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

Last updated on Apr 22, 2025

Latest Inverse of a Matrix MCQ Objective Questions

Inverse of a Matrix Question 1:

If A=[122130021], then A-1 is 

  1. [321112115]
  2. [326112225]
  3. [326110205]
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 5 : None of the above

Inverse of a Matrix Question 1 Detailed Solution

Concept:

The inverse of square matrix A exists only when |A| ≠ 0

For a square matrix A, A-1 1|A|adj(A), where adj(A) is the adjoint of A.

Solution:

Given, A=[122130021]

Determinant lAl = [122130021]

= 1(3 - 0) - 2(- 1 - 0) - 2(- 2 - 0)

= 9 

∴ Inverse of matrix A exists.

In order to find adjoint of A ,lets find minor of matrix of A

M[|3021||1001||1302||2221||1201||1202||2230||1210||1213|]

[312612625]

C0-factors of A is 

C[(1)1+1(3)(1)1+2(1)(1)1+3(2)(1)1+4(6)(1)1+5(1)(1)1+6(2)(1)1+7(6)(1)1+8(2)(1)1+9(5)]

[312612625]

Taking Transpose,

CA[366112225]=adj(A)

As we know that,

A-1 1|A|adj(A)

19[366112225] = [132323191929292959]

∴ The inverse of A is, A-1 [132323191929292959]

 

Inverse of a Matrix Question 2:

If A and B are invertible matrices of the same order, then

  1. (AB)-1 = (BA)-1
  2. (AB)-1 = B-1A-1
  3. (AB)-1 = A-1B-1
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : (AB)-1 = B-1A-1

Inverse of a Matrix Question 2 Detailed Solution

Given:

 A and B are invertible matrices of the same order

Concept:

 Matrix is not commutative always then AB and BA is not equal always.

Calculation:

If A and B are invertible matrices of the same order

Then det A ≠ 0 and det B ≠ 0

Now,

AB and BA exists .

and det(AB) = det A + det B ≠ 0

also det(BA) = det B + det A ≠ 0

Hence AB and BA are invertible .

But matrix is not commutative always then AB and BA is not equal always.

 Let AB = C

⇒  A-1AB = A-1C

⇒  IB = A-1C as the identity matrix I = A-1A

⇒  B-1B = B-1A-1C premultiply both sides by B-1

⇒  I = B-1A-1C as B-1B = I, the identity matrix

⇒  C-1=B-1A-1CC-1 post multiple both sides by C-1

⇒  C-1=B-1A-1 as CC-1 = I, the identity matrix

⇒  (AB)-1=B-1A-1

Hence the option (2) is correct.

Inverse of a Matrix Question 3:

For the matrix A=[001010100],A1 is equal to

  1. I
  2. A
  3. -A
  4. 2A

Answer (Detailed Solution Below)

Option 2 : A

Inverse of a Matrix Question 3 Detailed Solution

Formula Used:

Minor of an Element; Mij=det(Aij)

Cofactor of an Element;Cij=(1)i+jMij

adj(A)=CT

Where C is matrix of cofactors.

A1=1det(A)adj(A)

Explanation:

Given matrix A=[001010100]

⇒ det(A) = -1

Minors(A)=[001010100]

Cofactor Matrix of A=[001010100]

adj(A)=[001010100]

 A1=1det(A)adj(A)

A1=11[001010100]

A1=[001010100]

⇒ A-1 = A

Inverse of a Matrix Question 4:

If A=[122130021], then A-1 is 

  1. [321112115]
  2. [326112225]
  3. [326110205]
  4. none of these

Answer (Detailed Solution Below)

Option 4 : none of these

Inverse of a Matrix Question 4 Detailed Solution

Concept:

The inverse of square matrix A exists only when |A| ≠ 0

For a square matrix A, A-1 1|A|adj(A), where adj(A) is the adjoint of A.

Solution:

Given, A=[122130021]

Determinant lAl = [122130021]

= 1(3 - 0) - 2(- 1 - 0) - 2(- 2 - 0)

= 9 

∴ Inverse of matrix A exists.

