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

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

Edited By Kuldeep Maurya | 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

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## Determinants Excercise: 5.4

Determinants Exercise 5.4 Question 1

Answer: $\mathrm{x}=-6 \text { and } \mathrm{y}=-5$
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
$\mathrm{D}=\left|\begin{array}{cc} 1 & -2 \\ -3 & 5 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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} &\mathrm{D}_{1}=\left|\begin{array}{cc} 4 & -2 \\ -7 & 5 \end{array}\right| \\ &=(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.
$\mathrm{D}_{2}=\left|\begin{array}{cc} 1 & 4 \\ -3 & -7 \end{array}\right|$

Now,$\mathrm{x}=\frac{D_{1}}{D}=\frac{6}{-1}=-6$
$\mathrm{y}=\frac{D_{2}}{D}=\frac{5}{-1}=-5$
Hence,$\mathrm{x}=-6 \text { and } \mathrm{y}=-5$
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:$x=-3 \text { and } y=-7$
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
$\mathrm{D}=\left|\begin{array}{ll} 2 & -1 \\ 7 & -2 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$\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.

$\mathrm{D}_{1}=\left|\begin{array}{cc} 1 & -1 \\ -7 & -2 \end{array}\right|$

\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} &\mathrm{D}_{2}=\left|\begin{array}{cc} 2 & 1 \\ 7 & -7 \end{array}\right| \\ &=(2)(-7)-(7)(1) \\ &=-14-7 \\ &=-21 \end{aligned}
Now, $\mathrm{x}=\frac{D_{1}}{D}=\frac{-9}{3}=-3$
$\mathrm{y}=\frac{D_{2}}{D}=\frac{-21}{3}=-7$
$Hence, x = -3 and y= -7.$
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:$\mathrm{x}=7 \text { and } \mathrm{y}=-3$
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
$\mathrm{D}=\left|\begin{array}{cc} 2 & -1 \\ 3 & 5 \end{array}\right| \quad \because\left|\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$\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.

$\mathrm{D}_{1}=\left|\begin{array}{cc} 17 & -1 \\ 6 & 5 \end{array}\right|$

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

$\mathrm{y}=\frac{D_{2}}{D}=\frac{-39}{13}=-3$
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:$x=7 \text { and } y=-2$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: \begin{aligned} &3 \mathrm{x}+\mathrm{y}=19 \\ &3 \mathrm{x}-\mathrm{y}=23 \end{aligned}
Solution:
First D: determinant of the coefficient matrix
$\mathrm{D}=\left|\begin{array}{cc} 3 & 1 \\ 3 & -1 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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.
$\mathrm{D}_{1}=\left|\begin{array}{cc} 19 & 1 \\ 23 & -1 \end{array}\right|$
\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} \mathrm{D}_{2} &=\left|\begin{array}{ll} 3 & 19 \\ 3 & 23 \end{array}\right| \\ &=(3)(23)-(19)(3) \\ &=69-57 \\ &=12 \end{aligned}
\begin{aligned} &\text { Now, } \mathrm{x}=\frac{D_{1}}{D}=\frac{-42}{-6}=7 \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{12}{-6}=-2 \end{aligned}
Hence $x=7 \text { and } y=-2$
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 5

Answer:$\mathrm{x}=\frac{-5}{11} \text { and } \mathrm{y}=\frac{12}{11}$
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
$\mathrm{D}=\left|\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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.
$\mathrm{D}_{1}=\left|\begin{array}{cc} -2 & -1 \\ 3 & 4 \end{array}\right|$
\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.
$\mathrm{D}_{2}=\left|\begin{array}{cc} 2 & -2 \\ 3 & 3 \end{array}\right|$
\begin{aligned} &=(2)(3)-(3)(-2) \\ &=6+6 \\ &=12 \end{aligned}
Now, \begin{aligned} &\mathrm{x}=\frac{D_{1}}{D}=\frac{-5}{11} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{12}{11} \end{aligned}
Hence,$\mathrm{x}=\frac{-5}{11} \text { and } \mathrm{y}=\frac{12}{11}$

