For any invertible square matrix A, the determinant of its inverse is given by: |A⁻¹| = 1/|A| Since |A| = 4, we have: |A⁻¹| = 1/4 Thus, the correct answer is 1/4. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
For any invertible square matrix A, the determinant of its inverse is given by:
Starting with the differential equation: y (dy/dx) + x = C Rearrange to get a derivative alone on one side y (dy/dx) = C − x Multiply by dx y dy = (C − x) dx Integrate both sides ∫ y dy = ∫ (C − x) dx (1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration Multiply the whole equation by 2Read more
Starting with the differential equation:
y (dy/dx) + x = C
Rearrange to get a derivative alone on one side
y (dy/dx) = C − x
Multiply by dx
y dy = (C − x) dx
Integrate both sides
∫ y dy = ∫ (C − x) dx
(1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration
Multiply the whole equation by 2:
y² = 2Cx − x² + K, where K = 2k
Rewrite to bring together like terms:
x² + y² − 2Cx = −K
Complete the square for the x-terms:
(x − C)² + y² = C² − K
This is the equation of a circle with center at (C, 0) and radius √(C² − K).
Thus, the differential equation represents a family of circles.
For any square matrix A of order n, the determinant of its adjugate is given by: |adj A| = |A|^(n − 1) For a 2×2 matrix (n = 2): |adj A| = |A|^(2 − 1) = |A| Given |A| = −7, we have: |adj A| = −7 Thus, the correct answer is −7. Click here for more solution: https://www.tiwariacademy.com/ncert-solutioRead more
For any square matrix A of order n, the determinant of its adjugate is given by:
|adj A| = |A|^(n − 1)
For a 2×2 matrix (n = 2):
|adj A| = |A|^(2 − 1) = |A|
Among the given options, the differential equation y′ y″ + y = sin x is the only one that involves the second derivative (y″), making it a second order differential equation. Thus, the correct answer is: y′ y″ + y = sin x Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maRead more
Among the given options, the differential equation
y′ y″ + y = sin x
is the only one that involves the second derivative (y″), making it a second order differential equation.
Starting with the differential equation: 2x dx − 5y dy = 0 Separate variables: 2x dx = 5y dy Integrate both sides: ∫ 2x dx = ∫ 5y dy x² = (5/2)y² + C Rearrange the equation: x² − (5/2)y² = C This represents a family of conic sections. Since the equation is of the form: (x²) − (constant)·(y²) = C itRead more
Starting with the differential equation:
2x dx − 5y dy = 0
Separate variables:
2x dx = 5y dy
Integrate both sides:
∫ 2x dx = ∫ 5y dy
x² = (5/2)y² + C
Rearrange the equation:
x² − (5/2)y² = C
This represents a family of conic sections. Since the equation is of the form:
(x²) − (constant)·(y²) = C
it represents a family of hyperbolas (for nonzero C).
If A is any square matrix of order 3 x 3 such that |A| = 4, then the value of |A⁻¹| is
For any invertible square matrix A, the determinant of its inverse is given by: |A⁻¹| = 1/|A| Since |A| = 4, we have: |A⁻¹| = 1/4 Thus, the correct answer is 1/4. Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
For any invertible square matrix A, the determinant of its inverse is given by:
|A⁻¹| = 1/|A|
Since |A| = 4, we have:
|A⁻¹| = 1/4
Thus, the correct answer is 1/4.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
The differential equation y dy/dx + x = C represents
Starting with the differential equation: y (dy/dx) + x = C Rearrange to get a derivative alone on one side y (dy/dx) = C − x Multiply by dx y dy = (C − x) dx Integrate both sides ∫ y dy = ∫ (C − x) dx (1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration Multiply the whole equation by 2Read more
Starting with the differential equation:
y (dy/dx) + x = C
Rearrange to get a derivative alone on one side
y (dy/dx) = C − x
Multiply by dx
y dy = (C − x) dx
Integrate both sides
∫ y dy = ∫ (C − x) dx
(1/2)y² = Cx − (1/2)x² + k, where k is the constant of integration
Multiply the whole equation by 2:
y² = 2Cx − x² + K, where K = 2k
Rewrite to bring together like terms:
x² + y² − 2Cx = −K
Complete the square for the x-terms:
(x − C)² + y² = C² − K
This is the equation of a circle with center at (C, 0) and radius √(C² − K).
Thus, the differential equation represents a family of circles.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9
If A is any square matrix of order 2 x 2 such that |A| = -7, then the value of |adj. A| is
For any square matrix A of order n, the determinant of its adjugate is given by: |adj A| = |A|^(n − 1) For a 2×2 matrix (n = 2): |adj A| = |A|^(2 − 1) = |A| Given |A| = −7, we have: |adj A| = −7 Thus, the correct answer is −7. Click here for more solution: https://www.tiwariacademy.com/ncert-solutioRead more
For any square matrix A of order n, the determinant of its adjugate is given by:
|adj A| = |A|^(n − 1)
For a 2×2 matrix (n = 2):
|adj A| = |A|^(2 − 1) = |A|
Given |A| = −7, we have:
|adj A| = −7
Thus, the correct answer is −7.
Click here for more solution:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4
Which of the following is a second order differential equation?
Among the given options, the differential equation y′ y″ + y = sin x is the only one that involves the second derivative (y″), making it a second order differential equation. Thus, the correct answer is: y′ y″ + y = sin x Click here for more: https://www.tiwariacademy.com/ncert-solutions/class-12/maRead more
Among the given options, the differential equation
y′ y″ + y = sin x
is the only one that involves the second derivative (y″), making it a second order differential equation.
Thus, the correct answer is: y′ y″ + y = sin x
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9
Solution of the differential equation 2x dx – 5y dy = 0 represents
Starting with the differential equation: 2x dx − 5y dy = 0 Separate variables: 2x dx = 5y dy Integrate both sides: ∫ 2x dx = ∫ 5y dy x² = (5/2)y² + C Rearrange the equation: x² − (5/2)y² = C This represents a family of conic sections. Since the equation is of the form: (x²) − (constant)·(y²) = C itRead more
Starting with the differential equation:
2x dx − 5y dy = 0
Separate variables:
2x dx = 5y dy
Integrate both sides:
∫ 2x dx = ∫ 5y dy
x² = (5/2)y² + C
Rearrange the equation:
x² − (5/2)y² = C
This represents a family of conic sections. Since the equation is of the form:
(x²) − (constant)·(y²) = C
it represents a family of hyperbolas (for nonzero C).
Thus, the correct answer is a hyperbola.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9