Adjoint and Inverse of a Square Matrix MCQ Quiz - Objective Question with Answer for Adjoint and Inverse of a Square Matrix - Download Free PDF

Calculation: 

\(\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]\left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)

\(\mathrm{B}^{2}=\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\mathrm{i} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \mathrm{i} \\ 0 & 1 \end{array}\right] \)

\(\mathrm{B}^{3}=\left[\begin{array}{cc} 1 & -3 \mathrm{i} \\ 0 & 1 \end{array}\right]\)

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\(\mathrm{B}^{2023}=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right]\)

M = A^TBA

M² = M.M = A^TBA A^TBA = A^TB²A

M³ = M².M = A^TB²AA^TBA = A^TB³A

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M²⁰²³ = …………… A^TB²⁰²³A

AM²⁰²³A^T = AA^TB²⁰²³ AA^T =  B²⁰²³

\(=\left[\begin{array}{cc} 1 & -2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \)

Inverse of (AM²⁰²³A^T) is \(\left[\begin{array}{cc} 1 & 2023 \mathrm{i} \\ 0 & 1 \end{array}\right] \)

Hence, the correct answer is Option 4

Concept:

Determinant of Adjoint of a Matrix:

• If \( A \) is a square matrix of order \( n \), then:
• \( |\text{adj}(A)| = |A|^{n-1} \)
• \( |\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2} \)
• \( \text{adj}(A^2) = (\text{adj}(A))^2 \) if \( A \) is invertible
• Hence, \( |\text{adj}(\text{adj}(A^2))| = |\text{adj}((\text{adj}(A))^2)| = |\text{adj}(B^2)| \) where \( B = \text{adj}(A) \)
• Using \( |\text{adj}(B^2)| = |B^2|^{n-1} = (|B|^2)^{n-1} = |B|^{2(n-1)} \)
• Also, \( |B| = |\text{adj}(A)| = |A|^{n-1} \)

Matrix Determinant:

• \( |A| \) represents the scalar determinant value of matrix \( A \)
• If \( A \) is 3 × 3, then \( |\text{adj}(\text{adj}(A^2))| = |A|^{2(n-1)^2} = |A|^8 \)

Calculation:

Given,

\( A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix} \)

Let \( a = \log_x y,\ b = \log_x z,\ c = \log_y z \)

⇒ \( \log_y x = \frac{1}{a},\ \log_z x = \frac{1}{b},\ \log_z y = \frac{1}{c} \)

⇒ \( A = \begin{bmatrix} 1 & a & b \\ \frac{1}{a} & 2 & c \\ \frac{1}{b} & \frac{1}{c} & 3 \end{bmatrix} \)

⇒ \( |A| = 1 \cdot \begin{vmatrix} 2 & c \\ \frac{1}{c} & 3 \end{vmatrix} - a \cdot \begin{vmatrix} \frac{1}{a} & c \\ \frac{1}{b} & 3 \end{vmatrix} + b \cdot \begin{vmatrix} \frac{1}{a} & 2 \\ \frac{1}{b} & \frac{1}{c} \end{vmatrix} \)

⇒ \( |A| = (6 - 1) - a\left(\frac{3}{a} - \frac{c}{b}\right) + b\left(\frac{1}{ac} - \frac{2}{b}\right) \)

⇒ \( |A| = 5 - (3 - \frac{ac}{b}) + (\frac{b}{ac} - 2) \)

⇒ \( |A| = \frac{ac}{b} + \frac{b}{ac} \)

Using values: \( x = 2,\ y = 4,\ z = 8 \Rightarrow a = 2,\ b = 3,\ c = \frac{3}{2} \)

⇒ \( |A| = \frac{2 \times \frac{3}{2}}{3} + \frac{3}{2 \times \frac{3}{2}} = 1 + 1 = 2 \)

Matrix is of order \( n = 3 \)

⇒ \( |\text{adj}(\text{adj}(A^2))| = |A|^{2(n-1)^2} = 2^{2 \times 4} = 2^8 \)

∴ hence Option2 is the correct answer.

