第 12 类 RD Sharma 解决方案 - 第 6 章行列式练习前。 6.6 |设置 3

问题 38. 写出行列式的值$\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}$ .

解决方案：

 We have,

A = $\begin{bmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{bmatrix}$

|A| = $\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x\end{vmatrix}$

On taking 2x common from R₃ we get,

|A| = $2x\begin{vmatrix}2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4\end{vmatrix}$

As R₁ and R₃ are identical we get

|A| = 0

Therefore, the value of the determinant is 0.

问题 39. 如果 |A| = 2，其中 A 是 2 × 2 矩阵，求 |adj A|。

解决方案：

Given that |A| = 2 and the order of A matrix is 2 x 2

As we know that |adj A| = |A|^n-1

Here the value of n is 2

So, 

|adj A| = |2|^2-1

= 2

Therefore, the value of the |adj A| is 2.

问题 40. 行列式的值是多少$\begin{vmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{vmatrix}$ ?

解决方案：

We have,

A = $\begin{bmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{bmatrix}$

|A| = $\begin{vmatrix}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{vmatrix}$

= 0(18 – 20) – 2(12 – 16) + 0(10 – 12) 

= 0 + 8 + 0

= 8

Therefore, the value of the determinant is 8.

问题 41. x 的值是多少？ $\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}$单数？

解决方案：

As we know, a matrix is singular matrix, when the value of its determinant is 0.

Given that,

A = $\begin{bmatrix}6 - x & 4 \\ 3 - x & 1\end{bmatrix}$

So, 

|A| = $\begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix} = 0$

=> $\begin{vmatrix}6 - x & 4 \\ 3 - x & 1\end{vmatrix} = 0$

=> (6 – x) – 4(3 – x) = 0

=> 6 – x – 12 + 4x = 0

=> 3x – 6 = 0  

=> 3x = 6

=> x = 6/3

=> x = 2

Therefore, the value of x is 2.

问题 42. 3 × 3 阶矩阵 A 满足 |A| = 4. 求 |2 A| 的值。

解决方案：

We have,

A matrix A is of order 3 × 3 so the value of n is 3.

And |A| = 4.

As we know,

=> |K A| = Kⁿ |A|

So, 

|2 A| = 2³ (4)

= 8 (4)

= 32

Therefore, the value of |2 A| is 32.

问题 43. 评估： $\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}$ .

解决方案：

We have,

A = $\begin{bmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{bmatrix}$

|A| = $\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}$

= cos 15° cos 75° – sin 15° sin 75°

As cos A cos B – sin A sin B = cos (A + B), we get 

= cos (15° + 75°) 

= cos 90°

= 0

问题 44. 如果 A = $\begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}$ ，写出元素 a 的辅因子₃₂ 。

解决方案：

We have, 

A = $\begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}$

So, the minor of a₃₂ is,

M₃₂ = $\begin{vmatrix}5 & 8 \\ 2 & 1\end{vmatrix}$

= 5 – 16 

= -11  

Now, the cofactor of a₃₂ is,

A₃₂ = (−1)³⁺² M₃₂ 

= 11

Therefore, the cofactor of element a₃₂ is 11.

问题 45. 如果$\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}$ ，然后写出 x 的值。

解决方案：

We have,

$\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}$

On expanding the determinants of both sides, we get

=> (x + 1) (x + 2) – (x – 1) (x – 3) = 12 + 1

=> x² + 3x + 2 – x² + 4x – 3 = 13

=> 7x – 1 = 13

=> 7x = 14

=> x = 2 

Therefore, the value of x is 2.

问题 46. 如果$\begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}$ ，然后写出 x 的值。

解决方案：

We are given,

=> $\begin{vmatrix}2x & x + 3 \\ 2\left( x + 1 \right) & x + 1\end{vmatrix} = \begin{vmatrix}1 & 5 \\ 3 & 3\end{vmatrix}$

On expanding the determinants of both sides, we get

=> (2x) (x + 1) – 2 (x + 1) (x + 3) = 3 – 15  

=> (x + 1) (2x – 2x – 6) = -12 

=> -6x – 6 = – 12  

=> -6x = -6

=> x = 1

Therefore, the value of x is 1. 

问题 47. 如果$\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}$ ，求 x 的值。

解决方案：

We have,

=> $\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}$

On expanding the determinants of both sides, we get

=> 12x + 14 = 32 – 42  

=> 12x + 14 = -10

=> 12x = -24

=> x = -24/12  

=> x = -2

Therefore, the value of x is -2. 

问题 48. 如果$\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}$ ，写出 x 的值。

解决方案：

Here, we have,

=> $\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}$ 

On expanding the determinants of both sides, we get

=> 2x² – 40 = 18 + 14  

=> 2x² – 40 = 32

=> 2x² = 72  

=> x² = 72/2

=> x² = 36

=> x = ±6

Therefore, the value of x is ±6.

问题 49. 如果 A 是一个 3 × 3 矩阵，|A| ≠ 0 和 |3A| = k |A|然后写出k的值。

解决方案：

We are given,

A is a 3 × 3 matrix.

Also |A| ≠ 0 and |3A| = k |A|.

