\ ∴ Linear equations are inconsistent. \end{aligned}

Concept: Solving matrix of order 2x2 by solving linear equations
Note: When D = 0, there is either no solution or infinite solutions.

deteminants Exercise 5.4 Question 23

Answer:

x = \infty and y = \infty

Hint: Solving determinant gives zero.
Given:
$\begin{aligned} 3 x + y = 5 \\ - 6 x - 2 y = 9 \end{aligned}$
Solution:
$\begin{aligned} (3 x + y = 5) \times 2 \\ 6 x - 2 y = 10 \end{aligned}$ ....(1)
$\begin{matrix} 6 x + 2 y = 10 \\ - 6 x - 2 y = 9 \\ 0 \end{matrix}$
Hence, linear equations are inconsistent.
By Cramer’s rule:
Solving determinant,

\begin{aligned} | A | = | \begin{array}{cc} 3 & 1 \\ - 6 & - 2 \end{array} | = - 6 + 6 = 0 ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1}) \\ D = 0 \end{aligned}

\begin{aligned} D_{x} = | \begin{array}{cc} 5 & 1 \\ 9 & - 2 \end{array} | = - 10 - 9 = - 19 \neq 0 \\ D_{y} = | \begin{array}{cc} 3 & 5 \\ - 6 & 9 \end{array} | = 27 + 30 = 57 \neq 0 \\ By Cramer's rule, \\ \Rightarrow x = \frac{D_{x}}{D} = \frac{- 19}{0} = \infty \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{57}{0} = \infty \end{aligned}

Since,

D = 0

and

D_{x}

and

D_{y} \neq 0

∴

Linear equations are inconsistent.
Concept: Solving matrix of order 2x2 by solving linear equations
Note: When D = 0, there is either no solution or infinite solutions.

deteminants Exercise 5.4 Question 23

Answer:

x = \infty and y = \infty

\begin{aligned} | A | = | \begin{array}{cc} 3 & 1 \\ - 6 & - 2 \end{array} | = - 6 + 6 = 0 ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1}) \\ D = 0 \end{aligned}

\begin{aligned} D_{x} = | \begin{array}{cc} 5 & 1 \\ 9 & - 2 \end{array} | = - 10 - 9 = - 19 \neq 0 \\ D_{y} = | \begin{array}{cc} 3 & 5 \\ - 6 & 9 \end{array} | = 27 + 30 = 57 \neq 0 \\ By Cramer's rule, \\ \Rightarrow x = \frac{D_{x}}{D} = \frac{- 19}{0} = \infty \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{57}{0} = \infty \end{aligned}

Since,

D = 0

and

D_{x}

and

D_{y} \neq 0

∴

Linear equations are inconsistent.
Concept: Solving matrix of order 2x2 by solving linear equations
Note: When D = 0, there is either no solution or infinite solutions.

determinants Exercise 5.4 Question 24

Answer:

x = \infty, y = \infty and z = \infty

Hint: Solving determinant gives zero.
Given:
$\begin{aligned} 3 x - y + 2 z = 3 \\ 2 x + y + 3 z = 5 \\ x - 2 y - z = 1 \end{aligned}$
Solution:
By Cramer’s rule:
Solving determinant,

| A | = | \begin{array}{ccc} 3 & - 1 & 2 \\ 2 & 1 & 3 \\ 1 & - 2 & - 1 \end{array} |

Expanding along

1^{s t}

row,

\begin{aligned} = 3 (- 1 + 6) + 1 (- 2 - 3) + 2 (- 4 - 1) \\ = 3 (5) + 1 (- 5) + 2 (- 5) \\ = 15 - 5 - 10 \\ = 0 \end{aligned}

\begin{aligned} D & = 0 \\ D_{x} & = | \begin{array}{ccc} 3 & - 1 & 2 \\ 5 & 1 & 3 \\ 1 & - 2 & - 1 \end{array} | \\ = 3 (- 1 + 6) + 1 (- 5 - 3) + 2 (- 10 - 1) \\ = 3 (5) + 1 (- 8) + 2 (- 11) \\ = 15 - 8 - 22 \\ = - 15 \neq 0 \end{aligned}

