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  1. The given differential equation is:   2x dy - y dx = 0 Step 1: Rearrange the equation Put the equation into the separable variable form as:   (2x dy) = (y dx)   Now dividing both sides by x and changing,   dy/dx = y / (2x) Step 2: Solution through Separating Variables Write it,   dy/y = dx/(2x) InteRead more

    The given differential equation is:
     
    2x dy – y dx = 0
    Step 1: Rearrange the equation
    Put the equation into the separable variable form as:
     
    (2x dy) = (y dx)
     
    Now dividing both sides by x and changing,
     
    dy/dx = y / (2x)
    Step 2: Solution through Separating Variables
    Write it,
     
    dy/y = dx/(2x)
    Integrating on both sides:
     
    ∫ (1/y) dy = ∫ (1/2x) dx
     
    ln|y| = (1/2) ln|x| + C
    Step 3: Putting into Exponential Form
    Now, taking the exponent on both sides:

    y = e^C * x^(1/2)

    Let e^C = C’, then:

    y = C’ x^(1/2)

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-9

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  2. We are given that A is a square matrix of order 2, and |A| = |kA|. We have to compute the sum of all possible values of k. First recall that if A is any square matrix of order 2, then the determinant of kA where k is any scalar is given by |kA| = k² |A| This is because for a 2x2 matrix, multiplicatiRead more

    We are given that A is a square matrix of order 2, and |A| = |kA|. We have to compute the sum of all possible values of k.

    First recall that if A is any square matrix of order 2, then the determinant of kA where k is any scalar is given by
    |kA| = k² |A|

    This is because for a 2×2 matrix, multiplication of the matrix A by scalar k scales the determinant by k².

    We know that |A| = |kA|. Using the above formula, we substitute for the value of |kA|,

    |A| = k² |A|

    If |A| ≠ 0, we can divide both sides by |A| to get,

    1 = k²

    This gives two values for k

    k = 1 or k = -1

    Therefore, the sum of all possible values of k is:

    1 + (-1) = 0

    So, the correct answer is 0.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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    • 15
  3. We have the equation of a square matrix A as A² - A + I = O Rearrange the equation to get A² - A = -I Now we factor the left-hand side, A(A - I) = -I To get the inverse of A (A⁻¹), multiply both sides of the equation by A⁻¹, A⁻¹ * A(A - I) = A⁻¹ * (-I) This gives us, (A - I) = -A⁻¹ Thus we can writeRead more

    We have the equation of a square matrix A as

    A² – A + I = O

    Rearrange the equation to get
    A² – A = -I

    Now we factor the left-hand side,
    A(A – I) = -I

    To get the inverse of A (A⁻¹), multiply both sides of the equation by A⁻¹,

    A⁻¹ * A(A – I) = A⁻¹ * (-I)

    This gives us,
    (A – I) = -A⁻¹

    Thus we can write
    A⁻¹ = I – A

    Hence, the correct answer is option (c) I – A.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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    • 25
  4. We are given that A is a square matrix of order 3x3 and |adj(A)| = 25. We need to find the value of |A|. The formula for the relation between the determinant of a matrix A and its adjoint is given as under: |adj(A)| = |A|(n -1), where n is the order of the given matrix. When n = 3 for 3x3 matrix, itRead more

    We are given that A is a square matrix of order 3×3 and |adj(A)| = 25. We need to find the value of |A|.
    The formula for the relation between the determinant of a matrix A and its adjoint is given as under:
    |adj(A)| = |A|(n -1), where n is the order of the given matrix.
    When n = 3 for 3×3 matrix, it becomes:
    |adj(A)| = |A|²
    We are given that |adj(A)| = 25.
    |A|² = 25

    We find the square root of both sides of the equation
    |A| = ±5

    Since we know that |A| is non-positive we choose negative sign
    |A| = -5

    So, the correct answer is -5.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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  5. A square matrix A |A| = 5 find |AA T| Theorem on Determinants for Matrix multiplication: |AB| = |A||B| AAᵀ For matrix multiplication. Then, using this theorem. |AA ᵀ |= |A|. |A|ᵀ| AṀ A being a square Matrix | AṀ ṀT|= |AT | Hence, |Aᵀ| = |A|. And we get; |AA T |= | A||A | = | A|² Since, the value ofRead more

    A square matrix A |A| = 5 find |AA T|

    Theorem on Determinants for Matrix multiplication:
    |AB| = |A||B|
    AAᵀ For matrix multiplication. Then, using this theorem. |AA ᵀ |= |A|. |A|ᵀ|
    AṀ A being a square Matrix
    | AṀ ṀT|= |AT |
    Hence, |Aᵀ| = |A|.
    And we get;
    |AA T |= | A||A | = | A|²
    Since, the value of A = 5
    |AAᵀ| = 5² = 25

    Therefore, the correct value of |AAᵀ| is 25.

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    https://www.tiwariacademy.com/ncert-solutions/class-12/maths/#chapter-4

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