Let us considered A = $\begin{bmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{bmatrix}$  

3A = $\begin{bmatrix}3 a_1 & 3 a_2 & 3 a_3 \\ 3 b_1 & 3 b_2 & 3 b_3 \\ 3 c_1 & 3 c_2 & 3 c_3\end{bmatrix}$

|3A| = $\begin{vmatrix}3 a_1 & 3 a_2 & 3 a_3 \\ 3 b_1 & 3 b_2 & 3 b_3 \\ 3 c_1 & 3 c_2 & 3 c_3\end{vmatrix}$

On taking 3 common from each row we get,

= $3^3 \begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{vmatrix}$

= 27 |A|

Therefore, the value of k is 27. 

问题 50. 写出行列式的值$\begin{vmatrix}p & p + 1 \\ p - 1 & p\end{vmatrix}$ .

解决方案：

We have,

A = $\begin{bmatrix}p & p + 1 \\ p - 1 & p\end{bmatrix}$

|A| = $\begin{vmatrix}p & p + 1 \\ p - 1 & p\end{vmatrix}$

On expanding the determinant we have,

= p² – (p + 1) (p – 1)

= p² – (p² – 1)

= p² – p² + 1

= 1

Therefore, the value of the determinant is 1.

问题 51. 写出行列式的值$\begin{vmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}$ .

解决方案：

We have,

A = $\begin{bmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{bmatrix}$

|A| = $\begin{vmatrix}x + y & y + z & z + x \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}$

On applying R₁ -> R₁ + R₂ we get,

= $\begin{vmatrix}x + y + z & x + y + z & z + x + y \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}$

On taking x + y + z common from R₁ we have,

= $\left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}$ 

On applying R₃ -> R₃ + 3 R₁ we get,

= $\left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ - 3 & - 3 & - 3\end{vmatrix}$ 

= $\left( x + y + z \right)\begin{vmatrix}1 & 1 & 1 \\ z & x & y \\ 0 & 0 & 0\end{vmatrix}$

On expanding along the last row we get,

= 0

Therefore, the value of the determinant is 0.

问题 52. 如果 A = $\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}$ ，然后对于任何自然数，求 Det(A ⁿ ) 的值。

解决方案：

Given that, A = $\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}$

=> A² = $\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}$

= $\begin{bmatrix}\cos^2 \theta - \sin^2 \theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ - \sin\theta\cos\theta - \cos\theta\sin\theta & - \sin^2 \theta + \cos^2 \theta\end{bmatrix}$

= $\begin{bmatrix}\cos2\theta & \sin2\theta \\ - \sin2\theta & \cos2\theta\end{bmatrix}$

Similarly, Aⁿ = $\begin{bmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{bmatrix}$

So,  

|Aⁿ| = $\begin{vmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{vmatrix}$

= (cos nθ) (cos nθ) + (sin nθ) (sin nθ)

= cos² (nθ) + sin² (nθ)

= 1

Therefore, Det(Aⁿ) = 1.

问题 53. 求最大值$\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}$ .

解决方案：

We have,

A = $\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}$

On applying R₂ -> R₂ – R₁ and R₃ -> R₃ – R₁, we get

|A| = $\begin{vmatrix}1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ 0 & 0 & \cos\theta\end{vmatrix}$  

= sin θ cos θ

= (sin 2θ)/2

We know that −1 ≤ sin2θ ≤ 1.

So, the maximum value of |A| = (1/2) (1) 

= 1/2

Therefore, the maximum value is 1/2.

问题 54. 如果 x ∈ N 并且$\begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}$ = 8，然后求 x 的值。

解决方案：

Here we have,

A = $\begin{bmatrix}x + 3 & - 2 \\ - 3x & 2x\end{bmatrix}$

|A| = 8

On expansion we get,

=> (x + 3) (2x) – (-2) (-3x) = 8

=> 2 x² + 6x – 6x = 8

=> 2 x² = 8

=> 2x² – 8 = 0

=> x² – 4 = 0

=> x² = 4

As x ∈ N, we get

=> x = 2 

Therefore, the value of x is 2.

问题 55. 如果$\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8$ ，写出 x 的值。

解决方案：

We have,

A = $\begin{bmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{bmatrix}$

|A| = $\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8$

=> $\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8$

On expanding along R₁, we get

=> x (-x² – 1) – sin θ (-x sin θ – cos θ) + cos θ (-sin θ + x cos θ) = 8

=> -x³ – x + x sin² θ + sin θ cos θ – sin θ cos θ + x cos² θ = 8

=> -x³ – x + x (sin² θ + cos² θ) = 8

=> -x³ – x + x = 8

=> x³ + 8 = 0

=> x = -2

Therefore, the value of x is -2.

问题 56. 如果 A 是一个 3 × 3 可逆矩阵，那么如果 det(A ^–1 ) = (det A) ^k ， k 的值是多少。

解决方案：

Given that A is a 3 × 3 invertible matrix.

So, we know that

$A^{- 1} = \frac{Adj A}{\left| A \right|}$

Therefore we get,

=> $\left| A^{- 1} \right| = \frac{\left| Adj A \right|}{\left| A \right|}$

As we know that, $\left|adj\left( A \right) \right| = \left| A \right|^{n - 1}$

= $\frac{\left| A \right|^{3 - 1}}{\left| A \right|}$

= $\frac{\left| A \right|^2}{\left| A \right|}$

= |A|

|A^-1| = |A^k|, we get

=> k = 1

Therefore, the value of k is 1.