Concept:

A × A-1 = I, where I is an identity matrix

|A| = \(\rm 1\over {|A^{-1}|}\)

Calculation:

Given: \(\rm A=\begin{bmatrix} x & 2 \\\ 4 & 3 \end{bmatrix}\) and \(\rm A ^{-1}=\begin{bmatrix} {1\over8} & {-1\over 12} \\\ {-1\over 6}& {4\over 9} \end{bmatrix}\)

|A^-1| = \(\rm {4\over 72} - {1\over 72} = {3\over 72} = {1\over 24}\)

|A| = \(\rm {1 \over {|A^{-1}|}}\) = 24

⇒ 3x - 8 = 24

∴ x = \(\rm 32\over 3\)

Explanation:

Given 

⇒ \(adj(A) = \rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)

Now, A^-1 = \(\frac{1}{|A|} (Adj(A))\)

= \( \rm \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix}\)

∴ Option (a) is correct.

Explanation:

Given:

\(\rm A=\begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} \)

Now, |A| = 3(–3 + 4) –2(–3 + 4) + 0 = 3 – 2 = 1

A(adjA) = |A| I = I

∴ Option (d) is correct.

Concept:

For an invertible matrix A:

• A^-1 = \(\rm \frac{adj(A)}{|A|}\).
• |A-1| = |A|^-1 = \(\rm \frac{1}{|A|}\).

Calculation:

\(\rm A^{-1}=\begin{bmatrix} 1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{bmatrix}=\frac{adj(A)}{k}\)         -----(1)

From the definition of the inverse of a matrix, 

A-1 = \(\rm \frac{adj(A)}{|A|}\)              -----(2)

Comparing equation (1) & (2), we get

k = |A|  

Using the properties of the determinant of inverse of a matrix, we have:

k = |A| = \(\rm \frac{1}{|A^{-1}|}\)        ----(3)

We know, 

A.A^-1 = I

⇒ |A.A-1| = |I| = 1

⇒ |A| |A-1| = 1

⇒ |A| = 1/ |A-1|       ....(4)

Now,

|A^-1| = 1(24 - 3) + 2(9 - 12) + 3(2 - 12) = 21 - 6 - 30 = - 15.

|A-1| = -15

Therefore, from equation (3)

k = \(\rm - \frac1{15}\).

Mistake PointsNote that, we have A^-1 matrix, not an A matrix. So to find the value of k, don't you have to use relation |A| = 1/|A^-1|

Concept:

A × A-1 = I, where I is an identity matrix

|A| = \(\rm 1\over {|A^{-1}|}\)

Calculation:

Given: \(\rm A=\begin{bmatrix} x & 2 \\\ 4 & 3 \end{bmatrix}\) and \(\rm A ^{-1}=\begin{bmatrix} {1\over8} & {-1\over 12} \\\ {-1\over 6}& {4\over 9} \end{bmatrix}\)

|A^-1| = \(\rm {4\over 72} - {1\over 72} = {3\over 72} = {1\over 24}\)

|A| = \(\rm {1 \over {|A^{-1}|}}\) = 24

⇒ 3x - 8 = 24

∴ x = \(\rm 32\over 3\)

Concept:

Properties of Matrices Inverse:

If A and B are the non-singular matrices, then the inverse matrix should have the following properties

• (AB) - 1 = B - 1 A - 1
• (A - 1) - 1 = A
• (AT) - 1 = (A - 1)T
• (KA - 1) = \(\rm \frac{1}{k}\;{A^{ - 1}}\) for any K ≠ 0
• (An) - 1 = (A - 1)n
• AA - 1 = A - 1A = I

Calculation:

Given: A2 - 2A - I = 0

⇒ A.A - 2A = I

Post multiply by A^-1, we get

⇒ AAA-1 - 2AA-1 = IA-1

⇒ AI - 2I = A^-1             [∵ AA - 1 = A - 1A = I]

∴ A^-1 = A - 2

the inverse of A is A - 2

Concept:

For an invertible matrix A:

• A-1 = \(\rm \frac{adj(A)}{|A|}\).
• |A-1| = |A|-1 = \(\rm \frac{1}{|A|}\).

Calculation:

From the definition of the inverse of a matrix, \({{\rm{A}}^{ - 1}} = \frac{{{\rm{adj}}\left( {\rm{A}} \right)}}{{\left| {\rm{A}} \right|}}\).

Multiplying both sides by A, we get:

A(A^-1) = \(\rm \frac{A[adj(A)]}{|A|}\)

⇒ |A| I = A[adj(A)]

But it is given that A is a singular matrix, i.e. |A| = 0.

∴ A[adj(A)] = 0, or A[adj(A)] is a null matrix.