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 6

Answer:$\mathrm{x}=2 \text { and } \mathrm{y}=\frac{-2}{a}$
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} \mathrm{D} &=\left|\begin{array}{ll} 3 & a \\ 2 & a \end{array}\right| & \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right) \\ &=3 \mathrm{a}-2 \mathrm{a} \\ &=\mathrm{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} \mathrm{D}_{1} &=\left|\begin{array}{ll} 4 & a \\ 2 & a \end{array}\right| \\ &=4 \mathrm{a}-2 \mathrm{a} \\ &=2 \mathrm{a} \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
\begin{aligned} &\mathrm{D}_{2}=\left|\begin{array}{ll} 3 & 4 \\ 2 & 2 \end{array}\right|\\ &\text { (2) }\\ &\begin{aligned} &=(2)(3)-(4) \\ &=6-8 \\ &=-2 \end{aligned} \end{aligned}
Now,
\begin{aligned} \mathrm{x}=\frac{D_{1}}{D}=\frac{2 a}{a}=2 \\ \mathrm{y}=\frac{D_{2}}{D}=\frac{-2}{a} \end{aligned}
Hence,$\mathrm{x}=2 \text { and } \mathrm{y}=\frac{-2}{a}$
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:$x=\frac{16}{3} \text { and } y=\frac{-2}{9}$
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
$\mathrm{D}=\left|\begin{array}{ll} 2 & 3 \\ 1 & 6 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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} &\mathrm{D}_{1}=\left|\begin{array}{cc} 10 & 3 \\ 4 & 6 \end{array}\right| \\ &=60-12 \\ &=48 \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
\begin{aligned} \mathrm{D}_{2} &=\left|\begin{array}{ll} 2 & 10 \\ 1 & 4 \end{array}\right| \\ &=8-10 \\ &=-2 \end{aligned}
Now,
$\begin{gathered} \mathrm{x}=\frac{D_{1}}{D}=\frac{48}{9}=\frac{16}{3} \\ \mathrm{y}=\frac{D_{2}}{D}=\frac{-2}{9} \end{gathered}$
Hence,$x=\frac{16}{3} \text { and } y=\frac{-2}{9}$
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

deteminants Exercise 5.4 Question 8

Answer: $x=\frac{9}{2} \text { and } y=\frac{-7}{2}$
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
$\mathrm{D}=\left|\begin{array}{ll} 5 & 7 \\ 4 & 6 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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} \mathrm{D}_{1} &=\left|\begin{array}{rr} -2 & 7 \\ -3 & 6 \end{array}\right| \\ &=-12+21 \\ &=9 \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
\begin{aligned} &\mathrm{D}_{2}=\left|\begin{array}{ll} 5 & -2 \\ 4 & -3 \end{array}\right| \\ &\quad=-15+8 \\ &=-7 \\ &\text { Now, } \mathrm{x}=\frac{D_{1}}{D}=\frac{9}{2} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{-7}{2} \end{aligned}
Hence,$x=\frac{9}{2} \text { and } y=\frac{-7}{2}$

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 9

Answer:$\mathrm{x}=\frac{-10}{37} \text { and } \mathrm{y}=\frac{92}{37}$
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
$\mathrm{D}=\left|\begin{array}{cc} 9 & 5 \\ -2 & 3 \end{array}\right|$ $\because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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} \mathrm{D}_{1} &=\left|\begin{array}{cc} 10 & 5 \\ 8 & 3 \end{array}\right| \\ &=30-40 \\ &=-10 \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
\begin{aligned} &\mathrm{D}_{2}=\left|\begin{array}{cc} 9 & 10 \\ -2 & 8 \end{array}\right| \\ &=72+20 \\ &=92 \\ &\text { Now, } \mathrm{x}=\frac{D_{1}}{D}=\frac{-10}{37} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{92}{37} \end{aligned}
Hence
$y=92 / 37 \text { and } x=-10 / 37$
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 10

