NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1

Exercise (4.1)

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

Ans: Solving the determinant  $\begin{vmatrix}2 & 4 \\ -5& -1 \\\end{vmatrix}$ we have:

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

2. Evaluate the determinants.

 i. \[\left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|\] Ans: Solving the determinant

$\left| \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right|$

We have: $\Rightarrow \left| \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right|=\left( \cos \theta  \right)\left( \cos \theta  \right)-\left( -\sin \theta  \right)\left( \sin \theta  \right)$

\[\Rightarrow \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=\text{ }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \] We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=1\] ii. \[\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\]

Ans: Solving the determinant

$\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|$, 

We have: $\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{=}\left( {{\text{x}}^{\text{2}}}\text{-x+1} \right)\left( \text{x+1} \right)\text{-}\left( \text{x-1} \right)\left( \text{x+1} \right)$ 

$\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+x+}{{\text{x}}^{\text{2}}}\text{-x+1-}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$ 

So, $\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{+1-}{{\text{x}}^{\text{2}}}\text{+1}$

$\therefore \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1}  \\ \text{x+1} & \text{x+1}  \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+2}$.

3. If $\text{A=}$\[\left| \begin{matrix}1 & 2  \\   4 & 2  \\ \end{matrix} \right|\], then show that \[\left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

Ans: Given that,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\] Multiplying $\text{A}$ by $2$, we have: \[\Rightarrow \text{2A= 2}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\Rightarrow \text{2A=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\therefore \] L.H.S \[\text{=}\left| \text{2A} \right|\text{=}\left| \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{2A} \right|\text{=}2\times 4-4\times 8\] \[\Rightarrow \left| \text{2A} \right|\text{=}8-32\] \[\therefore \left| \text{2A} \right|\text{=}-24\] The value of determinant $\text{A}$ is \[\Rightarrow \left| \text{A} \right|\text{=}\left| \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2-8\] \[\therefore \left| \text{A} \right|\text{=}-6\] R.H.S is given as $\text{4}\left| \text{A} \right|$. \[\therefore \text{4}\left| \text{A} \right|\text{=4 }\!\!\times\!\!\text{ }\left( \text{-6} \right)\text{=-24}\] Hence, we have L.H.S $=$ R.H.S \[\therefore \left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

4. If \[\text{A=}\left[ \begin{matrix} 1 & 0 & 1  \\  0 & 1 & 2  \\ 0 & 0 & 4  \\  \end{matrix} \right]\], then show that \[\left| {\text{3A}} \right|\text{=27}\left| \text{A} \right|\].

Ans: Given, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right]\] Determining the value of determinant $\text{A}$, by expanding along the first column, i.e., \[\text{C1}\], we get: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{2} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{1} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( 4-0 \right)-0+0\] \[\Rightarrow \left| \text{A} \right|\text{=4}\] \[\therefore 27\left| \text{A} \right|\text{=27}\times \text{4}=\text{108}\] ……(1) The value of $\left| 3\text{A} \right|$ is obtained as: \[\begin{align} & \Rightarrow \text{3A=3}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right] \\ & \Rightarrow \text{3A=}\left[ \begin{matrix} \text{3} & \text{0} & \text{3} \\ \text{0} & \text{3} & \text{6} \\ \text{0} & \text{0} & \text{12} \\ \end{matrix} \right] \\ \end{align}\] \[\therefore \left| \text{3A} \right|\text{=3}\left| \begin{matrix} \text{3} & \text{6} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{3} & \text{6} \\ \end{matrix} \right|\] \[\Rightarrow \text{3}\left( \text{36-0} \right)\text{+0+0}\] \[\Rightarrow \left| \text{3A} \right|\text{=3}\times \text{36}\] Thus, \[\left| \text{3A} \right|\text{=}108\] ……(2) From equations (1) and (2), we have: \[\left| \text{3A} \right|\text{=27}\left| \text{A} \right|\]

Hence proved.