Concept:

If the matrix A is not an invertible matrix then | A | = 0

If the matrix A is the non-singular matrix then | A | ≠ 0 

Calculations:

Given, A = \(\begin{bmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{bmatrix}\)is not an invertible matrix

As we know, If the matrix A is non invertible matrix then | A | = 0

⇒ \(\begin{vmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{vmatrix}\) = 0

⇒ 1\(\rm (-8\lambda - 10)+3(2\lambda-20)+2(4+32)\) = 0

⇒ \(\rm -8\lambda - 10+6\lambda-60+72 = 0\)

⇒\(\rm -2\lambda +2 = 0\)

⇒\(\rm \lambda = 1\)

Hence, If \(\begin{bmatrix} 1 & -3 & 2 \\\ 2 & -8 & 5 \\\ 4 & 2 & λ \end{bmatrix}\) is not an invertible matrix, then the value of λ is 1.

Concept:

Determinants:

• For two invertible matrices A and B, we have: det(A × B) = det(A) × det(B), which can also be written as |A × B| = |A| × |B|.
• |adj(A)| = |A|n - 1, where n is the order of the square matrix A.

Calculation:

We know that |adj(A)| = |A|n - 1, where n is the order of the square matrix A.

Now, |A × adj(A)| = |A × |A|^{n - 1}| = |A|ⁿ.

The order of the given matrix A is n = 3 and |A| = 4.

∴ |A × adj(A)| = |A|n = 4³ = 64.

Concept:

Let A is an invertible matrix

As we know, AA^-1 = I

\( ⇒ {\rm{A}} × \left( {\frac{{{\rm{Adj\;A}}}}{{\det {\rm{A}}}}} \right) = {\rm{I}}\)

⇒ A (Adj A) = det A × I = |A|I

Calculation:

Given: A(adj A) \(=\begin{bmatrix} 10 & 0 \\\ 0 & 10 \end{bmatrix}\)

⇒ A(adj A) \(= 10\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix} = 10\rm I\)

As we know A (Adj A) = det A × I

∴ det A = |A| = 10

Concept:

Singular Matrix:

• A Singular Matrix is a matrix whose 'Multiplicative Inverse' does not exist. i.e. A × A^-1 ≠ I.
• A matrix is singular if and only if its determinant is zero. i.e. |A| = 0.

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​Calculation:

For the matrix to be singular, its determinant should be zero.

\(\rm |A|=\begin{vmatrix}1 & 0 & 2 \\ 5 & 1 & \rm x \\ 1 & 1 & 1 \end{vmatrix}=0\)

⇒ 1(1 × 1 - 1 × x) + 0(1 × x - 1 × 5) + 2(5 × 1 - 1 × 1) = 0

⇒ 1 - x + 0 + 8 = 0

⇒ x = 9.

Concept:

Consider a matrix A and let its inverse be A^-1

\(\rm {A^{ - 1}} = \frac{{{\rm{adj\;}}\left( {\rm{A}} \right){\rm{\;}}}}{{{\rm{det\;}}\left( {\rm{A}} \right)}}\)

Here; adj (A) is adjoint of matrix A and det (A) is determinant of matrix A.

⇒ If det (A) ≠ 0, so the inverse of a matrix exists.

⇒ If det (A) = 0, so inverse of a matrix does not exist.

Calculation:

Given A = \(\begin{bmatrix}3 & 1 & 2 \\ 4&2 & 1\\ 2 & a & 1 \end{bmatrix}\)

For A^-1 does not exist the |A| = 0

|A| = \(\begin{vmatrix}3 & 1 & 2 \\ 4&2 & 1\\ 2 & a & 1 \end{vmatrix}\) = 0

|A| = 3(2 - a) - 1(4 - 2) + 2(4a - 4)

|A| = 6 - 3a - 2 + 8a - 8

|A| = 5a - 4

|A| = 0

5a - 4 = 0

∴ a = \(\frac 4 5\)

Concept

If A is any matrix of order n and it’s inverse exists, then we can write

AA^-1 = A^-1A = I, where I = Identity matrix of order n

Calculation

Given: A is an identity matrix of order 3 i.e. A = I

Multiplying both sides by A^-1 we get

⇒ AA^-1 = IA^-1

⇒ I = A^-1 [∵ A matrix multiplied by the identity matrix is the matrix itself i.e. AI = A]