Answer:$x=\frac{7}{5} \text { and } y=\frac{-1}{5}$
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
$\mathrm{D}=\left|\begin{array}{ll} 1 & 2 \\ 3 & 1 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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} \mathrm{D}_{1} &=\left|\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right| \\ &=1-8 \\ &=-7 \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
\begin{aligned} \mathrm{D}_{2} &=\left|\begin{array}{ll} 1 & 1 \\ 3 & 4 \end{array}\right| \\ &=4-3 \\ &=1 \end{aligned}
Now,\begin{aligned} &\mathrm{x}=\frac{D_{1}}{D}=\frac{-7}{-5}=\frac{7}{5} \\ &\mathrm{y}=\frac{D_{2}}{D}=\frac{1}{-5}=\frac{-1}{5} \end{aligned}
Hence,$x=\frac{7}{5} \text { and } y=\frac{-1}{5}$
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:$x=-1, y=2 \text { and } z=3$
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.
$|\mathrm{A}|=\left|\begin{array}{ccc} 3 & 1 & 1 \\ 2 & -4 & 3 \\ 4 & 1 & -3 \end{array}\right|$$\therefore$(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.
$\mathrm{D}_{\mathbf{x}}=\left|\begin{array}{ccc} 2 & 1 & 1 \\ -1 & -4 & 3 \\ -11 & 1 & -3 \end{array}\right|$
\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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 3 & 2 & 1 \\ 2 & -1 & 3 \\ 4 & -11 & -3 \end{array}\right| \\ &=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} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 3 & 1 & 2 \\ 2 & -4 & -1 \\ 4 & 1 & -11 \end{array}\right| \\ &=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:$x=-1, y=-5 \text { and } z=8$
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.
$|\mathrm{A}|=\left|\begin{array}{ccc} 1 & -4 & -1 \\ 2 & -5 & 2 \\ -3 & 2 & 1 \end{array}\right|$ $\because$(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} &=\left|\begin{array}{ccc} 11 & -4 & -1 \\ 39 & -5 & 2 \\ 1 & 2 & 1 \end{array}\right| \\ &=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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 1 & 11 & -1 \\ 2 & 39 & 2 \\ -3 & 1 & 1 \end{array}\right| \\ &=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} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 1 & -4 & 11 \\ 2 & -5 & 39 \\ -3 & 2 & 1 \end{array}\right| \\ &=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: $\mathrm{x}=1, \mathrm{y}=2 \text { and } \mathrm{z}=1$
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.
$|\mathrm{A}|=\left|\begin{array}{ccc} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{array}\right|$$\because$(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.
$\mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4 \end{array}\right|$
\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.
$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 6 & 5 & -3 \\ 1 & 5 & -2 \\ 2 & 8 & 4 \end{array}\right|$
\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}=\left|\begin{array}{lll} 6 & 1 & 5 \\ 1 & 3 & 5 \\ 2 & 1 & 8 \end{array}\right| \\ &=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:$\mathrm{x}=3, \mathrm{y}=2 \text { and } \mathrm{z}=1$
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} |\mathrm{A}| &=\left|\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}\right| \\ &=1(1)-1(-1)+0(-1) \\ &=1+1 \end{aligned}$\because$(Taking first row for solving determinant)
$= 2$
Now for x, the x column is replaced with constant column i.e.
\begin{aligned} &\mathrm{D}_{\overline{\mathrm{y}}}=\left|\begin{array}{lll} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{array}\right| \\ &=5(1)-1(3-4)+0(0-4) \\ &=5+1 \\ &=6 \end{aligned}
\begin{aligned} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{lll} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{array}\right| \\ &=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} &\mathrm{D}_{z}=\left|\begin{array}{lll} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{array}\right| \\ &=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}
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 15

Answer:$\mathrm{x}=5, \mathrm{y}=-3 \text { and } \mathrm{z}=-2$
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} |\mathrm{A}| &=\left|\begin{array}{ccc} 0 & 2 & -3 \\ 1 & 3 & 0 \\ 3 & 4 & 0 \end{array}\right| \\ &=0(0)-2(0-0)-3(4-9) \\ &=-3(-5) \\ &=15 \end{aligned} $\because$(Taking first row for solving determinant)
Now for x, the x column is replaced with constant column i.e.
\begin{aligned} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 0 & 2 & -3 \\ -4 & 3 & 0 \\ 3 & 4 & 0 \end{array}\right| \\ &=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} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{ccc} 0 & 0 & -3 \\ 1 & -4 & 0 \\ 3 & 3 & 0 \end{array}\right| \\ &=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} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 0 & 2 & 0 \\ 1 & 3 & -4 \\ 3 & 4 & 3 \end{array}\right| \\ &=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}
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 16

Answer:$x=1, y=-1 \text { and } z=-1$
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.
$|\mathrm{A}|=\left|\begin{array}{ccc} 5 & -7 & 1 \\ 6 & -8 & -1 \\ 3 & 2 & -6 \end{array}\right|$ $\because$(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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 11 & -7 & 1 \\ 15 & -8 & -1 \\ 7 & 2 & -6 \end{array}\right| \\ &=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} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{ccc} 5 & 11 & 1 \\ 6 & 15 & -1 \\ 3 & 7 & -6 \end{array}\right| \\ &=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} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 5 & -7 & 11 \\ 6 & -8 & 15 \\ 3 & 2 & 7 \end{array}\right| \\ &=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}
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 17

Answer:$x=2, y=-3 \text { and } z=-4$
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.
$\mathrm{D}=\left|\begin{array}{ccc} 2 & -3 & -4 \\ -2 & 5 & -1 \\ 3 & -1 & 5 \end{array}\right|$ $\because$ (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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 29 & -3 & -4 \\ -15 & 5 & -1 \\ -11 & -1 & 5 \end{array}\right| \\ &=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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 2 & 29 & -4 \\ -2 & -15 & -1 \\ 3 & -11 & 5 \end{array}\right| \\ &=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} &\mathrm{D}_{z}=\left|\begin{array}{ccc} 2 & -3 & 29 \\ -2 & 5 & -15 \\ 3 & -1 & -11 \end{array}\right| \\ &=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}
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 18