In order to find adjoint of A ,lets find minor of matrix of A

M[|3021||1001||1302||2221||1201||1202||2230||1210||1213|]

[312612625]

C0-factors of A is 

C[(1)1+1(3)(1)1+2(1)(1)1+3(2)(1)1+4(6)(1)1+5(1)(1)1+6(2)(1)1+7(6)(1)1+8(2)(1)1+9(5)]

[312612625]

Taking Transpose,

CA[366112225]=adj(A)

As we know that,

A-1 1|A|adj(A)

19[366112225] = [132323191929292959]

∴ The inverse of A is, A-1 [132323191929292959]

The correct answer is Option 4.

Inverse of a Matrix Question 5:

If A=[334234011], then A-1 is 

  1. [110234233]
  2. [110232242]
  3. [110234233]
  4. none of these

Answer (Detailed Solution Below)

Option 4 : none of these

Inverse of a Matrix Question 5 Detailed Solution

Concept:

The inverse of square matrix A exists only when |A| ≠ 0

For a square matrix A, A-1 = 1|A|adj(A), where adj(A) is the adjoint of A.

Solution:

Given, A=[334234011]

∴ Determinant, IAI = 3(- 3 + 4) + 3(2 - 0) + 4(- 2 - 0)

= 1

∴ There exists inverse of matrix A

In order to find adjoint of A, lets find minor marrix of A

M[|3411||2401||2301||3411||3401||3301||3434||3424||3323|]

=[122133043]

Co-factors of A is

C[(1)1+1(1)(1)1+2(2)(1)1+3(2)(1)1+4(1)(1)1+5(3)(1)1+6(3)(1)1+7(0)(1)1+8(4)(1)1+9(3)]

[122133043]

Taking Transpose,

CAT = [110234233]= adj(A)

As we know that,

A-1 1|A|adj(A)

11[110234233] = [110234233]

∴ The inverse of matrix A, A-1 is [110234233]

The correct answer is Option 4.

Top Inverse of a Matrix MCQ Objective Questions

Which of the following is the inverse of the matrix A=[3012]

  1. [1301612]
  2. [0161612]
  3. [13161312]
  4. [130012]

Answer (Detailed Solution Below)

Option 1 : [1301612]

Inverse of a Matrix Question 6 Detailed Solution

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Concept:

Let A=[a11a12a13a21a22a23a31a32a33]

The inverse of a matrix:

The inverse of a matrix A is defined by the following formula.

A1=Adj(A)|A|

Where, Adj(A) = [cofactor(A)]T

cofactor(A) =

[+|a22a23a32a33||a21a23a31a33|+|a21a22a31a32||a12a13a32a33|+|a11a13a31a33||a11a12a31a32|+|a12a13a22a23||a11a13a21a23|+|a11a12a21a22|]

For a 2 × 2 matrix there is a short-cut formula for obtaining the inverse:

[a11a12a21a22]1=1a11a22a12a21[a22a12a21a11]

Calculation:

A=[3012]

A1=160[2013]=[1301612]

Note: Inverse exists only for a non-singular matrix (i.e. |A|0)

Let M4 = I, (where I denotes the identity matrix) and M ≠ I, M2 ≠ I and M3 ≠ I. Then, for any natural number k, M−1 equals:

  1. M4k + 1
  2. M4k + 2
  3. M4k + 3
  4. M4k

Answer (Detailed Solution Below)

Option 3 : M4k + 3

Inverse of a Matrix Question 7 Detailed Solution

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Explanation:

Let M4 = I      ---(i)

Multiplying by M-1 on both sides

M4 M-1 = I – M-1

M3 = M-1      ---(ii)

Multiplying M4 on both sides in equation (i)

M4 M4 = I M4

M8 = M4      ---(iii)

Multiplying M-1 on both sides, we get:

M8 – M-1 = M4 M-1

M7 = M3      ---(iv)

From (ii) and (iv)

M-1 = M3 = M7

Now again multiplying M4 on both sides in equation (iii)

M8 M4 = M4 M4

M12 = M8

M12 = M4 [from (iii)]

M12 = I      ---(v) [from (i)]

Multiplying M-1 on both sides we get

M11 = M-1      ---(vi)

From (ii), (iv) and (vi)

M-1 = M3 = M7 = M11 = …. & So on

3, 7, 11 ….. from general term 4k + 3 where k = 0, 1, 2

Hence, M-1 = M4k + 3

Option C is correct.

