RD Sharma Solutions Class 12 Mathematics Chapter 5 VSA

Kuldeep MauryaUpdated on 25 Jan 2022, 06:53 PM IST

RD Sharma Solutions are found to be a popular choice among students and teachers for the various benefits it offers. RD Sharma Solutions have been making NCERT Solutions for years now and have gained much popularity for their books. The RD Sharma Class 12th Exercise VSA is one of their NCERT Solution sets that has garnered the praise of several students who have appeared for the mathematics paper.

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

RD Sharma Class 12 Solutions Chapter 5 VSA Determinants - Other Exercise

Determinants Excercise: VSA

Determinants Exercise Very short question Question 1

Answer : $|A|=0$
Hint: Here we use basic concept of singular matrix
Given:$A \text { is singular matrix }$
Solution:
Any matrix becomes singular if it's determinate is zero
$\rightarrow \text { Here, } A \text { is singular So }|A| \text { must be zero. }$
So, $|A|=0$

Determinants Exercise Very short question Question 2

Answer :$x=3$
Hint: Here we use basic concept of singular matrix
Given :$\left|\begin{array}{cc} 5-x & x+1 \\ 2 & 4 \end{array}\right|$
Solution :
Let this matrix is singular So,
it's determinant should be zero
So now, $\left|\begin{array}{cc} 5-x & x+1 \\ 2 & 4 \end{array}\right|=0$
$\begin{aligned} &(5-x)(4)-2(x+1)=0 \\ &20-4 x-[2 x+2]=0 \\ &20-4 x-2 x-2=0 \\ &18-6 x=0 \end{aligned}$
So,
$x=\frac{18}{6}=3$

Determinants Exercise Very short question Question 10

Answer: $-8$
Hint: Here we use basic concept of determinant of matrix
Given: $\left|\begin{array}{ll} 4785 & 4787 \\ 4789 & 4791 \end{array}\right|$
Solution:
$\rightarrow$ let's perform some column operations
$\begin{aligned} &C_{1} \rightarrow C_{1}-C_{2} \\ &\left|\begin{array}{cc} -2 & 4787 \\ -2 & 4791 \end{array}\right| \end{aligned}$
$\rightarrow$ let's take $(-2)$ common from column $(1)$
$\begin{aligned} &(-2)\left|\begin{array}{ll} 1 & 4787 \\ 1 & 4791 \end{array}\right| \\ &=-2[1 \times 4791-4787] \\ &=-2[4] \\ &=-8 \end{aligned}$

Determinants Exercise Very short Answers Question 16

Answer:135
Hint: Here we use basic concept of determinant of matrix such that $(k A)=k^{n}|A|$
Given : $A \text { is } 3 \times 3 \text { matrix and }|A|=5$
Solution :
$|A|=5 \text { and } A \text { is } 3 \times 3 \text { matrix }$
So, $n = 3$
As,
$|k A|=k^{n}|A|$
$\rightarrow$ Here, we have to find $\left | 3A \right |$
So, $k=3$
$\begin{aligned} |3 A| &=3^{n}|A| \rightarrow[k=3] \\ |3 A| &=3^{3} \times 5 \\ |3 A| &=27 \times 5 \\ |3 A| &=135 \end{aligned}$

Determinants Exercise Very short Answers Question 19

Answer:0
Hint: Here we use basic concept of determinant of matrix
Given: $\begin{aligned} &A \text { is } 3 \times 3 \text { matrix }\\ &\mathrm{S} 0, \mathrm{n}=3,|A|=5 \text { and } \mathrm{C}_{i j} \text { is cofactor of } a_{i} \end{aligned}$
Solution :
$A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$
and cofactor $\mathrm{C}_{i j}=(-1)^{i+j}\left|m_{i j}\right|$
So, here
$a_{11} C_{21}+a_{12} C_{12}+a_{13} C_{23}$
So
$\begin{aligned} &C_{21}=(-1)^{2+1}\left|a_{12} a_{33}-a_{13} a_{32}\right| \\ &C_{22}=(-1)^{2+2}\left|a_{11} a_{33}-a_{13} a_{31}\right| \\ &C_{23}=(-1)^{2+3}\left|a_{11} a_{32}-a_{12} a_{31}\right| \end{aligned}$
$\begin{aligned} &\rightarrow a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{22} \\ &\rightarrow-a_{11} a_{12} a_{33}+a_{11} a_{13} a_{32}+a_{11} a_{12} a_{33}-a_{12} a_{13} a_{31}-a_{11} a_{13} a_{32}+a_{12} a_{13} a_{31} \\ &=0 \end{aligned}$

