Inverse Of A Matrix
For the two by two matrix A with rows (4, 7) and (2, 6), the inverse matrix A inverse, computed using the adjoint over determinant formula, has which top-left entry?
Select the correct option:
Solution
3/5
The inverse of a two by two matrix is the adjoint divided by the determinant, a standard JEE Advanced computation valid only when the determinant is non-zero. For A with rows (4,7),(2,6), the determinant is 4·6 - 7·2 = 24 - 14 = 10, which is non-zero so the inverse exists. The adjoint swaps the diagonal entries and negates the off-diagonal entries, giving adj(A) with rows (6, -7),(-2, 4). Dividing by the determinant 10, the inverse has rows (6/10, -7/10),(-2/10, 4/10). The top-left entry is 6/10 = 3/5 in simplest form. Option 2/5 wrongly takes the bottom-right entry 4/10 without swapping diagonals. Option -7/10 is the top-right off-diagonal entry, not the top-left. Option 1/4 ignores the determinant scaling and merely reciprocates the (1,1) entry of A. Hence the top-left entry is 3/5. Plausibility check: multiplying A by the proposed inverse should yield the identity, and the (1,1) entry 4·(6/10) + 7·(-2/10) = (24 - 14)/10 = 1 confirms the inverse is correct.
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About This Question
- Subject
- mathematics
- Chapter
- matrices and determinants
- Topic
- inverse of a matrix
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
3/5
The inverse of a two by two matrix is the adjoint divided by the determinant, a standard JEE Advanced computation valid only when the determinant is non-zero. For A with rows (4,7),(2,6), the determinant is 4·6 - 7·2 = 24 - 14 = 10, which is non-zero so the inverse exists. The adjoint swaps the diagonal entries and negates the off-diagonal entries, giving adj(A) with rows (6, -7),(-2, 4). Dividing by the determinant 10, the inverse has rows (6/10, -7/10),(-2/10, 4/10). The top-left entry is 6/10 = 3/5 in simplest form. Option 2/5 wrongly takes the bottom-right entry 4/10 without swapping diagonals. Option -7/10 is the top-right off-diagonal entry, not the top-left. Option 1/4 ignores the determinant scaling and merely reciprocates the (1,1) entry of A. Hence the top-left entry is 3/5. Plausibility check: multiplying A by the proposed inverse should yield the identity, and the (1,1) entry 4·(6/10) + 7·(-2/10) = (24 - 14)/10 = 1 confirms the inverse is correct.
This easy difficulty mathematics question is from the chapter matrices and determinants, covering the topic of inverse of a matrix. It appeared in the 2025 exam.
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