The inverse of (3175) is

  1. (5173)
  2. (5173)
  3. 18(5173)
  4. 8(5173)

Answer (Detailed Solution Below)

Option 3 : 18(5173)

Inverse of a Matrix Question 8 Detailed Solution

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Given: (3175)

Concept used:

The inverse of matrix (A) = adjoint of A / det(A) 

or (abcd)(-1) = (dbca)/ (Determinant A)

Calculations:

The inverse of (3175) = (5173)/ (Determinant of given matrix)

⇒ (5173)/ (15 - 7)

⇒ 1/8 ×  (5173)

Hence, The correct option is 3.

 

If a matrix is given by A = [123045001] , then the determinant of A-1 is:

  1. 3
  2. 13
  3. 14
  4. 4

Answer (Detailed Solution Below)

Option 3 : 14

Inverse of a Matrix Question 9 Detailed Solution

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Concept:

Determinant of (A-1)1Determinant of A

Where, Determinant of A = 1(4 × 1 - 5 × 0) - 2(0 - 0) + 3(0 - 0) = 4

Calculation:

Given:

Determinant of A = 4 

So, Determinant of (A-1) = 1Determinant of A = 14

Additional Information

A-1 = adjADeterminant of A

adj.A = Transverse of matrix [A11A12A13A21A22A23A31A32A33]

A11 = Aij = (-1)× (cross multiplication of remaining column and row) = (-1)n × (A22A23A32A33)

Where, n = (i + j)

We follow the same procedure for the A11, A12, A13, .......as so on. Also find the determinant of matrix A. Then applies the given formula after that we finally obtain the A-1.

Consider a 2 × 2 matrix M=[v1v2], where, v1 and v2 are the column vectors. Suppose M1=[u1Tu2T], where uT1 and uT2 are the row vectors. Consider the following statements.

Statement: uT1v1 = 1 and uT2v2 = 1

Statement: uT1v2 = 0 and uT2v1 = 0

Which of the following options is correct?

  1. Statement 1 is true and statement 2 is false
  2. Statement 2 is true and statement 1 is false
  3. Both the statements are true
  4. Both the statements are false

Answer (Detailed Solution Below)

Option 3 : Both the statements are true

Inverse of a Matrix Question 10 Detailed Solution

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Let matrix M=[abcd]

v1=[ac],v2=[bd]

M1=1adbc[dbca]

u1T=1adbc[db]

u2T=1adbc[ca]

u1Tv1=1adbc[db][ac]=1

u2Tv2=1adbc[ca][bd]=1

u1Tv2=1adbc[db][bd]=0

u2Tv1=1adbc[ca][ac]=0

Both the given statements are true.

 Let A and B be matrices of order 3. Which of the following is true?

  1. (AB)-1 = A-1B-1
  2. (AB)-1 = AB-1
  3. (BA)-1 = B-1
  4. (BA)-1 = A-1B-1

Answer (Detailed Solution Below)

Option 4 : (BA)-1 = A-1B-1

Inverse of a Matrix Question 11 Detailed Solution

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Concept:

Apply element-wise inversion i.e multiply by the inverse of the same element to make it an identity matrix until the desired objective is reached.

Calculation:

Let (BA)-1 = K

Multiply by the BA on both sides, in the same order

(BA)(BA)-1 = (BA)K

I = BAK   {∵ (A)(A-1) = I}

Multiply B-1 on both the sides

B-1I = B-1BAK

B-1 = AK

Multiply A-1 on both sides

A-1B-1 = A-1AK

A-1B-1 = K

so, (BA)-1 = A-1B-1  {∵ (BA)-1 = K}

A is an (m × n) matrix with m > n and ‘I’ is Identity matrix. Let A1 = (AT A)-1 AT, Then which of the following statement is false?