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 19}{0} = \infty \\ Since, D = 0 and D_{x}, D_{y} and D_{z} \neq 0 \end{aligned}

∴

Linear equations are inconsistent.
Concept: Solving matrix of order 3x3 by solving linear equations
Note: When D = 0, there is either no solution or infinite solutions.

determinants Exercise 5.4 Question 25

Answer:

x = - 3, y = - 1 and z = 7

Hint: Solving determinant gives zero.
Given:
$\begin{aligned} 3 x - y + 2 z = 6 \\ 2 x - y - z = 2 \\ 3 x + 6 y + 5 z = 20 \end{aligned}$
Solution:
By Cramer’s rule:
Solving determinant,

| A | = | \begin{array}{ccc} 3 & - 1 & 2 \\ 2 & - 1 & 1 \\ 3 & 6 & 5 \end{array} |

Expanding along

1^{s t}

row,

\begin{aligned} = 3 (- 5 - 6) + 1 (10 - 3) + 2 (12 + 3) \\ = - 33 + 7 + 30 \\ = 4 \end{aligned}

\begin{aligned} D_{\bar{x}} = | \begin{array}{ccc} 6 & - 1 & 2 \\ 2 & - 1 & 1 \\ 20 & 6 & 5 \end{array} | = - 12 \\ D_{y} = | \begin{array}{ccc} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{array} | = - 4 \end{aligned}

D_{z} = | \begin{array}{ccc} 3 & - 1 & 6 \\ 2 & - 1 & 2 \\ 3 & 6 & 20 \end{array} | = 28

By Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 12}{4} = - 3 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 4}{4} = - 1 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{28}{4} = 7 \end{aligned}

Concept: Solving matrix of order 3x3 by solving linear equations

determinants Exercise 5.4 Question 26

Answer :

x = \frac{5}{3}, y = k - \frac{4}{3} and z = k

Hint: Use Cramer’s rule for system of linear equations.
Given:
$\begin{aligned} x - y + z = 3 \\ 2 x + y - z = 2 \\ - x - 2 y + 2 z = 1 \end{aligned}$
Solution:

\begin{matrix} AX = B \\ [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2 \end{array}] [\begin{array}{l} x \\ y \\ z \end{array}] = [\begin{array}{l} 3 \\ 2 \\ 1 \end{array}] \\ Where A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 3 \\ 2 \\ 1 \end{array}] \end{matrix}

Solving determinant,

| A | = | \begin{array}{ccc} 1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2 \end{array} |

Expanding along

1^{s t}

row,

\begin{aligned} = 1 (2 - 2) + 1 (4 - 1) + 1 (- 4 + 1) \\ = 0 + 3 - 3 \end{aligned}

|A| = 0

\Rightarrow

System of linear equations have infinite number of solutions.
Let z = k

\Rightarrow x - y + k = 3

...(1)

\Rightarrow 2 x + y - k = 2

.....(2)
From (1) and (2),

\Rightarrow x - y = 3 - k

......(3)

\Rightarrow 2 x + y = 2 + k

.......(4)
Adding (3) and (4),

\begin{aligned} \Rightarrow 3 x = 5 \\ ∴ x = \frac{5}{3} \end{aligned}

From (3),

\Rightarrow \frac{5}{3} - y = 3 - k

\begin{aligned} y = \frac{5}{3} - 3 + k \\ y = \frac{5 - 9 + 3 k}{3} \\ y = \frac{3 k - 4}{3} \\ y = k - \frac{4}{3} \\ ∴ z = k \end{aligned}

Concept: Solving matrix of order 3x3 by solving linear equations
Note: When D = 0, there is either no solution or infinite solutions.

determinants Exercise 5.4 Question 27

Answer:

x = 5 - 2 k, y = k

Hint: Use Cramer’s rule for system of linear equations.
Given:
$\begin{aligned} x + 2 y = 5 \\ 3 x + 6 y = 15 \end{aligned}$
Solution:
Solving determinant,

| A | = | \begin{array}{ll} 1 & 2 \\ 3 & 6 \end{array} | ∵ | \begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array} | = (a_{1} b_{2} - a_{2} b_{1})

\begin{aligned} = 6 - 6 \\ = 0 \end{aligned}

|A| = 0

\Rightarrow

System of linear equations have infinite number of solutions.
Now for x, the x column is replaced with constant column i.e.

