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

Answer (76)

Sol.

A2(A - 4I) - 4(A - I) = 0 

A³ - 4A² - 4A + 4I = 0

Multiple by A

A4 = 4A3 + 4A2 - 4A

= 4(4A² + 4A – 4I) + 4A² – 4A 

= 20A² + 12A – 16I

Multiple again by A

⇒ A⁵ = 20A³ + 12A² – 16A

= 20(4A² + 4A – 4I) + 12A² – 16A

= 92A² + 64A – 80I = αA² + A + γI 

⇒ α = 92, β = 64, γ = -80 ⇒ α + β + γ = 76

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,

$A=\left[\begin{array}{ccc}1& {\log }_{x}y& {\log }_{x}z\\ {\log }_{y}x& 2& {\log }_{y}z\\ {\log }_{z}x& {\log }_{z}y& 3\end{array}\right]$

Let $a={\log }_{x}y,b={\log }_{x}z,c={\log }_{y}z$

⇒ ${\log }_{y}x=\frac{1}{a},{\log }_{z}x=\frac{1}{b},{\log }_{z}y=\frac{1}{c}$

⇒ $A=\left[\begin{array}{ccc}1& a& b\\ \frac{1}{a}& 2& c\\ \frac{1}{b}& \frac{1}{c}& 3\end{array}\right]$

⇒ $|A|=1\cdot \left|\begin{array}{cc}2& c\\ \frac{1}{c}& 3\end{array}\right|-a\cdot \left|\begin{array}{cc}\frac{1}{a}& c\\ \frac{1}{b}& 3\end{array}\right|+b\cdot \left|\begin{array}{cc}\frac{1}{a}& 2\\ \frac{1}{b}& \frac{1}{c}\end{array}\right|$

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

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

⇒ $|A|=\frac{ac}{b}+\frac{b}{ac}$

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| = $\frac{1}{|{\mathrm{A}}^{-1}|}$

Calculation:

Given: $\mathrm{A}=\left[\begin{array}{cc}x& 2\\ 4& 3\end{array}\right]$ and ${\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{1}{8}& \frac{-1}{12}\\ \frac{-1}{6}& \frac{4}{9}\end{array}\right]$

|A^-1| = $\frac{4}{72}-\frac{1}{72}=\frac{3}{72}=\frac{1}{24}$

|A| = $\frac{1}{|{\mathrm{A}}^{-1}|}$ = 24

⇒ 3x - 8 = 24

∴ x = $\frac{32}{3}$

Explanation:

Given 

⇒ $adj(A)=\left[\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}\right]$

Now, A^-1 = $\frac{1}{|A|}(Adj(A))$

= $\left[\begin{array}{ccc}1& -1& 0\\ -2& 3& -4\\ -2& 3& -3\end{array}\right]$

∴ Option (a) is correct.

Concept:

For an invertible matrix A:

• A^-1 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{|\mathrm{A}|}$.
• |A-1| = |A|^-1 = $\frac{1}{|\mathrm{A}|}$.

Calculation:

${\mathrm{A}}^{-1}=\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 3\\ 3& 1& 6\end{array}\right]=\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{\mathrm{k}}$         -----(1)

From the definition of the inverse of a matrix, 

A-1 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{|\mathrm{A}|}$              -----(2)

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

k = |A|  

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

k = |A| = $\frac{1}{|{\mathrm{A}}^{-1}|}$        ----(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 = $-\frac{1}{15}$.

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| = $\frac{1}{|{\mathrm{A}}^{-1}|}$

Calculation:

Given: $\mathrm{A}=\left[\begin{array}{cc}x& 2\\ 4& 3\end{array}\right]$ and ${\mathrm{A}}^{-1}=\left[\begin{array}{cc}\frac{1}{8}& \frac{-1}{12}\\ \frac{-1}{6}& \frac{4}{9}\end{array}\right]$

|A^-1| = $\frac{4}{72}-\frac{1}{72}=\frac{3}{72}=\frac{1}{24}$

|A| = $\frac{1}{|{\mathrm{A}}^{-1}|}$ = 24

⇒ 3x - 8 = 24

∴ x = $\frac{32}{3}$

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) = $\frac{1}{\mathrm{k}}{\mathrm{A}}^{-1}$ 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 = $\frac{\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})}{|\mathrm{A}|}$.
• |A-1| = |A|-1 = $\frac{1}{|\mathrm{A}|}$.

Calculation:

From the definition of the inverse of a matrix, ${\mathrm{A}}^{-1}=\frac{\mathrm{a}\mathrm{d}\mathrm{j}\left(\mathrm{A}\right)}{\left|\mathrm{A}\right|}$.

Multiplying both sides by A, we get:

A(A^-1) = $\frac{\mathrm{A}[\mathrm{a}\mathrm{d}\mathrm{j}(\mathrm{A})]}{|\mathrm{A}|}$

⇒ |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 = $\left[\begin{array}{ccc}1& -3& 2\\ 2& -8& 5\\ 4& 2& \lambda \end{array}\right]$is not an invertible matrix

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

⇒ $\left|\begin{array}{ccc}1& -3& 2\\ 2& -8& 5\\ 4& 2& \lambda \end{array}\right|$ = 0

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

⇒ $-8\lambda -10+6\lambda -60+72=0$

⇒$-2\lambda +2=0$

⇒$\lambda =1$

Hence, If $\left[\begin{array}{ccc}1& -3& 2\\ 2& -8& 5\\ 4& 2& \lambda \end{array}\right]$ 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

$\Rightarrow \mathrm{A}\times \left(\frac{\mathrm{A}\mathrm{d}\mathrm{j}\mathrm{A}}{det\mathrm{A}}\right)=\mathrm{I}$

⇒ A (Adj A) = det A × I = |A|I

Calculation:

Given: A(adj A) $=\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]$

⇒ A(adj A) $=10\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=10\mathrm{I}$

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.

$|\mathrm{A}|=\left|\begin{array}{ccc}1& 0& 2\\ 5& 1& \mathrm{x}\\ 1& 1& 1\end{array}\right|=0$

⇒ 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

${\mathrm{A}}^{-1}=\frac{\mathrm{a}\mathrm{d}\mathrm{j}\left(\mathrm{A}\right)}{\mathrm{d}\mathrm{e}\mathrm{t}\left(\mathrm{A}\right)}$

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 = $\left[\begin{array}{ccc}3& 1& 2\\ 4& 2& 1\\ 2& a& 1\end{array}\right]$

For A^-1 does not exist the |A| = 0

|A| = $\left|\begin{array}{ccc}3& 1& 2\\ 4& 2& 1\\ 2& a& 1\end{array}\right|$ = 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]