Answer: $x=-2, y=3 \text { and } z=-4$
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.
$|\mathbf{A}|=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{array}\right|$ $\because$(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} \mathrm{D}_{\mathrm{x}} &=\left|\begin{array}{ccc} 1 & 1 & 0 \\ -6 & 0 & 1 \\ 3 & -1 & -2 \end{array}\right| \\ &=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} \mathrm{D}_{\mathrm{y}} &=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & -6 & 1 \\ 1 & 3 & -2 \end{array}\right| \\ &=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} \mathrm{D}_{z} &=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -6 \\ 1 & -1 & 3 \end{array}\right| \\ &=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:$\mathrm{x}=\frac{-(c-d)(d-b)}{(a-b)(c-a)}, \mathrm{y}=\frac{-(a-d)(d-c)}{(a-b)(b-c)} \text { and } \mathrm{z}=\frac{-(b-d)(d-a)}{(b-c)(c-a)}$
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} &|\mathrm{A}|=\left|\begin{array}{lll} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{array}\right| \\ &\quad C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}
$=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & c-a \\ a^{2} & b^{2}-a^{2} & c^{2}-a^{2} \end{array}\right|$
Now taking (b-a) and (c-a) from $C_{2}$ and $C_{3}$ respectively,
$=(b-a)(c-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b+a & c+a \end{array}\right|$
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.
$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ d & b & c \\ d^{2} & b^{2} & c^{2} \end{array}\right|$
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}
$D_{x}$$=-\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & b-d & c-d \\ d^{2} & b^{2}-d^{2} & c^{2}-d^{2} \end{array}\right|$
Now taking (b-d) and (c-d) from $C_{2}$ and $C_{3}$ respectively,
$=-(b-d)(c-d)\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & b+d & c+d \end{array}\right|$
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.
$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{lll} 1 & 1 & 1 \\ a & d & c \\ a^{2} & d^{2} & c^{2} \end{array}\right|$
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}
$\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & d-a & c-a \\ a^{2} & d^{2}-a^{2} & c^{2}-a^{2} \end{array}\right|$
Now taking (d-a) and (c-a) from $C_{2}$ and $C_{3}$ respectively,
$=-(d-a)(c-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ d & 1 & 1 \\ d^{2} & d+a & c+a \end{array}\right|$
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.
$\mathrm{D}_{z}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & d \\ a^{2} & b^{2} & d^{2} \end{array}\right|$
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1} \\ &C_{3} \rightarrow C_{3}-C_{1} \end{aligned}
\begin{aligned} &\mathrm{D}_{z}\\ &=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & d-a \\ a^{2} & b^{2}-a^{2} & d^{2}-a^{2} \end{array}\right| \end{aligned}
Now taking (b-a) and (d-a) from $C_{2}$and $C_{3}$ respectively,$=-(b-a)(d-a)\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & b+a & d+a \end{array}\right|$