\Rightarrow D_{x} = | \begin{array}{cc} 5 & 2 \\ 15 & 6 \end{array} | = 30 - 30 = 0

If we are solving for y, the y column is replaced with constant column i.e.

\Rightarrow D_{x} = | \begin{array}{cc} 1 & 5 \\ 3 & 15 \end{array} | = 15 - 15 = 0

Since, D = D_{x} = D_{x} = 0

Let y = k, then we have:

$\begin{aligned} \Rightarrow x + 2 k = 5 \\ \Rightarrow x = 5 - 2 k \end{aligned}$ are the infinitive solutions of the given system.
Concept: Solving matrix of order 2x2 by Cramer’s rule.
Note: When D = 0, there is either no solution or infinite solutions.

deteminants Exercise 5.4 Question 28

Answer:

x = k, y = 2 k and z = 3 k

Hint: Use Cramer’s rule for system of linear equations.
Given:
$\begin{aligned} x + y - z = 0 \\ x - 2 y + z = 0 \\ 3 x + 6 y - 5 z = 0 \end{aligned}$
Solution:
Solving determinant,

\begin{aligned} | A | & = | \begin{array}{ccc} 1 & 1 & - 1 \\ 1 & - 2 & 1 \\ 3 & 6 & - 5 \end{array} | \\ = 1 (10 - 6) - 1 (- 5 - 3) - 1 (6 + 6) \\ = 4 + 8 - 12 \\ = 0 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\Rightarrow D_{x} = | \begin{array}{ccc} 0 & 1 & - 1 \\ 0 & - 2 & 1 \\ 0 & 6 & - 5 \end{array} | = 0

If we are solving for y, the y column is replaced with constant column i.e.

\Rightarrow D_{y} = | \begin{array}{ccc} 1 & 0 & - 1 \\ 1 & 0 & 1 \\ 3 & 0 & - 5 \end{array} | = 0

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} \Rightarrow D_{z} = | \begin{array}{ccc} 1 & 1 & 0 \\ 1 & - 2 & 0 \\ 3 & 6 & 0 \end{array} | = 0 \\ So, D = D_{x} = D_{y} = D_{z} = 0 \end{aligned}

The given system has either infinite solutions or it is inconsistent.

\begin{aligned} x + y = z \\ x - 2 y = - z \end{aligned}

Using Cramer’s rule,

\begin{aligned} D = | \begin{array}{cc} 1 & 1 \\ 1 & - 2 \end{array} | = - 2 - 1 = - 3 \\ D_{x} = | \begin{array}{cc} z & 1 \\ - z & - 2 \end{array} | = - 2 z + z = - z \end{aligned}

\begin{aligned} D_{y} = | \begin{array}{cc} 1 & z \\ 1 & - z \end{array} | = - z - z = - 2 z \\ \Rightarrow x = \frac{D_{x}}{D} = \frac{- z}{- 3} = \frac{z}{3} \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 2 z}{- 3} = \frac{2 z}{3} \end{aligned}

Let z = 3k, then x = k and y = 2k
Concept: Solving matrix of order 3x3 by Cramer’s rule.
Note: When D = 0, there is either no solution or infinite solutions.

deteminants Exercise 5.4 Question 29

Answer:

x = \frac{3 k + 6}{5}, y = \frac{8 + 4 k}{5} and z = k

Hint: Use Cramer’s rule for system of linear equations.
Given:
$\begin{aligned} 2 x + y - 2 z = 4 \\ x - 2 y + z = - 2 \\ 5 x - 5 y + z = - 2 \end{aligned}$
Solution:
Solving determinant,

\begin{aligned} | A | = | \begin{array}{ccc} 2 & 1 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 5 & 1 \end{array} | \\ = 2 (- 2 + 5) - 1 (1 - 5) - 2 (- 5 + 10) \\ = 6 + 4 - 10 \\ = 0 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\Rightarrow D_{x} = | \begin{array}{ccc} 4 & 1 & - 2 \\ - 2 & - 2 & 1 \\ - 2 & - 5 & 1 \end{array} |

