Q.Find the inverse of the following matrix, if it exists: A=24−71−12301
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The Inverse Matrix Method
Many problems reduce to a system of linear equations, for example
2x+3yx− y=8=−1.
The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.
Writing the system as AX=B
Collect the coefficients, the unknowns and the constants:
A=(213−1),X=(xy),B=(8−1),
so the whole system becomes AX=B.
The idea
For numbers, ax=b gives x=a−1b provided a=0. The same works for matrices: if A is invertible, multiply AX=B on the left by A−1:
A−1(AX)=A−1B⇒IX=A−1B⇒X=A−1B.
X=A−1B
Multiplying on the left matters — matrix products do not commute, so BA−1 would be wrong.
When it works
The inverse A−1 exists only when detA=0, so:
- detA=0: the system is consistent with the unique solution X=A−1B.
- detA=0: no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.
Worked steps
For the system above, detA=(2)(−1)−(3)(1)=−5=0, and
A−1=−51(−1−1−32).
Then …
Use A−1=detA1adj(A) for A=24−71−12301.
Determinant (expand along row 1): 2(−1−0)−1(4−0)+3(8−7)=−2−4+3=−3=0, so A−1 exists.
Cofactors: C11=−1, C12=−4, C13=1; C21=5, C22=23, C23=−11; C31=3, C32=12, C33=−6.
Adjoint (transpose of the cofactor matrix): adj(A)=−1−41523−11312−6.
Divide by detA=−3: …
detA=−3=0, so the inverse exists; A−1=3134−31−35−323311−1−42.
The inverse of a 3×3 matrix is A−1=detA1adj(A), where adj(A) is the transpose of the cofactor matrix. It exists only when detA=0.
1. Determinant
Expanding along row 1:
detA=2−1201−14−701+34−7−12=2(−1)−1(4)+3(1)=−3.
Since −3=0, A is invertible.
2. Cofactors Cij=(−1)i+jMij
C11=+−1201=−1,C12=−4−701=−4,C13=+4−7−12=1,
C21=−1231=5,C22=+2−731=23,C23=−2−712=−11,
C31=+1−130=3,C32=−2430=12,C33=+241−1=−6.
3. Adjoint
Transpose the cofactor matrix: …
Method: Inverse of a 3×3 Matrix via the Adjoint
This method finds the inverse of any square matrix whose determinant is non-zero, using cofactors and the adjoint — the standard NCERT route for a 3×3 inverse.
Steps
Step 1: Compute detA and confirm it is non-zero
The inverse of a matrix exists only when detA=0 (a singular matrix has no inverse). Expand the determinant along any convenient row or column — pick one with the most zeros to minimise the arithmetic.
A−1 exists⟺detA=0
Step 2: Compute all nine cofactors
For each entry aij, delete its row and column to get the minor Mij, then apply the checkerboard sign:
Cij=(−1)i+jMij
Work systematically row by row (C11,C12,C13, then C21,…) so no entry is skipped, keeping the alternating +,−,+ sign pattern in front of you rather than re-deriving it each time.
Step 3: Form the adjoint by transposing the cofactor matrix …
Common Mistakes
Mistake 1: Forgetting to transpose the cofactor matrix into the adjoint
Why it's wrong: the adjoint is defined as the transpose of the cofactor matrix, adj(A)=[Cij]T — using the cofactor matrix unchanged silently swaps every off-diagonal pair (e.g. C12 and C21), giving a wrong inverse that still "looks" plausible. Correct approach: write the cofactor matrix out fully first, then explicitly transpose it as its own step before dividing by detA.
Mistake 2: A sign slip in the checkerboard pattern when computing cofactors …
- GUJCET 2026Set x1 markMCQQ.If inverse matrix of A=[213−4] is A−1=[a111113b], then a+b= ______ (A) 112 (B) 116 (C) −112 (D) −116
›Reveal solutionSolution
Use A−1=detA1adj(A) and read off a and b.
detA=(2)(−4)−(3)(1)=−11.
A−1=−111[−4−1−32]=[114111113−112]. …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.If A=[2134], then A−1 = ____.(a) 51[−231−4](b) 51[4132](c) 51[4−1−32](d) 51[−4−1−3−2]
›Reveal solutionSolution
A−1=∣A∣1adj(A).
∣A∣=2(4)−3(1)=8−3=5.
For a 2×2 matrix [acbd], adj=[d−c−ba]=[4−1−32].
…
- GUJCET 2024Set 131 markMCQQ.If A=[2−3−46] then A−1= __________. (A) Does not exist (B) 241[−234−6] (C) 241[−634−2] (D) 241[6342]
›Reveal solutionSolution
The determinant of A is zero, so the inverse does not exist.
Concept. A matrix is invertible iff its determinant is non-zero.
Steps. For A=[2−3−46], …
- GUJCET 2022Set 081 markMCQQ.If A=121−1111−31, 10B=4−5120−22α3 and B is inverse of A then α= ______. (A) 10 (B) 9 (C) 3 (D) 5
›Reveal solutionSolution
B = A⁻¹ means A·(10B) = 10I; comparing the appropriate entries fixes α.
Concept. If B=A−1 then AB=I, so A(10B)=10I.
Solution. With M=10B, compute the third column of AM (the only place α appears), where A=121−1111−31, M's third column =(2,α,3)T: …
- GUJCET 2020Set 071 markMCQQ.If A=013121231 and inverse of A is 211−8x−16−31−21 then x= ________. (A) 5 (B) 3 (C) 2 (D) 4
›Reveal solutionSolution
Multiply A by the given inverse; matching one entry of AA−1=I fixes x=5.
Concept. If A−1=21M, then AM=2I. We only need one convenient entry.
With A=013121231 and M=1−8x−16−31−21, take the (1,1) entry of AM (row 1 of A · column 1 of M): …
- GUJCET 2019Set 171 markMCQQ.If the inverse of the matrix A=122212221 is 51−3222−322α−3 then, α=. (A) 4 (B) 2 (C) 3 (D) −2
›Reveal solutionSolution
[!TLDR]
Using τ=pEsinθ with the given values yields 1.73×10−4 Nm.
Concept
An electric dipole of moment p in a uniform field E experiences a torque τ=pEsinθ, where θ is the angle between the dipole axis and the field.
Solution
Given p=4×10−9 C·m, E=5×104 NC−1, θ=60∘ (so sin60∘=0.866): …
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