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:$\mathrm{x}=-2, \mathrm{y}=3, \mathrm{z}=\frac{3}{2} \text { and } \mathrm{w}=\frac{-1}{2}$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
\begin{aligned} &\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}=2 \\ &\mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+2 \mathrm{w}=-6 \\ &2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}+2 \mathrm{w}=-5 \\ &3 \mathrm{x}-\mathrm{y}+3 \mathrm{z}-3 \mathrm{w}=-3 \end{aligned}
Solution:
Solving determinant,
$|\mathbf{A}|=\left|\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -2 & 2 & 2 \\ 2 & 1 & -2 & 2 \\ 3 & -1 & 3 & -3 \end{array}\right|$
\begin{aligned} C_{2} & \rightarrow C_{2}-C_{1} \\ C_{3} & \rightarrow C_{3}-C_{1} \\ C_{4} & \rightarrow C_{4}-C_{1} \end{aligned}
$=\left|\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & -3 & 1 & 1 \\ 2 & -1 & -4 & 0 \\ 3 & -4 & 0 & -6 \end{array}\right|$
Expanding along $R_{1}$,
$=1\left|\begin{array}{ccc} -3 & 1 & 1 \\ -1 & -4 & 0 \\ -4 & 0 & -6 \end{array}\right|$
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{3} \\ &C_{1} \rightarrow C_{1}+3 C_{3} \end{aligned}
\begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -1 & -4 & 0 \\ -22 & 6 & -6 \end{array}\right| \\ &=1(-6-88) \end{aligned}
$=-94$
$\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cccc} 2 & 1 & 1 & 1 \\ -6 & -2 & 2 & 2 \\ -5 & 1 & -2 & 2 \\ -3 & -1 & 3 & -3 \end{array}\right|=188$
$\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cccc} 1 & 2 & 1 & 1 \\ 1 & -6 & 2 & 2 \\ 2 & -5 & -2 & 2 \\ 3 & -3 & 3 & -3 \end{array}\right|=-282$
$\mathrm{D}_{z}=\left|\begin{array}{cccc} 1 & 1 & 2 & 1 \\ 1 & -2 & -6 & 2 \\ 2 & 1 & -5 & 2 \\ 3 & -1 & -3 & -3 \end{array}\right|=-141$
$\mathrm{D}_{\mathrm{w}}=\left|\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 1 & -2 & 2 & -6 \\ 2 & 1 & -2 & -5 \\ 3 & -1 & 3 & -3 \end{array}\right|=47$
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: $\mathrm{x}=1, \mathrm{y}=\frac{2}{7}, \mathrm{z}=\frac{2}{7} \text { and } \mathrm{w}=\frac{-1}{7}$
Hint: Use Cramer’s rule to solve a system of linear equations
Given:
\begin{aligned} &2 \mathrm{x}-3 \mathrm{z}+\mathrm{w}=1 \\ &\mathrm{x}-\mathrm{y}+2 \mathrm{w}=1 \\ &-3 \mathrm{y}+\mathrm{z}+\mathrm{w}=1 \\ &\mathrm{x}+\mathrm{y}+\mathrm{z}=1 \end{aligned}
Solution:
Solving determinant,
$|\mathrm{A}|=\left|\begin{array}{cccc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right|$
\begin{aligned} C_{2} & \rightarrow C_{2}-C_{1} \\ C_{3} & \rightarrow C_{3}-C_{1} \end{aligned}
$=\left|\begin{array}{cccc} 2 & -2 & -5 & 1 \\ 1 & -2 & -1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{array}\right|$
Expanding along $R_{1}$,
$=-1\left|\begin{array}{ccc} -2 & -5 & 1 \\ -2 & -1 & 2 \\ -3 & 1 & 1 \end{array}\right|$
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{3} \\ &C_{1} \rightarrow C_{1}+3 C_{3} \end{aligned}
\begin{aligned} &=-1\left|\begin{array}{ccc} 1 & -6 & 1 \\ 4 & -3 & 2 \\ 0 & 0 & 1 \end{array}\right| \\ &=-1(-3+24) \\ &=-21 \end{aligned}
\begin{aligned} &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cccc} 1 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 1 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right|=-21 \\ &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cccc} 2 & 1 & 1 & 0 \end{array}\right|=-6 \\ &\mathrm{D}_{z}=\left|\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}\right|=-6 \\ &\mathrm{D}_{\mathrm{w}}=\left|\begin{array}{cccc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 1 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array}\right|=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:$\mathrm{x}=\infty \text { and } \mathrm{y}=\infty$
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,
$|\mathrm{A}|=\left|\begin{array}{ll} 2 & -1 \\ 4 & -2 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\begin{aligned} &=-4+4 \\ &\mathrm{D}=0 \end{aligned}
\begin{aligned} &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ll} 5 & -1 \\ 7 & -2 \end{array}\right|=-10+7=-3 \neq 0 \\ &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ll} 2 & 5 \\ 4 & 7 \end{array}\right|=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\\ &\text { Since, } \mathrm{D}=0 \text { and } \mathrm{D}_{x} \text { and } \mathrm{D}_{\mathrm{y}} \neq 0\\ &\therefore \text { 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:$x=\infty \text { and } y=\infty$
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)
$\begin{gathered} 6 x+2 y=10 \\ -6 x-2 y=9 \\ \hline 0 \end{gathered}$
