Q.Let A=[3275] and B=[6789]. Verify that (AB)−1=B−1A−1.
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Inverse of a Product: The "Socks and Shoes" Principle
You put on your socks first, then your shoes. To take them off, you can't remove the socks while the shoes are still on — you must reverse the order: shoes off first, then socks.
That's exactly the inverse of a product of matrices. If you apply transformation A first, then B, the combined effect is BA (read right-to-left: A acts first, then B). To undo it, undo B first, then A:
(AB)−1=B−1A−1
The order flips — forced by the logic of undoing.
Why the order must reverse
Check that B−1A−1 is the inverse of AB. We need (AB)(B−1A−1)=I and (B−1A−1)(AB)=I:
(AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I
B and B−1 cancel first, leaving A and A−1 to cancel. The other check works the same way:
(B−1A−1)(AB)=B−1(A−1A)B=B−1IB=B−1B=I
If you tried (AB)−1=A−1B−1 instead:
(AB)(A−1B−1)=A(BA−1)B−1
and BA−1 is not I — the matrices are in the wrong order. So the reversal is essential.
A common mistake is writing (AB)−1=A−1B−1. This is false unless A and B commute (which they almost never do). Always flip the order.
A concrete example with numbers
Let A=(1021) and B=(1101), with inverses:
A−1=(10−21),B−1=(1−101)
Then:
AB=(1021)(1101)=(3121),(AB)−1=(1−1−23)
Now compute B−1A−1:
B−1A−1=(1−101)(10−21)=(1−1−23)
They match. Try A−1B−1 and you'll get a different matrix — the wrong answer.
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For invertible A,B: (AB)−1=B−1A−1. Verify by computing both sides.
AB: [3275][6789]=[67478761].
(AB)−1: det(AB)=67⋅61−87⋅47=4087−4089=−2, so (AB)−1=−21[61−47−8767]=[−261247287−267].
A−1,B−1: detA=1⇒A−1=[5−2−73]; detB=−2⇒B−1=−21[9−7−86]=[−29274−3]. …
Both (AB)−1 and B−1A−1 equal [−261247287−267], so (AB)−1=B−1A−1 is verified.
The rule (AB)−1=B−1A−1 reverses the order (the "socks and shoes" idea), because (AB)(B−1A−1)=A(BB−1)A−1=AA−1=I. Let us check it on the given matrices.
1. Compute AB
AB=[3275][6789]=[18+4912+3524+6316+45]=[67478761].
2. Invert AB
For [acbd] the inverse is ad−bc1[d−c−ba].
det(AB)=67⋅61−87⋅47=4087−4089=−2,
(AB)−1=−21[61−47−8767]=[−261247287−267].
3. Invert A and B separately
detA=15−14=1, so A−1=[5−2−73].
detB=54−56=−2, so B−1=−21[9−7−86]=[−29274−3].
4. Multiply B−1A−1 (reverse order) …
Method: Verifying (AB)−1=B−1A−1
This method verifies the "reversal rule" for the inverse of a product by computing both sides independently and checking they agree.
Steps
Step 1: Compute the product AB
Multiply the two given matrices in the order given — order matters, since matrix multiplication does not commute.
Step 2: Invert AB directly
For a 2×2 matrix (acbd), use
(acbd)−1=ad−bc1(d−c−ba)
Check det(AB)=0 before proceeding.
Step 3: Invert A and B separately, using the same 2×2 inverse formula …
Common Mistakes
Mistake 1: Computing A−1B−1 instead of B−1A−1
Why it's wrong: matrix multiplication does not commute, so (AB)−1=B−1A−1 — the order must reverse, not stay the same. Multiplying in the original order gives a matrix that, in general, does not equal (AB)−1. Correct approach: always write the inverse of a product with the factors in reverse order, and verify the "socks and shoes" logic ((AB)(B−1A−1)=A(BB−1)A−1=AA−1=I) if unsure which order is right.