Taking 2 common from

C_{1}

= - 2 | \begin{array}{ccc} - 2 & 1 & - 2 \\ 1 & - 2 & 1 \\ 1 & - 5 & 1 \end{array} | = 0 (∵ C_{1} = C_{3})

If we are solving for y, the y column is replaced with constant column i.e.

\begin{array}{r} \Rightarrow D_{x} = | \begin{array}{ccc} 2 & 4 & - 2 \\ 1 & - 2 & 1 \\ 5 & - 2 & 1 \end{array} | \\ C_{2} \to C_{2} + 2 C_{3} \end{array}

= | \begin{array}{ccc} 2 & 0 & - 2 \\ 1 & 0 & 1 \\ 5 & 0 & 1 \end{array} | = 0 (∵ Expanding along R_{1})

If we are solving for z, the z column is replaced with constant column i.e.

\begin{aligned} \Rightarrow D_{z} = | \begin{array}{ccc} 2 & 1 & 4 \\ 1 & - 2 & - 2 \\ 5 & - 5 & - 2 \end{array} | Expanding along R_{1}, \\ = & 2 (4 - 10) - 1 (- 2 + 10) + 4 (- 5 + 10) \\ = & - 12 - 8 + 20 \\ = & 0 \end{aligned}

So,

D = D_{x} = D_{x} = D_{2} = 0

The given system has either infinite solutions or it is inconsistent.

\begin{aligned} x - 2 y = - 2 - z \\ 5 x - 5 y = - 2 - z \end{aligned}

Using Cramer’s rule,

\begin{aligned} D = | \begin{array}{ll} 1 & - 2 \\ 5 & - 5 \end{array} | = - 5 + 10 = 5 \\ D_{x} = | \begin{array}{ll} - 2 - z & - 2 \\ - 2 - z & - 5 \end{array} | = 10 + 5 z - 2 z - 4 = 3 z + 6 \\ D_{x} = | \begin{array}{ll} 1 & - 2 - z \\ 5 & - 2 - z \end{array} | = - 2 - z + 10 + 5 z = 8 + 4 z \\ \Rightarrow x = \frac{D_{x}}{D} = \frac{3 z + 6}{5} \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{8 + 4 z}{5} \\ Let z = k, then x = \frac{3 z + 6}{5} and y = \frac{8 + 4 z}{5} \end{aligned}

Concept: Solving matrix of order 3x3 by Cramer’s rule.
Note: When D = 0, there is either no solution or infinite solutions.

determinants Exercise 5.4 Question 30

Answer: $x = \frac{7 - 3 z}{2}, y = \frac{3 z - 5}{2} and z = k$
Hint: Use Cramer’s rule for system of linear equations.
Given:
$\begin{aligned} x - y + 3 z = 6 \\ x + 3 y - 3 z = - 4 \\ 5 x + 3 y + 3 z = 10 \end{aligned}$
Solution:
Solving determinant,

\begin{aligned} | A | & = | \begin{array}{ccc} 1 & - 1 & 3 \\ 1 & 3 & - 3 \\ 5 & 3 & 3 \end{array} | \\ = 1 (9 + 9) + 1 (3 + 15) + 3 (3 - 15) \\ = 18 + 18 - 36 \\ = 0 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\Rightarrow D_{x} = | \begin{array}{ccc} 6 & - 1 & 3 \\ - 4 & 3 & - 3 \\ 10 & 3 & 3 \end{array} |

\begin{aligned} R_{1} \to R_{1} + R_{2} \\ R_{3} \to R_{3} + R_{2} \end{aligned}

= | \begin{array}{ccc} 2 & 2 & 0 \\ - 4 & 3 & - 3 \\ 6 & 6 & 0 \end{array} | = 0 (∵ Expanding along R_{1})

If we are solving for y, the y column is replaced with constant column i.e.

