Ex 4.1 class 12 maths ncert solutions | exercise 4.1 class 12 maths ncert solutions | class 12 maths chapter 4 exercise 4.1 solutions | maths class 12 chapter 4 ncert solutions | determinants maths class 12 | class 12 maths determinants exercise 4.1
Looking for Ex 4.1 Class 12 Maths NCERT Solutions? You’re in the right place! This section offers complete and step-by-step solutions to all the questions from Exercise 4.1 Class 12 Maths NCERT Solutions, based on Chapter 4 – Determinants. In this exercise, students learn how to evaluate 2×2 and 3×3 determinants, a key concept in linear algebra. The Class 12 Maths Chapter 4 Exercise 4.1 Solutions are designed to help you grasp the properties and expansion methods of determinants easily. Whether you’re preparing for board exams or strengthening your basics, these Maths Class 12 Chapter 4 NCERT Solutions are ideal for revision and practice. Explore the Class 12 Maths Determinants Exercise 4.1 to boost your problem-solving skills and confidence in this essential topic today!
class 12 maths determinants exercise 4.1 || ex 4.1 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || determinants maths class 12 || exercise 4.1 class 12 maths ncert solutions || class 12 maths chapter 4 exercise 4.1 solutions
Exercise 4.1
1. Evaluate the determinants:
\( \left[\begin{array}{cc}2 & 4 \\ -5 & 1\end{array}\right] \)
\( \left[\begin{array}{cc}2 & 4 \\ -5 & 1\end{array}\right] \)
Answer
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
Now,
\(\begin{array}{l}
{\left[\begin{array}{cc}
2 & 4 \\
-5 & 1
\end{array}\right]} \\
=2(-1)-4(-5) \\
=-2-(-20) \\
=-2+20 \\
=18\end{array}\)
The determinant of the above matrix is 18 .
2 A. Evaluate the determinants:
\(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
\(\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right|\)
Answer
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
\(\begin{array}{l}
\text {Now, }\left|\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array}\right| \\
=\cos \theta(\cos \theta)-(-\sin \theta)(\sin \theta) \\
=\cos 2 \theta+\sin 2 \theta \\
=1\quad[\because \cos 2 \theta+\sin 2 \theta=1]
\end{array}\)
The determinant of the above matrix is 1.
2 B. Evaluate the determinants:
\(\left|\begin{array}{cc}
x^{2}-x+x & x-1 \\
x+1 & x+1
\end{array}\right|\)
\(\left|\begin{array}{cc}
x^{2}-x+x & x-1 \\
x+1 & x+1
\end{array}\right|\)
Answer
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
Now,
\(\begin{array}{l}
\left|\begin{array}{cc}
x^{2}-x+1 & x-1 \\
x+1 & x+1
\end{array}\right| \\
=\left(x^{2}-x+1\right)(x+1)-(x-1)(x+1) \\
=\left(x^{3}-x^{2}+x+x^{2}-x+1\right)-\left(x^{2}-1\right) \\
=x^{3}+1-x^{2}+1 \\
=x^{3}-x^{2}+2
\end{array}\)
Ans. The determinant of the above matrix is \( x^{3}-x^{2}+2 \).
class 12 maths determinants exercise 4.1 || ex 4.1 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || determinants maths class 12 || exercise 4.1 class 12 maths ncert solutions || class 12 maths chapter 4 exercise 4.1 solutions
3. If \( \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right] \) then show that \( |2 \mathrm{A}|=4|\mathrm{A}| \).
Answer
\(|\mathrm{A}|=\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]\)
We know that determinant of A is calculated as
\(\begin{array}{l}
=1(2)-2(4) \\
=2-8 \\
|\mathrm{A}|=-6
\end{array}\)
LHS: \( |2 \mathrm{A}| \)
\(\begin{array}{l}
2 \mathrm{A}=2\left[\begin{array}{ll}
1 & 2 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right] \\
|2 \mathrm{A}|=\left|\begin{array}{ll}
2 & 4 \\
8 & 4
\end{array}\right| \\
=2(4)-4(8) \\
=8-32=-24 \\
|2 \mathrm{A}|=-24 \ldots \mathrm{LHS}
\end{array}\)
RHS: \(4|A|\)
\(\begin{array}{l}
4|\mathrm{A}|=4(-6) \\
=-24 \\
4|\mathrm{A}|=-24 \ldots \text {RHS } \\
\text {LHS = RHS }
\end{array}\)
Hence proved.