Hence, linear equations are inconsistent.
By Cramer’s rule:
Solving determinant,
\begin{aligned} &|\mathrm{A}|=\left|\begin{array}{cc} 3 & 1 \\ -6 & -2 \end{array}\right|=-6+6=0 \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right) \\ &\mathrm{D}=0 \end{aligned}
\begin{aligned} &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cc} 5 & 1 \\ 9 & -2 \end{array}\right|=-10-9=-19 \neq 0\\ &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cc} 3 & 5 \\ -6 & 9 \end{array}\right|=27+30=57 \neq 0\\ &\text { 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 $D_{y}\neq 0$
$\therefore$ 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:$x=\infty \text { and } y=\infty$
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)
$\begin{gathered} 6 x+2 y=10 \\ -6 x-2 y=9 \\ \hline 0 \end{gathered}$
Hence, linear equations are inconsistent.
By Cramer’s rule:
Solving determinant,
\begin{aligned} &|\mathrm{A}|=\left|\begin{array}{cc} 3 & 1 \\ -6 & -2 \end{array}\right|=-6+6=0 \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right) \\ &\mathrm{D}=0 \end{aligned}
\begin{aligned} &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cc} 5 & 1 \\ 9 & -2 \end{array}\right|=-10-9=-19 \neq 0\\ &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cc} 3 & 5 \\ -6 & 9 \end{array}\right|=27+30=57 \neq 0\\ &\text { 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 $D_{y}\neq 0$
$\therefore$ 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: $\mathrm{x}=\infty, \mathrm{y}=\infty \text { and } \mathrm{z}=\infty$
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,
$|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & -1 \end{array}\right|$
Expanding along $1^{st}$ 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} \mathrm{D} &=0 \\ \mathrm{D}_{x} &=\left|\begin{array}{ccc} 3 & -1 & 2 \\ 5 & 1 & 3 \\ 1 & -2 & -1 \end{array}\right| \\ &=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\\ &\text { Since, } \mathrm{D}=0 \text { and } \mathrm{D}_{x}, \mathrm{D}_{\mathrm{y}} \text { and } \mathrm{D}_{z} \neq 0 \end{aligned}
$\therefore$ 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:$x=-3, y=-1 \text { and } z=7$
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,
$|\mathrm{A}|=\left|\begin{array}{ccc} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 6 & 5 \end{array}\right|$
Expanding along $1^{st}$ row,
\begin{aligned} &=3(-5-6)+1(10-3)+2(12+3) \\ &=-33+7+30 \\ &=4 \end{aligned}
\begin{aligned} &\mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 6 & -1 & 2 \\ 2 & -1 & 1 \\ 20 & 6 & 5 \end{array}\right|=-12 \\ &D_{y}=\left|\begin{array}{ccc} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{array}\right|=-4 \end{aligned}
$D_{z}=\left|\begin{array}{ccc} 3 & -1 & 6 \\ 2 & -1 & 2 \\ 3 & 6 & 20 \end{array}\right|=28$
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 :$\mathrm{x}=\frac{5}{3}, \mathrm{y}=k-\frac{4}{3} \text { and } \mathrm{z}=\mathrm{k}$
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:
$\begin{gathered} \mathrm{AX}=\mathrm{B} \\ \qquad\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \\ \text { Where } \mathrm{A}=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right], \mathrm{X}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l} 3 \\ 2 \\ 1 \end{array}\right] \end{gathered}$
Solving determinant,
$|\mathbf{A}|=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array}\right|$
Expanding along $1^{st}$ 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)
$\Rightarrow 2 \mathrm{x}+\mathrm{y}-\mathrm{k}=2$ .....(2)
From (1) and (2),
$\Rightarrow x-y=3-k$ ......(3)
$\Rightarrow 2 \mathrm{x}+\mathrm{y}=2+\mathrm{k}$ .......(4)
\begin{aligned} &\Rightarrow 3 \mathrm{x}=5 \\ &\therefore 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} \\ &\therefore 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:$\mathrm{x}=5-2 \mathrm{k}, \mathrm{y}=\mathrm{k}$
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,
$|\mathrm{A}|=\left|\begin{array}{ll} 1 & 2 \\ 3 & 6 \end{array}\right| \quad \because\left|\begin{array}{ll} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\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.
$\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{cc} 5 & 2 \\ 15 & 6 \end{array}\right|=30-30=0$
If we are solving for y, the y column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{cc} 1 & 5 \\ 3 & 15 \end{array}\right|=15-15=0$
$\text { Since, } \mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{x}}=0$
Let y = k, then we have:

