Q.Find inverse of the given matrices, by using Adjugate (matrix) method:
Concept understanding — Matrix Inversion
What is Matrix Inversion?
Imagine you have a number, say 5. Its multiplicative inverse is 51, because 5×51=1. That "1" is the identity for multiplication — the number that does nothing when you multiply by it.
Matrix inversion is the exact same idea, but for matrices. For a square matrix A, its inverse A−1 is the matrix that, when multiplied with A, gives the identity matrix I:
A×A−1=A−1×A=I
The identity matrix I is the matrix equivalent of the number 1 — it has 1s on the main diagonal and 0s everywhere else. For a 2×2 matrix, I=[1001].
Only square matrices can have inverses. A 3×2 matrix cannot be inverted — it's like asking for the reciprocal of a number that isn't there.
Why Does This Matter?
In algebra, if you have 5x=20, you solve for x by multiplying both sides by 51: x=51×20=4.
In matrix algebra, if you have Ax=b, you solve for x by multiplying both sides by A−1:
x=A−1b
This is how you solve systems of linear equations — the heart of everything from engineering to economics.
The Precise Definition
Let A be an n×n square matrix. If there exists an n×n matrix B such that:
AB=BA=In
then B is called the inverse of A, written A−1. If such a B exists, A is called invertible or non-singular. If no such B exists, A is singular.
Not every square matrix has an inverse. A matrix with determinant zero is singular — it collapses space into a lower dimension, and you cannot "undo" that collapse.
How to Find the Inverse (for 2×2)
For a 2×2 matrix A=[acbd]:
- Compute the determinant: det(A)=ad−bc
- If det(A)=0, stop — no inverse exists.
- If det(A)=0, the inverse is:
A−1=ad−bc1[d−c−ba]
The pattern is: swap the diagonal entries a and d, negate the off-diagonals b and c, then divide everything by the determinant.
Example
Find the inverse of A=[2513].
det(A)=(2)(3)−(1)(5)=6−5=1=0, so invertible.
A−1=11[3−5−12]=[3−5−12]
Check: A×A−1=[2513][3−5−12]=[6−515−15−2+2−5+6]=[1001] — it works.
The Big Picture
Matrix inversion is the tool that lets you "divide" by a matrix. It undoes a linear transformation. If A rotates and stretches space, A−1 rotates and stretches it back. That's why it only exists when the transformation doesn't collapse anything — when the determinant is non-zero.
The inverse of a matrix A is the unique matrix A−1 such that AA−1=A−1A=I.
The adjugate method finds the inverse as A−1=detA1adj A, where the adjoint is the transpose of the matrix of cofactors.
- A−1=111[2−314],
- A−1=−214−22−8−121−2124.
Using A−1=detA1adj A with adj A=[cofactor]T.
A−1=detA1adj A, valid when detA=0.
(i) A=[43−12]
- detA=(4)(2)−(−1)(3)=8+3=11.
- adj A=[2−314].
- A−1=111[2−314].
(ii) A=243−10−2427
- detA=2(0⋅7−2⋅(−2))−(−1)(4⋅7−2⋅3)+4(4⋅(−2)−0⋅3) =2(4)+1(22)+4(−8)=8+22−32=−2.
- Cofactors: C11=4,C12=−22,C13=−8; C21=−1,C22=2,C23=1; C31=−2,C32=12,C33=4.
- adj A=4−22−8−121−2124 (transpose of the cofactor matrix).
- A−1=−214−22−8−121−2124=−211421−1−211−6−2.
- Check (row 1 of A times col 1 of A−1): 2(−2)+(−1)(11)+4(4)=−4−11+16=1. ✓
A(i)−1=111[2−314], A(ii)−1=−21141/2−1−1/21−6−2.
- CBSE 2025Set 465/W1XZY/41 markMCQQ.The inverse of matrix A=[42−11] is (A) 61[−4−12−1] (B) [313261−61] (C) [61−316132] (D) [−32−3161−61]
›Reveal solutionSolution
detA=6 and adjA=[1−214], so A−1=[1/6−1/31/62/3].
For A=[acbd]: A−1=ad−bc1[d−c−ba].
- Compute the determinant: detA=(4)(1)−(−1)(2)=4+2=6 (non-zero, so the inverse exists).
- Form the adjoint by swapping diagonal entries and negating off-diagonal: adjA=[1−214].
- Multiply by detA1: A−1=61[1−214]=[61−316132].
✓Final answerA−1=[61−316132] — option (C).
- CBSE 2025Set 465/S/WXYZ/41 markMCQQ.If A is an invertible matrix, then which of the following is not true ? (A) ∣A−1∣=∣A∣−1 (B) (A2)−1=(A−1)2 (C) (A′)−1=(A−1)′ (D) ∣A∣=0
›Reveal solutionSolution
Options (A), (C) and (D) are fundamental invertibility properties that always hold; the official marking scheme identifies (B) as the "not true" option.
For an invertible matrix A: ∣A−1∣=∣A∣−1, (AB)−1=B−1A−1, (A′)−1=(A−1)′, and ∣A∣=0.
- (A) ∣A−1∣=∣A∣−1: since AA−1=I, taking determinants gives ∣A∣∣A−1∣=1, so ∣A−1∣=∣A∣−1. True.
- (C) (A′)−1=(A−1)′: transposing AA−1=I gives (A−1)′A′=I, so (A′)−1=(A−1)′. True.
- (D) ∣A∣=0: an invertible matrix is non-singular by definition. True.
- (B) is the option the question asks for ("not true"); the official key designates (B) as the answer. I note that, as written, (A2)−1=(A−1)2 is itself a correct identity ((AA)−1=A−1A−1), so (B) is the designated choice rather than a genuine falsehood among the four.
✓Final answer(B) (A2)−1=(A−1)2 (official key)
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.If A=[2xx0x] and A−1=[1−102], then the value of x is : (A) 1 (B) 21 (C) −21 (D) 2
›Reveal solutionSolution
AA−1=I forces 2x=1, so x=21.
By definition of inverse, AA−1=I=[1001]; multiply and match entries.
- Multiply: [2xx0x][1−102]=[2x+0x−x0+00+2x]=[2x002x].
- Set equal to I: [2x002x]=[1001].
- Therefore 2x=1⇒x=21.
✓Final answer(B) 21
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