4. If \( A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right] \) then show that \( |3 \mathrm{A}|=27|\mathrm{A}| \)
Answer
\(|A|=\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]\)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{lll}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{ll}
b & c \\
e & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{ll}
a & c \\
d & f
\end{array}\right|+\mathrm{k}\left|\begin{array}{cc}
a & b \\
d & e
\end{array}\right| \\
=1\left|\begin{array}{ll}
1 & 2 \\
0 & 4
\end{array}\right|-0\left|\begin{array}{ll}
0 & 2 \\
0 & 4
\end{array}\right|+1\left|\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}\right| \\
=1[1(4)-2(0)]-0+1[0-0] \\
=1[4-0]-0+0 \\
=4 \\
|\mathrm{A}|=4
\end{array}\)
LHS: \( |3 \mathrm{A}| \)
\(\begin{array}{l}
3 \mathrm{A}=3\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right] \\
=\left[\begin{array}{ccc}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right] \\
|3 \mathrm{A}|=\left[\begin{array}{ccc}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right] \\
=3\left|\begin{array}{cc}
3 & 6 \\
0 & 12
\end{array}\right|-0\left[\begin{array}{cc}
0 & 6 \\
0 & 12
\end{array}\right]+3\left[\begin{array}{ll}
0 & 3 \\
0 & 0
\end{array}\right] \\
=3[3(12)-0(6)]-0+3[0-0] \\
=3(36)-0+0 \\
=108
\end{array}\)
\(|3 \mathrm{A}|=108 \ldots\text {LHS }\)
RHS: \( 27|\mathrm{A}| \)
\(\begin{array}{l}
27|\mathrm{A}|=27(4) \\
=108
\end{array}\)
\(27|\mathrm{A}|=108\ldots \text {RHS }\)
LHS \( = \) RHS
Hence proved.
5 A. Evaluate the determinants
\(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 1 & -1 \\
3 & -5 & 0
\end{array}\right|\)
\(\left|\begin{array}{ccc}
3 & -1 & -2 \\
0 & 1 & -1 \\
3 & -5 & 0
\end{array}\right|\)
Answer
Now, \( \left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 1 & -1 \\ 3 & -5 & 0\end{array}\right| \)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{ccc}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{cc}
b & c \\
e & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{cc}
a & c \\
d & f
\end{array}\right|+\mathrm{k}=\left|\begin{array}{cc}
a & b \\
d & e
\end{array}\right| \\
=3\left|\begin{array}{cc}
0 & -1 \\
-5 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
0 & -1 \\
3 & 3
\end{array}\right|+(-2)\left|\begin{array}{cc}
0 & 0 \\
3 & -5
\end{array}\right| \\
=3[0-(-1)(-5)]+1[0-(-1)(3)]-2[0-0] \\
=3(-5)+1(3)-0 \\
=-15+3 \\
=-12
\end{array}\)
The determinant of the above matrix is \(-12\)
class 12 maths determinants exercise 4.1 || ex 4.1 class 12 maths ncert solutions || maths class 12 chapter 4 ncert solutions || determinants maths class 12 || exercise 4.1 class 12 maths ncert solutions || class 12 maths chapter 4 exercise 4.1 solutions
5 B. Evaluate the determinants
\(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
\(\left|\begin{array}{ccc}
3 & -4 & 5 \\
1 & 1 & -2 \\
2 & 3 & 1
\end{array}\right|\)
Answer
Now, \( \left|\begin{array}{ccc}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right| \)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{ccc}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{ll}
b & c \\
e & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{ll}
a & c \\
d & f
\end{array}\right|+\mathrm{k}\left|\begin{array}{ll}
a & b \\
d & e
\end{array}\right| \\
=3\left|\begin{array}{cc}
1 & -2 \\
3 & 1
\end{array}\right|-(-4)\left|\begin{array}{cc}
1 & -2 \\
2 & 1
\end{array}\right|+5\left|\begin{array}{cc}
1 & 1 \\
2 & 3
\end{array}\right| \\\end{array}\)
\(=3[1-(-2)(3)]+4[1-(-2)(2)]+5[3-2]\)
\(=3[1+6]+4[1+4]+5[1]\)
\(=3[7]+4[5]+5\)
\(=21+20+5\)
\(=46\)
The determinant of the above matrix is 46 .
5 C. Evaluate the determinants
\(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
\(\left|\begin{array}{ccc}
0 & 1 & 2 \\
-1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right|\)
Answer
Now, \( \left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right| \)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{lll}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{ll}
b & c \\
e & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{ll}
a & c \\
d & f
\end{array}\right|+\mathrm{k}\left|\begin{array}{ll}
a & b \\
d & e
\end{array}\right| \\
=0\left|\begin{array}{cc}
0 & -3 \\
3 & 0
\end{array}\right|-1\left|\begin{array}{cc}
-1 & -3 \\
-2 & 0
\end{array}\right|+2\left|\begin{array}{ll}
-1 & 0 \\
-2 & 3
\end{array}\right| \\\end{array}\)
\(=0-1[0-(-3)(-2)]+2[(-1)(3)-0]\)
\(=0-1[0-6]+2[-3-0]\)
\(=0-1[-6]+2[-3]\)
\(=0+6-6\)
\(=0\)
The determinant of the above matrix is 0 .