\begin{aligned} &\Rightarrow \mathrm{x}+2 \mathrm{k}=5 \\ &\Rightarrow \mathrm{x}=5-2 \mathrm{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:$x=k, y=2 k \text { and } z=3k$
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| &=\left|\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{array}\right| \\ &=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.
$\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 6 & -5 \end{array}\right|=0$
If we are solving for y, the y column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{y}=\left|\begin{array}{ccc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 3 & 0 & -5 \end{array}\right|=0$
If we are solving for z, the z column is replaced with constant column i.e.
\begin{aligned} &\Rightarrow D_{z}=\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & -2 & 0 \\ 3 & 6 & 0 \end{array}\right|=0 \\ &\text { 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} &\mathrm{D}=\left|\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right|=-2-1=-3 \\ &\mathrm{D}_{x}=\left|\begin{array}{cc} z & 1 \\ -z & -2 \end{array}\right|=-2 \mathrm{z}+z=-z \end{aligned}
\begin{aligned} &\mathrm{D}_{\mathrm{y}}=\left|\begin{array}{cc} 1 & z \\ 1 & -z \end{array}\right|=-\mathrm{z}-\mathrm{z}=-2 \mathrm{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:$x=\frac{3 k+6}{5}, y=\frac{8+4 k}{5} \text { and } z=k$
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|=\left|\begin{array}{ccc} 2 & 1 & -2 \\ 1 & -2 & 1 \\ 5 & -5 & 1 \end{array}\right| \\ &=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.
$\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1 \end{array}\right|$
Taking 2 common from $C_{1}$,
$=-2\left|\begin{array}{ccc} -2 & 1 & -2 \\ 1 & -2 & 1 \\ 1 & -5 & 1 \end{array}\right|=0 \quad\left(\because C_{1}=C_{3}\right)$
If we are solving for y, the y column is replaced with constant column i.e.
$\begin{array}{r} \Rightarrow \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1 \end{array}\right| \\ \quad \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+2 C_{3} \end{array}$
$=\left|\begin{array}{ccc} 2 & 0 & -2 \\ 1 & 0 & 1 \\ 5 & 0 & 1 \end{array}\right|=0 \quad\left(\because \text { Expanding along } \mathrm{R}_{1}\right)$
If we are solving for z, the z column is replaced with constant column i.e.
\begin{aligned} & \Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2 \end{array}\right| \quad \text { Expanding along } \mathrm{R}_{1}, \\ =& 2(4-10)-1(-2+10)+4(-5+10) \\ =&-12-8+20 \\ =& 0 \end{aligned}
So,$\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{2}=0$
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} &\mathrm{D}=\left|\begin{array}{ll} 1 & -2 \\ 5 & -5 \end{array}\right|=-5+10=5\\ &\mathrm{D}_{x}=\left|\begin{array}{ll} -2-z & -2 \\ -2-z & -5 \end{array}\right|=10+5 z-2 z-4=3 z+6\\ &\mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ll} 1 & -2-z \\ 5 & -2-z \end{array}\right|=-2-\mathrm{z}+10+5 \mathrm{z}=8+4 \mathrm{z}\\ &\Rightarrow x=\frac{D_{x}}{D}=\frac{3 z+6}{5}\\ &\Rightarrow y=\frac{D_{y}}{D}=\frac{8+4 z}{5}\\ &\text { Let } z=k, \text { then } x=\frac{3 z+6}{5} \text { 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: $x=\frac{7-3 z}{2}, y=\frac{3 z-5}{2} \text { and } z=k$
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| &=\left|\begin{array}{ccc} 1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3 \end{array}\right| \\ &=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.
$\Rightarrow \mathrm{D}_{x}=\left|\begin{array}{ccc} 6 & -1 & 3 \\ -4 & 3 & -3 \\ 10 & 3 & 3 \end{array}\right|$
\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2} \\ &R_{3} \rightarrow R_{3}+R_{2} \end{aligned}
$=\left|\begin{array}{ccc} 2 & 2 & 0 \\ -4 & 3 & -3 \\ 6 & 6 & 0 \end{array}\right|=0 \quad\left(\because \text { Expanding along } R_{1}\right)$
If we are solving for y, the y column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ccc} 1 & 6 & 3 \\ 1 & -4 & -3 \\ 5 & 10 & 3 \end{array}\right|$
\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2} \\ &R_{3} \rightarrow R_{3}+R_{2} \end{aligned}
$=\left|\begin{array}{ccc} 2 & 2 & 0 \\ 1 & -4 & -3 \\ 6 & 6 & 0 \end{array}\right|=0 \quad\left(\because \text { Expanding along } R_{1}\right)$
If we are solving for z, the z column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 1 & -1 & 6 \\ 1 & 3 & -4 \\ 5 & 3 & 10 \end{array}\right|$
\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-5 R_{1} \end{aligned}
\begin{aligned} &=\left|\begin{array}{ccc} 1 & -1 & 6 \\ 0 & 4 & -10 \\ 0 & 8 & -20 \end{array}\right|\\ &=0 \quad\left(\because \text { Expanding along } \mathrm{R}_{1}\right) \end{aligned}
So, $\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{z}=0$