5. Evaluate the determinants

i. \[\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\]

Ans:Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the second row, we have: \[\Rightarrow \left| \text{A} \right|\text{=}-0\left| \begin{matrix} -1 & -2 \\ -5 & 0 \\ \end{matrix} \right|+0\left| \begin{matrix} 3 & -2 \\ 3 & 0 \\ \end{matrix} \right|-\left( -1 \right)\left| \begin{matrix} 3 & -1 \\ 3 & -5 \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}\left( \text{-15+3} \right)\] \[\therefore \left| \text{A} \right|\text{=-12}\]

ii.\[\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \\ \end{matrix} \right|\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-4} & \text{5} \\ \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{3} & \text{1} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{3} & \text{1} \\ \end{matrix} \right|\text{+4}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{2} & \text{1} \\ \end{matrix} \right|\text{+5}\left| \begin{matrix} \text{1} & \text{1} \\ \text{2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{1+6} \right)\text{+4}\left( \text{1+4} \right)\text{+5}\left( \text{3-2} \right)\] \[\Rightarrow \left| \text{A} \right|\text{=21+20+5}\] \[\therefore \left| \text{A} \right|\text{=}46\]

iii. \[\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=0}\left| \begin{matrix} \text{0} & \text{-3} \\ \text{3} & \text{0} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{-1} & \text{-3} \\ \text{-2} & \text{0} \\ \end{matrix} \right|\text{+2}\left| \begin{matrix} \text{-1} & \text{0} \\ \text{-2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=0}\left( 9 \right)-\left( -6 \right)+2\left( -3 \right)\] \[\therefore \left| \text{A} \right|\text{=0}\]

iv. $\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$

Ans: Let $\text{A=}\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$ Determining the value of $\text{A}$ by expanding along the first column, we have: \[\Rightarrow \left| \text{A} \right|\text{=2}\left| \begin{matrix} \text{2} & \text{-1} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{+3}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{2} & \text{-1} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2\left( -5 \right)-0+3\left( 5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}-\text{10+15}\] \[\therefore \left| \text{A} \right|\text{=5}\]

6. If \[\text{A=}\left[ \begin{matrix} 1 & 1 & -2  \\   2 & 1 & -3  \\ 5 & 4 & -9  \\ \end{matrix} \right]\], find \[\left| \text{A} \right|\].

Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{1} & \text{-3} \\ \text{5} & \text{4} & \text{-9} \\ \end{matrix} \right]\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{-3} \\ \text{4} & \text{-9} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{2} & \text{-3} \\ \text{5} & \text{-9} \\ \end{matrix} \right|\text{-2}\left| \begin{matrix} \text{2} & \text{1} \\ \text{5} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( -9+12 \right)-1\left( -18+15 \right)-2\left( 8-5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}3+3-6\] \[\therefore \left| \text{A} \right|\text{=}0\]

7. Find values of \[x\] , if

i. \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( 2\times 1 \right)-\left( 5\times 4 \right)=\left( 2x\times x \right)-\left( 6\times 4 \right)\] \[\Rightarrow 2-20=2{{x}^{2}}-24\] \[\Rightarrow -18+24=2{{x}^{2}}\] \[\Rightarrow 3={{x}^{2}}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x = }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\] ii. \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 4} \right)\text{=}\left( \text{x }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 2x} \right)\] \[\Rightarrow \text{10}-\text{12=5x}-\text{6x}\] \[\Rightarrow -\text{2=}-\text{x}\] Multiplying by $\left( -1 \right)$ on both the sides, we obtain: \[\Rightarrow \text{x = 2}\]

8. If \[\left| \begin{matrix} x & 2  \\ 18 & x  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} 6 & 2 \\  18 & 6  \\ \end{matrix} \right|\], then \[x\] is equal to

1. \[\text{6}\]

2. \[\text{ }\!\!\pm\!\!\text{ 6}\]

3. \[\text{-6}\]

4. \[\text{0}\]

Ans: Given,\[\left| \begin{matrix} \text{x} & \text{2} \\ \text{18} & \text{x} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{6} & \text{2} \\ \text{18} & \text{6} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36 = 36}-\text{36}\] \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36=0}\] \[\Rightarrow {{\text{x}}^{\text{2}}}\text{=36}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x= }\!\!\pm\!\!\text{ 6}\] Hence, B. \[\text{ }\!\!\pm\!\!\text{ 6}\] is the correct answer.

Conclusion

Vedantu's NCERT Solutions for Class 12 Maths Chapter 4 - Determinants, Exercise 4.1, provides a complete understanding of determinants and their properties. It's important to focus on the basic concepts and methods of calculating determinants, as these are fundamental for solving more complex problems. Practice each question thoroughly to build a strong foundation. Pay special attention to the different types of determinant problems to ensure you're well-prepared for the exams.

Class 12 Maths Chapter 4: Exercises Breakdown

CBSE Class 12 Maths Chapter 4 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

Related Links for NCERT Class 12 Maths in Hindi

Explore these essential links for NCERT Class 12 Maths in Hindi, providing detailed solutions, explanations, and study resources to help students excel in their mathematics exams.

Important Related Links for NCERT Class 12 Maths