NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths 4 Determinants Ex 4.1.
- Determinants Class 12 Ex 4.2
- Determinants Class 12 Ex 4.3
- Determinants Class 12 Ex 4.4
- Determinants Class 12 Ex 4.5
- Determinants Class 12 Ex 4.6
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Determinants |
| Exercise | Ex 4.1 |
| Number of Questions Solved | 8 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1
Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
Solution:
\(\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}\)
= 2x(-1)-(-5)x(4)
=-2+20
=18
Ex 4.1 Class 12 Maths Question 2.
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
(ii) \(\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}\)
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1
Ex 4.1 Class 12 Maths Question 3.
If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A|=|4A|
Solution:
\(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\)
=> \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
L.H.S = |2A|
= \(2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
= – 24
Ex 4.1 Class 12 Maths Question 4.
\(A=\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right] \) , then show that |3A| = 27|A|
Solution:
3A = \(3\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right] \)
= \(3\left[ \begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix} \right] \)
Ex 4.1 Class 12 Maths Question 5.
Evaluate the following determinant:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
(ii) \(\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} \right| \)
(iii) \(\left| \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right| \)
(iv) \(\left| \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
Solution:
(i) \(\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right| \)
>
Ex 4.1 Class 12 Maths Question 6.
If \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \), find |A|
Solution:
|A| = \(\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right] \)
= 1(-9+12)-1(-18+15)-2(8-5)
= 0
Ex 4.1 Class 12 Maths Question 7.
Find the values of x, if
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
(ii)\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
Solution:
(i) \(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii)\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}\)
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2
Ex 4.1 Class 12 Maths Question 8.
If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\), then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\)
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6
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