Q.The inverse of matrix A=[42−11] is (A) 61[−4−12−1] (B) [313261−61] (C) [61−316132] (D) [−32−3161−61]
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🔒 Start your 14-day free trial to unlock the full solution →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 inverse of a 2×2 matrix is detA1 times the adjoint (swap the diagonal entries, negate the off-diagonal). With detA=6, this gives 61[1−214]=[1/6−1/31/62/3]. …
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]. …
- 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]. …
- 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. …
- 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]. …
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