\Rightarrow D_{y} = | \begin{array}{ccc} 1 & 6 & 3 \\ 1 & - 4 & - 3 \\ 5 & 10 & 3 \end{array} |

\begin{aligned} R_{1} \to R_{1} + R_{2} \\ R_{3} \to R_{3} + R_{2} \end{aligned}

= | \begin{array}{ccc} 2 & 2 & 0 \\ 1 & - 4 & - 3 \\ 6 & 6 & 0 \end{array} | = 0 (∵ Expanding along R_{1})

If we are solving for z, the z column is replaced with constant column i.e.

\Rightarrow D_{z} = | \begin{array}{ccc} 1 & - 1 & 6 \\ 1 & 3 & - 4 \\ 5 & 3 & 10 \end{array} |

\begin{aligned} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - 5 R_{1} \end{aligned}

\begin{aligned} = | \begin{array}{ccc} 1 & - 1 & 6 \\ 0 & 4 & - 10 \\ 0 & 8 & - 20 \end{array} | \\ = 0 (∵ Expanding along R_{1}) \end{aligned}

So,

D = D_{x} = D_{y} = D_{z} = 0

The given system has either infinite solutions or it is inconsistent.

\begin{aligned} x - y = 6 - 3 z \\ x + 3 y = - 4 + 3 z \end{aligned}

Using Cramer’s rule,

\begin{aligned} D = | \begin{array}{cc} 1 & - 1 \\ 1 & 3 \end{array} | = 3 + 1 = 4 \\ D_{x} = | \begin{array}{cc} 6 - 3 z & - 1 \\ - 4 + 3 z & 3 \end{array} | = 18 - 9 z - 4 + 3 z = 14 - 6 z \\ D_{x} = | \begin{array}{lc} 1 & 6 - 3 z \\ 1 & - 4 + 3 z \end{array} | = - 4 + 3 z - 6 + 3 z = 6 z - 10 \\ \Rightarrow x = \frac{D_{x}}{D} = \frac{14 - 6 z}{4} = \frac{7 - 3 z}{2} \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{6 z - 10}{4} = \frac{3 z - 5}{2} \\ Let z = k, then x = \frac{7 - 3 z}{2} and y = \frac{3 z - 5}{2} \end{aligned}

Concept: Solving matrix of order 3x3 by Cramer’s rule.
Note: When D = 0, there is either no solution or infinite solutions.

determinants Exercise 5.4 Question 31

Answer: $x = 2, y = 4 and z = 11$
Hint: Use Cramer’s rule for system of linear equations.
Given:

Months	sale of unit			Total commission drawn
	A	B	C
Jan	90	100	20	800
Feb	130	50	40	900
March	60	100	30	850

Solution:
To form linear equation, let the rates of commissions on items A, B and C be x, y and z respectively. This can be expressed as a system of linear equations

\begin{aligned} 90 x + 100 y + 20 z = 800 \\ 130 x + 50 y + 40 z = 900 \\ 60 x + 100 y + 30 z = 850 \end{aligned}

By Cramer’s rule, solving determinant:

D = | \begin{array}{ccc} 90 & 100 & 20 \\ 130 & 50 & 40 \\ 60 & 100 & 30 \end{array} |

\begin{aligned} R_{1} \to R_{1} - 2 R_{2} \\ R_{3} \to R_{3} - 2 R_{2} \end{aligned}

= | \begin{array}{ccc} - 170 & 0 & - 60 \\ 130 & 50 & 40 \\ - 200 & 0 & - 50 \end{array} |

\begin{aligned} = - 170 (- 2500) - 60 (10000) \\ = 425000 - 600000 \\ = - 175000 \end{aligned}

Now for x, the x column is replaced with constant column i.e.