Determinants Exercise Very short Answers Question 20

Answer : 1
Hint: Here we use basic concept of determinant of matrix
Given: $\left[\begin{array}{cc} \sin 20^{\circ} & -\cos 20^{\circ} \\ \sin 70^{\circ} & \cos 70^{\circ} \end{array}\right]$
Solution :
So let's find determinate
$\left|\begin{array}{cc} \sin 20^{0} & -\cos 20^{\circ} \\ \sin 70^{0} & \cos 70^{0} \end{array}\right|$
$\begin{aligned} &=\sin 20^{\circ} \times \cos 70^{\circ}-\left(-\cos 20^{\circ} \times \sin 70^{\circ}\right) \\ &=\left(\sin 20^{\circ} \times \cos 70^{\circ}\right)+\left(\cos 20^{\circ} \times \sin 70^{\circ}\right) \end{aligned}$
$\rightarrow$ using the formula
$\begin{aligned} &\sin (A+B)=\sin A \cos B+\cos A \sin B \\ &\text { Here } \mathrm{A}=20^{\circ} \\ &\mathrm{B}=70^{\circ} \end{aligned}$
So,
$\begin{aligned} &\sin (20+70)^{0}=\left(\sin 20^{\circ} \times \cos 70^{\circ}\right)+\left(\cos 20^{\circ} \times \sin 70^{\circ}\right) \\ &=\sin 90^{\circ} \\ &=1 \end{aligned}$

Determinants Exercise Very Short Answer Question 23

Answer:0
Hint: Here we use basic concept of determinant of matrix
Given: A is skew –symmetric matrix
Solution:
$\rightarrow \text { Let } \mathrm{A} \text { is } 3 \times 3 \text { skew-symmetric matrix }$
So, $\mathrm{A}=\left[\begin{array}{ccc} 0 & a_{12} & a_{13} \\ a_{21} & 0 & a_{23} \\ a_{31} & a_{32} & 0 \end{array}\right]$
$\rightarrow$Here all diagonal entries are zero
$\rightarrow$ So, we know that it’s determinant zero

Determinants Exercise Very Short Answer Question 24

Answer:-4
Hint: Here we use basic concept of determinant of matrix
Given:
$\rightarrow \mathrm{A} \text { is } 3 \times 3 \text { matrix }$
So, $n=3$
and $\left | A \right |=4$
Solution :
$\left | -A \right |$ for what we use below formula $|k A|=k^{n}|A|$
when k is constant
n is order of matrix
$\rightarrow$ In $|-A|,-1$ is constant
So,
$\begin{array}{ll} |k A|=k^{n}|A| \\ |-1 A|=(-1)^{n}|A| & {[k=-1]} \\ |-1 A|=(-1)^{3}|A| & {[n=3]} \end{array}$
$\begin{aligned} &|-1 A|=(-1)^{3} \times 4 \quad[|A|=4] \\ &|-1 A|=-4 \end{aligned}$

