# NCERT Solutions for Exercise 4.1 Class 12 Maths Chapter 4 - Determinants

In this article, you will get NCERT Solutions for Class 12 Maths chapter 4 exercise 4.1 consists of questions related to finding determinants of matrices of a different order. Determinant of the matrix is the factor by which its volume blows up. There are a lot of applications of determinants like calculating the value of square matrices, solving a set of linear equations, check whether a matrix has an inverse, evaluate the inverse of the matrix using cofactors, etc.

In exercise 4.1 Class 12 Maths, there are 8 questions related to finding determinants of matrices of a different order. Class 12th Maths chapter 4 exercise 4.1 questions are solved with necessary steps. Only knowing the answer is not enough, you need to know how to write solutions in board exams to get good marks. Class 12 Maths chapter 4 exercise 4.1 solutions are designed in a descriptive manner which you will understand easily. Check for NCERT solutions here.

Also, see

• Determinants Exercise 4.2
• Determinants Exercise 4.3
• Determinants Exercise 4.4
• Determinants Exercise 4.5
• Determinants Exercise 4.6
• Determinants Miscellaneous Exercise

## Determinants Exercise:4.1

Question:1 Evaluate the following determinant- $\begin{vmatrix} 2 & 4\\ -5 & -1\end{vmatrix}$

The determinant is evaluated as follows

$\begin{vmatrix} 2 & 4\\ -5 & -1\end{vmatrix} = 2(-1) - 4(-5) = -2 + 20 = 18$

Question:2(i) Evaluate the following determinant- $\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix}$

The given two by two determinant is calculated as follows

$\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix} = cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2\theta + \sin ^2 \theta = 1$

Question:2(ii) Evaluate the following determinant- $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

We have determinant $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}$

So, $\dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix} = (x^2-x+1)(x+1) - (x-1)(x+1)$

$= (x+1)(x^2-x+1-x+1) = (x+1)(x^2-2x+2)$

$=x^3-2x^2+2x +x^2-2x+2$

$= x^3-x^2+2$

Question:3 If $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ , then show that $\dpi{100} | 2 A |=4|A|$

Given determinant $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then we have to show that $\dpi{100} | 2 A |=4|A|$,

So, $\dpi{100} A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix}$ then, $\dpi{100} 2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}$

Hence we have $\dpi{100} \left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24$

So, L.H.S. = |2A| = -24

then calculating R.H.S. $\dpi{100} 4\left | A \right |$

We have,

$\dpi{100} \left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6$

hence R.H.S becomes $\dpi{100} 4\left | A \right | = 4\times(-6) = -24$

Therefore L.H.S. =R.H.S.

Hence proved.

Question:4 If $A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$ then show that $\dpi{100} |3A|=27|A|$

Given Matrix$A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}$

Calculating $3A =3\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} = \begin{bmatrix} 3 &0 &3 \\ 0& 3& 6\\ 0& 0 &12 \end{bmatrix}$

So, $\left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108$

calculating $\dpi{100} 27|A|$,

$\dpi{100} |A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4$

So, $\dpi{100} 27|A| = 27(4) = 108$

Therefore $\dpi{100} |3A|=27|A|$.

Hence proved.

Question:5(i) Evaluate the determinants.

$\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$

Given the determinant $\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix}$;

now, calculating its determinant value,

$\dpi{100} \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix} = 3\begin{vmatrix} 0 &-1 \\ -5& 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 &0 \\ 3& -5 \end{vmatrix}$

$\dpi{100} = 3(0-5)+1(0+3) -2(0-0) = -15+3-0 = -12$.

Question:5(ii) Evaluate the determinants.

$\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$

Given determinant $\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}$;

Now calculating the determinant value;

$\dpi{100} \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix} = 3\begin{vmatrix} 1 &-2 \\ 3&1 \end{vmatrix} -(-4)\begin{vmatrix} 1 &-2 \\ 2& 1 \end{vmatrix}+5\begin{vmatrix} 1 & 1\\ 2& 3 \end{vmatrix}$

$= 3(1+6) +4(1+4) +5(3-2) = 21+20+5 = 46$.