  1. AA1A = A
  2. (AA1)2 = AA1
  3. AA1 = I
  4. AA1A = A1

Answer (Detailed Solution Below)

Option 4 : AA1A = A1

Inverse of a Matrix Question 12 Detailed Solution

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A1 = (AT A)-1 AT

From the property, (AB)-1 = B-1A-1

⇒ A1 = A-1 (AT)-1 AT

⇒ A1 = A-1 I

⇒ A1 = A-1

1. AA1A = AA-1A = A

2. (AA1)2 = (AA-1)2 = I = AA-1 = AA1

3. AA1 = AA-1 = I

4. AA1A = AA-1A = A ≠ A1

The inverse of the matrix [234431124] is

  1. [2459534514511565]
  2. [2459534514511565]
  3. [104915414516]
  4. [104915414516]

Answer (Detailed Solution Below)

Option 2 : [2459534514511565]

Inverse of a Matrix Question 13 Detailed Solution

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Let,

A=[234431124] 

A1=1|A|×(adjointA)

CofactorofMatrix=[101554419146]

Adj(A)=[104915414516]

|A|=|234431124|=2[122]3[161]+4[83]

⇒ |A| = 20 – 45 + 20 = -5

A1=Adj(A)|A|=15[104915414516]

A1=[2459534514511565] 

If A = [101211232], then the top row of A-1 is

  1. [2 0 -1]
  2. [2 -1 12]
  3. [5 6 4]
  4. [5 -3 1]

Answer (Detailed Solution Below)

Option 4 : [5 -3 1]

Inverse of a Matrix Question 14 Detailed Solution

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Given:

[101211232]

Concept Used:

A-1 = adjA|A|

Here, adj A = transpose of cofactors matrix.

Solution:

Let C is the cofactors matrix we can calculate cofactors matrix by minors 

C = [a11a12a13a21a22a23a31a32a33]

Here, a11 = (- 1 )1 + 1 minor a11 = 1 x (2 + 3) = 5, {where minor is determinant after excluding its row and column.

similarly, a12 = (- 1)1 + 2 minor a12 = (- 1) x (4 + 2) = - 6

a13 = (- 1)1 + 3 (6 - 2) = 4

a21 = (- 1)2 + 1 (0 + 3) = - 3

a22 = (-1)2 + 2 (2 +2) = 4

a23 = (- 1)2 + 3 (3 - 0) = - 3

a31 = (- 1)3 + 1 (0 + 1) = 1

a32 = (- 1)3 + 2 (- 1 + 2) = - 1

a33 = (- 1)3 + 3 (1 + 0) = 1

So, the cofactors matrix will be given by

C = [564343111]

Adj A = [531641431]

|A| = 1 (2 + 3) - 0 (4 + 2) - 1 (6 - 2)

|A| = 5 -4 

|A| = 1

Hence, A-1 = [531641431]

Top row of A-1 = [ 5     - 3     1]

 Option 4 is correct.

 

The inverse of the matrix [3+2iii32i]is

  1. 112[3+2iii32i]
  2. 112[32iii3+2i]
  3. 114[3+2iii32i]
  4. 114[32iii3+2i]

Answer (Detailed Solution Below)

Option 2 : 112[32iii3+2i]

Inverse of a Matrix Question 15 Detailed Solution

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Concept:

For a 2 x 2 matrix, there is a short-cut formula for inverse is given below

[abcd]1=1(adbc)[dbca]

Calculation:

We know that inverse of the matrix [abcd]1=1(adbc)[dbca]

a = 3 + 2i, b = -i

c = i, d = 3 - 2i

⇒ [3+2iii32i]1=1(3+2i)(32i)1[32iii3+2i]

⇒ [3+2iii32i]1=14+91[32iii3+2i]

⇒ [3+2iii32i]1=112[32iii3+2i]

Hence, option 2 is correct.

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