\Rightarrow D_{\bar{x}} = | \begin{array}{ccc} 800 & 100 & 20 \\ 900 & 50 & 40 \\ 850 & 100 & 30 \end{array} |

\begin{aligned} R_{1} \to R_{1} - 2 R_{2} \\ R_{3} \to R_{3} - 2 R_{2} \end{aligned}

= | \begin{array}{ccc} - 1000 & 0 & - 60 \\ 900 & 50 & 40 \\ - 950 & 0 & - 50 \end{array} |

\begin{aligned} = - 1000 (- 2500) - 60 (47500) \\ = 2500000 - 2850000 \\ = - 350000 \end{aligned}

If we are solving for y, the y column is replaced with constant column i.e.

\Rightarrow D_{y} = | \begin{array}{ccc} 90 & 800 & 20 \\ 130 & 900 & 40 \\ 60 & 850 & 30 \end{array} |

\begin{aligned} R_{2} \to R_{2} - 2 R_{1} \\ R_{3} \to R_{3} - \frac{3}{2} R_{1} \end{aligned}

\begin{aligned} = | \begin{array}{ccc} 90 & 800 & 20 \\ - 50 & - 700 & 0 \\ - 75 & - 350 & 0 \end{array} | \\ = 20 (17500 - 52500) \\ = - 700000 \end{aligned}

If we are solving for z, the z column is replaced with constant column i.e.

\Rightarrow D_{z} = | \begin{array}{ccc} 90 & 100 & 800 \\ 130 & 50 & 900 \\ 60 & 100 & 850 \end{array} |

\begin{aligned} R_{1} \to R_{1} - 2 R_{2} \\ R_{3} \to R_{3} - 2 R_{2} \end{aligned}

= | \begin{array}{ccc} - 170 & 0 & - 1000 \\ 130 & 50 & 900 \\ - 200 & 0 & - 950 \end{array} |

\begin{aligned} = - 170 (- 47500) - 1000 (10000) \\ = 8075000 - 10000000 \\ = - 1925000 \end{aligned}

Using Cramer’s rule,

\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{- 350000}{- 175000} = 2 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{- 700000}{- 175000} = 4 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{- 1925000}{- 175000} = 11 \end{aligned}

∴

The rates of commission of items A, B and C are 2%, 4% and 11% respectively.
Concept: Solving matrix of order 3x3 by Cramer’s rule.

determinants Exercise 5.4 Question 32

Answer: $x = 2, y = 3 and z = 4$
Hint: Use Cramer’s rule for system of linear equations.
Given:
Solution:
To form linear equation, let the rates of commissions on items A, B and C be x, y and z respectively. This can be expressed as a system of linear equations.
$\begin{aligned} 2 x + 3 y + 4 z = 29 \\ x + y + 2 z = 13 \\ 3 x + 2 y + z = 16 \end{aligned}$
Where x, y and z are number of cars $C_{1}, C_{2} a n d C_{3}$ respectively.
By Cramer’s rule, solving determinant:
$\begin{aligned} D = & | \begin{array}{lll} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{array} | \\ R_{1} \to R_{1} - 4 R_{3} \\ R_{2} \to R_{2} - 2 R_{3} \\ = | \begin{array}{ccc} - 10 & - 5 & 0 \\ - 5 & - 3 & 0 \\ 3 & 2 & 1 \end{array} | \\ = & 30 - 25 \\ = & 5 \end{aligned}$
Now for x, the x column is replaced with constant column i.e.
$\begin{aligned} \Rightarrow D_{\bar{x}} = & | \begin{array}{ccc} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{array} | \\ \begin{aligned} R_{1} & \to R_{1} - 4 R_{3} \\ R_{2} & \to R_{2} - 2 R_{3} \end{aligned} \\ = & | \begin{array}{lll} - 35 & - 5 & 0 \\ - 19 & - 3 & 0 \\ 16 & 2 & 1 \end{array} | \\ = & 190 - 175 \\ = & 15 \end{aligned}$
If we are solving for y, the y column is replaced with constant column i.e.
$\begin{array}{r} \Rightarrow D_{y} = | \begin{array}{lll} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{array} | \\ R_{1} \to R_{1} - 4 R_{3} \\ R_{2} \to R_{2} - 2 R_{3} \end{array}$
$\begin{aligned} = | \begin{array}{ccc} - 10 & - 35 & 0 \\ - 5 & - 19 & 0 \\ 3 & 16 & 1 \end{array} | \\ = 190 - 175 \end{aligned}$
$= 15$
If we are solving for z, the z column is replaced with constant column i.e.
$\begin{array}{r} \Rightarrow D_{z} = | \begin{array}{lll} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} | \\ R_{1} \to R_{1} - (R_{2} + R_{3}) \end{array}$
$= | \begin{array}{ccc} - 2 & 0 & 0 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} |$
$\begin{aligned} = - 2 (16 - 26) \\ = 20 \end{aligned}$
Using Cramer’s rule,
$\begin{aligned} \Rightarrow x = \frac{D_{x}}{D} = \frac{10}{5} = 2 \\ \Rightarrow y = \frac{D_{y}}{D} = \frac{15}{5} = 3 \\ \Rightarrow z = \frac{D_{z}}{D} = \frac{20}{5} = 4 \end{aligned}$
$∴$ The number of cars produced of type $C_{1}, C_{2} a n d C_{3}$ are 2, 3 and 4 respectively.
Concept: Solving matrix of order 3x3 by Cramer’s rule.