Question:5(iii) Evaluate the determinants.

$\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$

Given determinant $\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}$;

Now calculating the determinant value;

$\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix} = 0\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} -1\begin{vmatrix} -1 &-3 \\ -2& 0 \end{vmatrix}+2\begin{vmatrix} -1 &0 \\ -2& 3 \end{vmatrix}$

$= 0 - 1(0-6)+2(-3-0) = 6 -6 =0$

Question:5(iv) Evaluate the determinants.

$\begin{vmatrix}2 &-1 &2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}$

Given determinant: $\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix}$,

We now calculate determinant value:

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

$=2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5$

Question:6 If $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ , then find $|A|$.

Given the matrix $A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix}$ then,

Finding the determinant value of A;

$|A| = 1\begin{vmatrix} 1 &-3 \\ 4& -9 \end{vmatrix} -1\begin{vmatrix} 2 &-3 \\ 5& -9 \end{vmatrix}-2\begin{vmatrix} 2 &1 \\ 5& 4 \end{vmatrix}$

$= 1(-9+12)-1(-18+15)-2(8-5) =3+3-6 =0$

Question:7(i) Find values of x, if

$\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

Given that $\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}$

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

$\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =2-20 = -18$ and $\dpi{100} \begin{vmatrix}2x &4 \\6 &x \end{vmatrix} = 2x(x)-24 = 2x^2-24$

So, we have then,

$\dpi{100} -18= 2x^2-24$ or $\dpi{100} 3= x^2$ or $\dpi{100} x= \pm \sqrt{3}$

Question:7(ii) Find values of x, if

$\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$

Given $\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}$;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

$\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}= 10 - 12 = -2$

Similarly R.H.S. determinant value;

$\dpi{100} \begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x$

So, we have then;

$\dpi{100} -2 = -x$ or $\dpi{100} x =2$.

Question:8 If $\dpi{100} \begin{vmatrix}x &2 \\18 &x \end{vmatrix}=\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix}$ , then $\dpi{100} x$ is equal to

(A) $\dpi{100} 6$ (B) $\dpi{100} \pm 6$ (C) $\dpi{100} -6$ (D) $\dpi{100} 0$

Solving the L.H.S. determinant ;

$\dpi{100} \begin{vmatrix}x &2 \\18 &x \end{vmatrix}= x^2 - 36$

and solving R.H.S determinant;

$\dpi{100} \begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} = 36-36 = 0$

So equating both sides;

$\dpi{100} x^2 - 36 =0$ or $\dpi{100} x^2 = 36$ or $\dpi{100} x = \pm 6$

Hence answer is (B).

More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.1:-

The first exercise of this chapter of NCERT book is based on the basics of determinants and calculating determinants of a different order of the square matrices. There are 5 solved examples given before the Class 12 Maths ch 4 ex 4.1 that you can solve to get clarity. There are 7 long-answer types of questions and 1 multiple-type question in Class 12th Maths chapter 4 exercise 4.1. There are few questions in this exercise that could be solved using the properties of determinants.

Also Read| Determinants Class 12 Chapter 4 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter Exercise 4.1:-

• NCERT Solutions for Class 12 Maths Chapter Exercise 4.1 are very helpful for the students to get conceptual clarity.
• Exercise 4.1 Class 12 Maths solutions are provided in a very detailed manner, so you can understand them very easily.
• Class 12 Maths chapter 4 exercise 4.1 solutions can be used for quick revision before the final exam.

Also see-

Class 10 NSTSE 5-Year Analysis: Which Chapters Get More Weightage And Why 9 min read Apr 11, 2022 Read More How To Ace Class 8 NSTSE: 5-Year Analysis, Pattern, Syllabus, Topics And Weightage 8 min read Apr 05, 2022 Read More

• NCERT Solutions for Class 12 Maths Chapter 4

• NCERT Exemplar Solutions Class 12 Maths Chapter 4

## NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!