Mistake 2: An arithmetic slip in the 2×2 matrix multiplication, especially with fractions …
- AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQQ.Let B=21−160141−1 and C=−112010132. If a matrix A is such that BAC=I, then A−1= (A) −302−5925146 (B) −302−50145916 (C) −302−5914−626 (D) −302−5914−526
›Reveal solutionSolution
From BAC=I we get A−1=CB directly — no need to invert B or C separately. The answer is (D).
Concept and Intuition
If BAC=I, then A is invertible and A=B−1IC−1=B−1C−1. Taking inverse of both sides of A=B−1C−1 gives A−1=(B−1C−1)−1=CB. This is a shortcut: instead of computing B−1 and C−1 (expensive), we only need the product CB, which is a plain matrix multiplication.
Step-by-Step Solution
- From BAC=I, multiply on the left by B−1 and on the right by C−1: A=B−1C−1.
- Then A−1=(B−1C−1)−1=CB (since inverse of a product reverses order: (XY)−1=Y−1X−1).
- Compute CB where C=−112010132, B=21−160141−1.
- Row 1 of CB: (−1)(2)+0(1)+1(−1)=−3; (−1)(6)+0(0)+1(1)=−5; (−1)(4)+0(1)+1(−1)=−5. So row 1 = (−3,−5,−5).
- Row 2 of CB: 1(2)+1(1)+3(−1)=0; 1(6)+1(0)+3(1)=9; 1(4)+1(1)+3(−1)=2. Row 2 = (0,9,2). …
- AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQQ.Let A=[1−221] and B−1=[1012]. If (AB−1)−1=[acbd], then 2b+5c+10d= (A) 0 (B) 1 (C) −1 (D) 2
›Reveal solutionSolution
Uses (XY)−1=Y−1X−1 to write (AB−1)−1=BA−1, then plain 2×2 inversion and multiplication; answer is 1.
Concept and Intuition
Inverting a product reverses the order: (XY)−1=Y−1X−1. Here X=A, Y=B−1, so (AB−1)−1=(B−1)−1A−1=BA−1. This avoids ever needing B−1's inverse-of-inverse gymnastics — we just need B (the inverse of the given B−1) and A−1.
Step-by-Step Solution
- B−1=[1012] has determinant 2, so B=21[20−11]=[10−1/21/2].
- A=[1−221] has determinant 1−(−4)=5, so A−1=51[12−21]. …
- AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQQ.If A=2−14121303, B=313221130 then det(2B−1A−1)= (A) 61 (B) 24−1 (C) 31 (D) 6−1
›Reveal solutionSolution
det(2B−1A−1)=detA⋅detB23=−488=−61.
Concept and Intuition
For an n×n matrix, det(cM)=cndet(M), and det(M−1)=1/det(M). Combining these avoids ever computing the inverse matrices explicitly.
Step-by-Step Solution
- detA: expanding A=2−14121303 along row 1: 2(6−0)−1(−3−0)+3(−1−8)=12+3−27=−12.
- detB: expanding B=313221130 along row 1: 3(0−3)−2(0−9)+1(1−6)=−9+18−5=4. …
- AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQQ.Let A, B, C, D and E be n×n matrices, each with non-zero determinant. If ABCDE=I, then C−1= (A) E−1D−1B−1A−1 (B) DEAB (C) [AMBIGUOUS] (D) A−1B−1D−1E−1
›Reveal solutionSolution
Isolating C from ABCDE=I and then inverting the resulting product gives C−1=DEAB.
Concept and Intuition
For invertible matrices, (XY)−1=Y−1X−1 — the inverse of a product reverses the order and inverts each factor. Solving a matrix equation like ABCDE=I for one letter in the middle is done exactly like solving for an unknown sandwiched between known invertible factors: multiply on the left and right by the appropriate inverses.
Step-by-Step Solution
- Start with ABCDE=I.
- Left-multiply both sides by (AB)−1=B−1A−1: CDE=(AB)−1I=B−1A−1.
- Right-multiply both sides by (DE)−1=E−1D−1: C=B−1A−1E−1D−1.
- Take the inverse of both sides. Using (WXYZ)−1=Z−1Y−1X−1W−1 with W=B−1,X=A−1,Y=E−1,Z=D−1: C−1=DEAB. …
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