Class 12, mathematics chapter 5, Determinants, has around five exercises. RD Sharma class 12th exercise 5.4, gets deeper into the topic of determinants. This exercise covers concepts like Cramer's rule, Systems of linear equations has an infinite number of equations, Inconsistent Linear Equations, and Application-based questions on determinants. There are 32 questions in this exercise, including the subparts and the word problems. Hence, scads of time are required to solve the problems without a guide. Here is where the RD Sharma Class 12 Chapter 5 Exercise 5.4 comes to the rescue.

Students can use RD Sharma Class 12th exercise 5.4 solution with confidence as the answers in this book are provided by educational experts. It follows the NCERT pattern making it beneficial for the CBSE board students to use it. Especially to score more marks in the Determinants chapter, the Class 12 RD Sharma Chapter 5 Exercise 5.4 Solution will be of great help. As you start practicing with this book, you will soon see yourself crossing your benchmark scores. You can use it while solving homework, assignments, and even while preparing for exams.

Even though the RD Sharma book offers many benefits, the students need not pay even a single penny to own a copy. It can be downloaded for free from the Career360 website. The RD Sharma Class 12 Solutions Determinants Ex 5.4 will help you solve all your doubts while doing your homework and assignments. Moreover, there is a high probability that the question for the public exam will be asked from this book. Therefore, using it as your reference guide and working out the practice questions would certainly help you during the exam.

Many students have benefitted from using the RD Sharma Class 12 Solutions Chapter 5 ex 5.4 to understand the concepts better. They can download a copy of RD Sharma Class 12th exercise 5.4 Solutions for all the subjects and chapters and prepare for the exams every day by referring to the solutions given in these books. Using the best solution book as the guide, students will get higher scores than their previous performances.

Chapter-wise RD Sharma Class 12 Solutions

Frequently Asked Questions (FAQs)

1. Where should I check for the correct answers for the Class 12 mathematics Determinants chapter?

You can refer to the RD Sharma Class 12th exercise 5.4 solutions book to clarify your doubts on the Determinants concept. The solutions given by the experts will clear all your queries.

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The RD Sharma Class 12th exercise 5.4 solution books are available free of cost on the Career360 website. Anyone can download it for free from this website.

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Visit the official Career360 website and select class 12. Then, search for the Subject mathematics and find the RD Sharma Class 12 Chapter 5 Exercise 5.4 solution. Then, download it and refer to it from your device at any time.

4. Is it enough if I depend on the RD Sharma books for the public exam preparation?

Yes, the RD Sharma solution books are enough for the class 12 students to prepare for the public examinations. It consists of answers for every question and numerous practice questions for the students to work out.

5. How many exercises are given in the RD Sharma class 12 mathematics chapter 5 solutions?

There are five exercises, ex 5.1 to 5.5, given in the textbook. The RD Sharma Class 12-chapter, five mathematics books, consists of solutions for all these exercises.

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