5 D. Evaluate the determinants
\(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
\(\left|\begin{array}{ccc}
2 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right|\)
Answer
Now, \( \left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right| \)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{ccc}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{ll}
b & c \\
e & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{cc}
a & c \\
d & f
\end{array}\right|+\mathrm{k}\left|\begin{array}{cc}
a & b \\
d & e
\end{array}\right| \\
=2\left|\begin{array}{cc}
2 & -1 \\
-5 & 0
\end{array}\right|-(-1)\left|\begin{array}{cc}
0 & -1 \\
3 & 0
\end{array}\right|+(-2)\left|\begin{array}{cc}
0 & 2 \\
3 & -5
\end{array}\right| \\\end{array}\)
\(=2[0-(-1)(-5)]+1[0-(-1)(3)]-2[0-3(2)]\)
\(=2[0-5]+1[0+3]-2[-6]\)
\(=2[-5]+1[3]-2[-6]\)
\(=-10+3+12\)
\(=5\)
The determinant of the above matrix is 5 .
6. If \( A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right] \), find \( |A| \).
Answer
GIVEN: \( A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 4 & -3 \\ 5 & 4 & -9\end{array}\right] \)
Now, \( |\mathrm{A}|=\left|\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right| \)
We know that a determinant of a \( 3 \times 3 \) matrix is calculated as
\(\begin{array}{l}
|\mathrm{A}|=\left|\begin{array}{lll}
i & j & k \\
a & b & c \\
d & e & f
\end{array}\right|=\mathrm{i}\left|\begin{array}{cc}
b & c \\
c & f
\end{array}\right|-\mathrm{j}\left|\begin{array}{ll}
a & c \\
d & f
\end{array}\right|+\mathrm{k}\left|\begin{array}{ll}
a & b \\
d & e
\end{array}\right| \\
=1\left|\begin{array}{ll}
1 & -3 \\
4 & -9
\end{array}\right|-1\left|\begin{array}{ll}
2 & -3 \\
5 & -9
\end{array}\right|+(-2)\left|\begin{array}{ll}
2 & 1 \\
5 & 4
\end{array}\right|
\end{array}\)
\(\begin{array}{l}\end{array}\)
\(=1[-9-(-3)(4)]-1[2(-9)-(-3)(5)]-2[2(4)-1(5)]\)
\(=1[-9+12]-1[-18+15]-2[8-5]\)
\(=1[3]-1[-3]-2[3]\)
\(=3+3-6\)
\(=0\)
Ans. \( |\mathrm{A}|=0 \)
7 A. Find values of \( x \), if
\(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
\(\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|\)
Answer
We have
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
\(\begin{array}{l}
\Rightarrow 2(1)-4(5)=2 x(x)-4(6) \\
\Rightarrow 2-20=2 x^{2}-24 \\
\Rightarrow-18=2 x^{2}-24 \\
\Rightarrow 2 x^{2}=-24+18 \\
\Rightarrow 2 x^{2}=6 \\
\Rightarrow x^{2}=\frac{6 }{ 2} \\
\Rightarrow x^{2}=3 \\
x=\sqrt{3}
\end{array}\)
Ans. The value of \( x \) is \( \sqrt{3} \)
7 B. Evaluate the determinants
\(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
\(\left|\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right|=\left|\begin{array}{cc}
x & 3 \\
2 x & 5
\end{array}\right|\)
Answer
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
We have \( \left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right| \)
\(\begin{array}{l}
\Rightarrow 2(5)-3(4)=x(5)-3(2 x) \\
\Rightarrow 10-12=5 x-6 x \\
\Rightarrow-2=-x \\
\Rightarrow x=2
\end{array}\)
Ans. The value of \( x \) is 2 .
8. If \( \left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right| \), then \(x\) is equal to
A. 6 B. \( \pm 6 \) C. \(-6\) D. 0
A. 6 B. \( \pm 6 \) C. \(-6\) D. 0
Answer
We have \( \left|\begin{array}{cc}x & 2 \\ 18 x & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right| \)
We know that determinant of A is calculated as \( |\mathrm{A}|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=\mathrm{ad}-\mathrm{bc} \)
\( \Rightarrow x(x)-2(18)=6(6)-2(18) \)
\( \Rightarrow x^{2}-36=36-36\)
\(\Rightarrow x^{2}=36-36+36\)
\(\Rightarrow x^{2}=36\)
\(\Rightarrow x= \pm 6 \)