Determinants Exercise VSQ Question 26

Answer: 0
Hint: Here we use basic concept of determinant of matrix
Given: $\left|\begin{array}{ccc} 243 & 156 & 300 \\ 81 & 52 & 100 \\ -3 & 0 & 4 \end{array}\right|$
Solution :
Using the property that if the equimultiples of corresponding elements of other rows are added to every element of any row of determinant then the value of determinant remains the same.
$\rightarrow$ Using row transformation $\mathrm{R}_{1} \rightarrow R_{1}-3 R_{2}$
We get,
$\Delta=\left|\begin{array}{ccc} 243-81 \times 3 & 156-52 \times 3 & 300-100 \times 3 \\ 81 & 52 & 100 \\ -3 & 0 & 4 \end{array}\right|$
$\Delta=\left|\begin{array}{ccc} 0 & 0 & 0 \\ 81 & 52 & 100 \\ -3 & 0 & 4 \end{array}\right|$
$\rightarrow$ Here 1st whole column is zero
So $\Delta =0$

Determinants Exercise VSQ Question 27

Answer: 0
Hint: Here we use basic concept of determinant of matrix
Given: $\left|\begin{array}{ccc} 2 & -3 & 5 \\ 4 & -6 & 10 \\ 6 & -9 & 15 \end{array}\right|$
Solution :
$\rightarrow$ Using column operation, $\mathrm{C}_{2} \rightarrow C_{2}+C_{3}$
We get,
$\Delta=\left|\begin{array}{ccc} 2 & -3+5 & 5 \\ 4 & -6+10 & 10 \\ 6 & -9+15 & 15 \end{array}\right|$
$\Delta=\left|\begin{array}{ccc} 2 & 2 & 5 \\ 4 & 4 & 10 \\ 6 & 6 & 15 \end{array}\right|$
$\rightarrow$Here 1st and 2nd column are similar
$\rightarrow$According to determinant if two or more columns are similar
It's determinant must be zero
So $\Delta = 0$

Determinants Exercise VSQ Question 28

Answer: -4
Hint: Here we use basic concept of determinant and singular matrix
Given: $\left|\begin{array}{cc} 5 x & 2 \\ -10 & 1 \end{array}\right|$ is similar
Solution :
If matrix is singular then it's determinant always zero
So, $\left|\begin{array}{cc} 5 x & 2 \\ -10 & 1 \end{array}\right|=0$
$\begin{aligned} &5 x(1)-[-10(2)]=0 \\ &5 x+20=0 \\ &5 x=-20 \\ &x=\frac{-20}{5} \\ &x=-4 \end{aligned}$

Determinants Exercise VSQ Question 31

Answer: non-singular
Hint: Here we use basic concept of determinant of matrix
Given: Here A and B non singular matrix
So,
$\begin{aligned} &|A| \neq 0 \\ &|B| \neq 0 \end{aligned}$
Solution : So now in $\left | AB \right |,$
$\begin{aligned} &|A B|=|A| \times|B| \\ &\rightarrow \text { Here }|A| \neq 0 \text { and }|B| \neq 0 \end{aligned}$
then multiplication of 2 non zero numbers, answer always be non zero
$\begin{aligned} &\text { So, }|A B| \neq 0\\ &\rightarrow|A B| \end{aligned}$ is non singular

RD Sharma Class 12 Chapter 5 Exercise VSA is a crucial section in the NCERT book because it covers questions from the entire fifth chapter of the mathematics book. Chapter 5 of the book deals with Inverse Determinants, a critical Chapter that works with linear equations. RD Sharma Class 12th Exercise VSA has 58 questions that require short answers. RD Sharma Class 12th Exercise VSA will have questions from concepts like the area of triangles, adjoint, and inverse of a matrix, applications of Determinants, and matrices.

The RD Sharma Class 12 Solutions Chapter 5 Ex VSA will answer all the questions present in the NCERT books. The Class 12 RD Sharma Chapter 5 Exercise VSA has Expert-created Solutions compiled by highly skilled individuals with a lot of knowledge in mathematics. Their new and improved ways of solving mathematical problems would help you to answer questions. These unique methods are not always taught in Class by school teachers or even tuition teachers.

By using RD Sharma Class 12th Exercise VSA, students will be able to improve their knowledge of Determinants through self-practice. The Solutions will help confirm their answers and enable them to focus on their weak points. Students can use the questions in the book to practice at home and test their knowledge before appearing for their school and board Exams. They might end up finding common questions from these Exercises as well.

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