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JEE Maths
Inverse Matrix Questions

Inverse Matrix Questions 

Inverse matrices are an essential concept in linear algebra used to solve systems of linear equations, especially in competitive exams like JEE and board-level mathematics. A matrix has an inverse only if it is square and non-singular (i.e., its determinant is non-zero). Understanding how to compute and apply the inverse of a matrix helps in areas such as cryptography, transformations, and computational mathematics. In this article, we explore key Inverse Matrix Questions and Answers for deeper understanding and practice.

1.0What Is an Inverse Matrix?

If A is a square matrix, and there exists another matrix A^{-1}such that:

A⋅A−1=A−1⋅A=I

Then A−1is called the inverse of matrix A, and II is the identity matrix of the same order.

Not all matrices have inverses. A matrix must be square and non-singular (i.e., det(A)=0).

Inverse of a 2×2 Matrix

IfA=[ac​bd​]Then,A−1=ad−bc1​[d−c​−ba​],if ad−bc=0

2.0Inverse of a 3×3 Matrix 

To find the inverse of a 3×3 matrix, we follow these steps:

Let A=​adg​beh​cfi​​

The inverse of A, denoted as A−1, exists only if det(A)=0, and is given by:

A−1=det(A)1​⋅adj(A) 

Where:

  • det(A) = determinant of matrix A
  • adj(A) = adjugate (or adjoint) of A, which is the transpose of the cofactor matrix

3.0Steps to Find Inverse of 3×3 Matrix

  1. Calculate the Determinant det(A)
  2. Find the Matrix of Minors
  3. Turn it into a Matrix of Cofactors (apply sign pattern)
  4. Transpose the Cofactor Matrix → get the Adjugate
  5. Divide the Adjugate Matrix by the Determinant

4.0Inverse Matrix Questions and Answers

Example 1: Find the inverse of the matrix

Solution:

Example 1: Find the inverse of the matrix A=[21​34​]Solution:Determinant:det(A)=(2)(4)−(3)(1)=8−3=5A−1=51​[4−1​−32​]Answer:A−1=​54​−51​​−53​52​​​

Example 2: If A=[12​24​], does exist?

Solution:

det(A) = (1)(4) - (2)(2) = 4 - 4 = 0 

Since the determinant is 0, the inverse does not exist.

Answer:
No, A−1 does not exist. The matrix is singular.

Example 3: Solve the system using inverse matrix:

x + 2y = 5  

3x + 4y = 6 

Solution:

Write as matrix equation:AX=B,where A=[13​24​],X=[xy​],B=[56​]FindA−1:det(A)=1⋅4−2⋅3=4−6=−2A−1=−21​[4−3​−21​]=[−21.5​1−0.5​]Now:X=A−1⋅B= [−21.5​1−0.5​][56​]⇒X= [(−2)(5)+(1)(6)(1.5)(5)+(−0.5)(6)​][−10+67.5−3​]=[−44.5​]Answer:x=−4,y=29​

Example 4: Find the inverse of the matrix

 A=​211​012​113​​

Solution: 

Step 1: Find the Determinant |A|We’ll use the first row for cofactor expansion:∣A∣=2​12​13​​−0​11​13​​+1​11​12​​⇒∣A∣=2(1⋅3−1⋅2)+1(1⋅2−1⋅1)⇒∣A∣=2(3−2)+(2−1)⇒∣A∣=2+1=3So,∣A∣=3,and the inverse exists.M= ​​12​13​​​02​13​​​01​11​​​​11​13​​​21​13​​​21​11​​​​11​12​​​21​02​​​21​01​​​​M= ​3−20−20−1​3−16−12−1​2−14−02−0​​​1−2−1​251​142​​

Step 3: Matrix of Cofactors (Apply Signs)

Use the checkerboard pattern:

C= ​+−+​−+−​+−+​​⋅M= ​12−1​−25−1​1−42​​ 

Step 4: Find the Adjugate Matrix (Transpose of Cofactor Matrix)

adj(A)=CT= ​1−21​25−4​−1−12​​ 

Step 5: Final Inverse

A−1=∣A∣1​⋅adj(A)=31​ ​1−21​25−4​−1−12​​ 

Final Answer:

A−1= ​31​−32​31​​32​35​−34​​−31​−31​32​​​ 

Example 5: Find the inverse of​105​216​340​​

Solution: 

Step 1: Find the Determinant |A|We expand along the first row: ∣A∣=1⋅ ​ 16​40​ ​ −2⋅ ​ 05​40​ ​ +3⋅ ​ 05​16​ ​Now compute each 2×2 determinant:​   1  6​40  ​   ​=(1)(0)−(4)(6)=−24​     0  5​40  ​   ​=(0)(0)−(4)(5)=−20​     0  5​16  ​   ​=(0)(6)−(1)(5)=−5Now plug in:∣A∣=1(−24)−2(−20)+3(−5)=−24+40−15=1 Since the determinant is non-zero, the inverse exists.

Step 2: Find the Cofactor Matrix

Let’s compute the matrix of minors, then apply the sign pattern.

Minor matrix:

Entry

Minor

Value

C11​

​ 16​40​ ​

-24

C12​

​ 05​40​ ​

-20

C13​

​ 05​16​ ​

-5

C21​

​ 26​30​ ​

-18

C22​

​15​30​ ​

-15

C23​

​ 15​26​ ​

-4

C31​

​ 21​34​ ​

5

C32​

​ 10​34​ ​

4

C33​

​ 10​21​ ​

1

Now apply sign pattern (+ - + / - + - / + - +):

Cofactor Matrix= ​−24185​20−15−4​−541​​

Step 3: Find Adjugate Matrix (Transpose of Cofactor Matrix)

adj(A)=Transpose of Cofactor Matrix=​−2420−5​18−154​5−41​​

Step 4: Final Inverse

Since |A| = 1, the inverse is just:

A−1=11​⋅adj(A)=​−2420−5​18−154​5−41​​

Final Answer:

A−1=​−2420−5​18−154​5−41​​ 

Example 6: Find the inverse of A= ​211​130​120​​ 

Solution:  

Step 1: Compute the Determinant∣A∣=2(3⋅0−2⋅0)−1(1⋅0−2⋅1)+1(1⋅0−3⋅1)∣A∣=2(0)−1(−2)+1(−3)∣A∣=0+2−3=−1So the inverse exists.

Step 2: Find Adjugate of A

After computing cofactors and transposing them (details skipped here for brevity), the adjugate is:

adj(A)= ​0−23​02−1​1−15​​

Step 3: Inverse Formula

A−1=−11​adj(A)=−1⋅adj(A)A−1= ​02−3​0−21​−11−5​​ 

Example 7: Solve the system of equations using matrices:

x + 2y + 3z = 14  

2x + 3y + z = 13  

3x + y + 2z = 13 

Solution: 

Write it in matrix form:AX=B,where  A= ​123​231​312​​,B= ​141313​​Now find A−1 and use X=A−1B.

Solution: After computing A−1, multiply it by B to get X= ​xyz​​

The result will be:

X= ​123​​

5.0Practice Inverse Matrix Questions

  1. Find the inverse of A=  [  3 2​14 ​]
  2. Check if A=  [ 02​20​]is invertible.
  3. Solve: 

2x + 3y = 8  

4x + y = 10 

using inverse matrices.

  1. If A⋅B=I, find B given A.
  2. Prove: (A−1)−1=A

Table of Contents


  • 1.0What Is an Inverse Matrix?
  • 2.0Inverse of a 3×3 Matrix 
  • 3.0Steps to Find Inverse of 3×3 Matrix
  • 4.0Inverse Matrix Questions and Answers
  • 5.0Practice Inverse Matrix Questions

Frequently Asked Questions

When its determinant is zero (i.e., the matrix is singular).

Compute its determinant. If it’s non-zero, the inverse exists.

No. Only square matrices can have true inverses.

A square matrix with 1’s on the diagonal and 0’s elsewhere. It's the multiplicative identity in matrix algebra.

The matrix is non-invertible or singular, and no inverse exists.

For exams, using row operations (Gauss-Jordan method) is faster than adjoint-cofactor method.

Yes, scientific calculators can compute 3×3 inverses, but understanding the process is crucial for theory-based questions.

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