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=\left|\begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array}\right|=3+1=4 \\ &D_{x}=\left|\begin{array}{cc} 6-3 z & -1 \\ -4+3 z & 3 \end{array}\right|=18-9 z-4+3 z=14-6 z \\ &D_{x}=\left|\begin{array}{lc} 1 & 6-3 z \\ 1 & -4+3 z \end{array}\right|=-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} \\ &\text { Let } z=k, \text { then } x=\frac{7-3 z}{2} \text { 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:$\mathrm{x}=2, \mathrm{y}=4 \text { and } \mathrm{z}=11$
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 \mathrm{x}+100 \mathrm{y}+20 \mathrm{z}=800 \\ &130 \mathrm{x}+50 \mathrm{y}+40 \mathrm{z}=900 \\ &60 \mathrm{x}+100 \mathrm{y}+30 \mathrm{z}=850 \end{aligned}
By Cramer’s rule, solving determinant:
$D=\left|\begin{array}{ccc} 90 & 100 & 20 \\ 130 & 50 & 40 \\ 60 & 100 & 30 \end{array}\right|$
\begin{aligned} &R_{1} \rightarrow R_{1}-2 R_{2} \\ &R_{3} \rightarrow R_{3}-2 R_{2} \end{aligned}
$=\left|\begin{array}{ccc} -170 & 0 & -60 \\ 130 & 50 & 40 \\ -200 & 0 & -50 \end{array}\right|$
\begin{aligned} &=-170(-2500)-60(10000) \\ &=425000-600000 \\ &=-175000 \end{aligned}
Now for x, the x column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{\bar{x}}=\left|\begin{array}{ccc} 800 & 100 & 20 \\ 900 & 50 & 40 \\ 850 & 100 & 30 \end{array}\right|$
\begin{aligned} &R_{1} \rightarrow R_{1}-2 R_{2} \\ &R_{3} \rightarrow R_{3}-2 R_{2} \end{aligned}
$=\left|\begin{array}{ccc} -1000 & 0 & -60 \\ 900 & 50 & 40 \\ -950 & 0 & -50 \end{array}\right|$
\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.
$\Rightarrow \mathrm{D}_{\mathrm{y}}=\left|\begin{array}{ccc} 90 & 800 & 20 \\ 130 & 900 & 40 \\ 60 & 850 & 30 \end{array}\right|$
\begin{aligned} &R_{2} \rightarrow R_{2}-2 R_{1} \\ &R_{3} \rightarrow R_{3}-\frac{3}{2} R_{1} \end{aligned}
\begin{aligned} &=\left|\begin{array}{ccc} 90 & 800 & 20 \\ -50 & -700 & 0 \\ -75 & -350 & 0 \end{array}\right| \\ &=20(17500-52500) \\ &=-700000 \end{aligned}
If we are solving for z, the z column is replaced with constant column i.e.
$\Rightarrow \mathrm{D}_{z}=\left|\begin{array}{ccc} 90 & 100 & 800 \\ 130 & 50 & 900 \\ 60 & 100 & 850 \end{array}\right|$
\begin{aligned} &R_{1} \rightarrow R_{1}-2 R_{2} \\ &R_{3} \rightarrow R_{3}-2 R_{2} \end{aligned}
$=\left|\begin{array}{ccc} -170 & 0 & -1000 \\ 130 & 50 & 900 \\ -200 & 0 & -950 \end{array}\right|$
\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}
$\therefore$ 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:$\mathrm{x}=2, \mathrm{y}=3 \text { and } z=4$
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 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z}=29 \\ &\mathrm{x}+\mathrm{y}+2 \mathrm{z}=13 \\ &3 \mathrm{x}+2 \mathrm{y}+z=16 \end{aligned}
Where x, y and z are number of cars $C_{1}, C_{2}\: and \: C_{3}$ respectively.
By Cramer’s rule, solving determinant:
\begin{aligned} D=&\left|\begin{array}{lll} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{array}\right| \\ & R_{1} \rightarrow R_{1}-4 R_{3} \\ & R_{2} \rightarrow R_{2}-2 R_{3} \\ &=\left|\begin{array}{ccc} -10 & -5 & 0 \\ -5 & -3 & 0 \\ 3 & 2 & 1 \end{array}\right| \\ =& 30-25 \\ =& 5 \end{aligned}
Now for x, the x column is replaced with constant column i.e.
\begin{aligned} \Rightarrow D_{\bar{x}}=&\left|\begin{array}{ccc} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{array}\right| \\ & \quad \begin{aligned} R_{1} & \rightarrow R_{1}-4 R_{3} \\ R_{2} & \rightarrow R_{2}-2 R_{3} \end{aligned} \\ =&\left|\begin{array}{lll} -35 & -5 & 0 \\ -19 & -3 & 0 \\ 16 & 2 & 1 \end{array}\right| \\ =& 190-175 \\ =& 15 \end{aligned}
If we are solving for y, the y column is replaced with constant column i.e.
$\begin{array}{r} \Rightarrow \mathrm{D}_{\mathrm{y}}=\left|\begin{array}{lll} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{array}\right| \\ \quad R_{1} \rightarrow R_{1}-4 R_{3} \\ R_{2} \rightarrow R_{2}-2 R_{3} \end{array}$
\begin{aligned} &=\left|\begin{array}{ccc} -10 & -35 & 0 \\ -5 & -19 & 0 \\ 3 & 16 & 1 \end{array}\right| \\ &=190-175 \end{aligned}
$=15$
If we are solving for z, the z column is replaced with constant column i.e.
$\begin{array}{r} \Rightarrow \mathrm{D}_{z}=\left|\begin{array}{lll} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array}\right| \\ R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right) \end{array}$
$=\left|\begin{array}{ccc} -2 & 0 & 0 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array}\right|$
\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}
$\therefore$ The number of cars produced of type $C_{1}, C_{2}\: and \: C_{3}$ 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.

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Chapter-wise RD Sharma Class 12 Solutions

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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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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