ilovepdf_merged (6).pdf). Content covers sections 22.1–22.x.For a square matrix A, a matrix B is an inverse of A if AB = BA = I. When it exists it is unique and written A⁻¹. This lesson links inverses to determinants and adjoints and uses them to solve linear systems.
A is invertible ⇔ det A ≠ 0 ⇔ A is non-singular. If det A = 0, no inverse exists.
If A is invertible, X = A⁻¹ B. Always verify by multiplication when numbers are small.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Inverse of a Matrix and its Applications.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: INVERSE OF A MATRIX AND ITS
Memorise and apply: Abhinav spends Rs. 120 in buying 2 pens and 5 note books whereas Shantanu spends Rs.
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style (AB)⁻¹, Board-style 2×2 inverse, Board-style existence, Board-style solve system
PYQ1. Find the inverse of A=[[2,1],[5,3]].
Model Answer
1) A⁻¹ = [[3, −1], [−5, 2]]
Explanation
1) det A=6−5=1≠
0. A⁻¹=(1/det)[[d,−b],[−c,a]]=[[3,−1],[−5,2]].
PYQ2. Does B=[[1,2],[2,4]] have an inverse? Justify.
Model Answer
1) No; det B=0 so B is singular
Explanation
1) det B=4−4=
0. Inverse exists only when determinant is non-zero.
PYQ3. Solve 2x+y=5 and 5x+3y=13 using A⁻¹ where A=[[2,1],[5,3]].
Model Answer
1) x=2, y=1
Explanation
1) A⁻¹=[[3,−1],[−5,2]]. X=A⁻¹B with B=[[5],[13]] gives [[2],[1]]. Check: 4+1=5, 10+3=13.
PYQ4. If A and B are invertible, simplify (AB)⁻¹A.
Model Answer
1) B⁻¹
Explanation
1) (AB)⁻¹A = B⁻¹A⁻¹A = B⁻¹I = B⁻¹.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
Find the inverse of A=[[2,1],[5,3]].
1) det A = 6−5 = 1 ≠ 0, so inverse exists.
2) A⁻¹ = (1/1)[[3,−1],[−5,2]] = [[3,−1],[−5,2]].
Answer: [[3, −1], [−5, 2]]
For a 2×2 matrix, swap diagonals, change signs of off-diagonals, divide by det.
det=1; swap 2 and 3, negate 1 and 5.
Verify AA⁻¹=I.
If det=0, stop — no inverse.
Does B=[[1,2],[2,4]] have an inverse?
1) det B=4−4=0 ⇒ B is singular ⇒ no inverse.
Answer: No inverse (det=0)
A square matrix has an inverse if and only if its determinant is non-zero.
Rows proportional → det 0 → no inverse.
adj formula would divide by zero.
Do not attempt A⁻¹ formula when det=0.
If A=[[1,2],[0,1]], find adj A and A⁻¹.
1) C₁₁=1, C₁₂=0, C₂₁=−2, C₂₂=1 → cofactor matrix [[1,0],[−2,1]].
2) adj A = transpose = [[1,−2],[0,1]].
3) det A=1; A⁻¹=[[1,−2],[0,1]].
Answer: adj A=[[1,−2],[0,1]]; A⁻¹ same since det=1
Adjoint is the transpose of the cofactor matrix; divide by determinant.
Build cofactors, transpose, divide by det (here 1).
For upper triangular matrices, det is product of diagonals.
Transpose step is easy to forget.
Solve 2x+y=5, 5x+3y=13 using inverse of A=[[2,1],[5,3]].
1) From earlier, A⁻¹=[[3,−1],[−5,2]], B=[[5],[13]].
2) X=A⁻¹B = [[3·5+(−1)·13],[−5·5+2·13]] = [[15−13],[−25+26]] = [[2],[1]].
3) So x=2, y=1.
Answer: x=2, y=1
The unique solution of AX=B is X=A⁻¹B when A is invertible.
Multiply the inverse by the constants column.
Check in original equations: 4+1=5, 10+3=13.
Column vector order matters.
If A and B are invertible of the same order, simplify (AB)⁻¹A.
1) (AB)⁻¹A = B⁻¹A⁻¹A = B⁻¹I = B⁻¹.
Answer: B⁻¹
Using (AB)⁻¹=B⁻¹A⁻¹ and A⁻¹A=I.
Inverse of product reverses order; then A cancels.
Similarly A(AB)⁻¹=A B⁻¹ A⁻¹ (not as simple).
Do not write A⁻¹B⁻¹ for (AB)⁻¹.
Verify that [[3,−1],[−5,2]] is the inverse of [[2,1],[5,3]] by multiplication.
1) [[2,1],[5,3]][[3,−1],[−5,2]] = [[6−5, −2+2],[15−15, −5+6]] = [[1,0],[0,1]]=I.
Answer: Product is I₂, so they are inverses
Two matrices are inverses if their product (both ways) is the identity; here one side is shown I.
Multiply and get the identity matrix.
Both AB and BA should be I for two-sided inverse.
One successful product is strong evidence; check BA if required.
Find the inverse of A = [[1, 1], [0, 1]].
1) det A = 1.
2) A⁻¹ = [[1, −1], [0, 1]].
3) Check: A A⁻¹ = I.
Answer: [[1, −1], [0, 1]]
Swap diagonal, negate off-diagonal, divide by det.
Shear matrix inverse is the opposite shear.
det must be non-zero.
Check each algebraic step carefully.
Solve x + y = 5, y = 2 using matrices (find x).
1) From y=2 and x+y=5 ⇒ x=3.
2) In matrix form A=[[1,1],[0,1]], B=[[5],[2]], X=[[3],[2]].
Answer: x = 3, y = 2
Unique solution when A is invertible.
Second equation gives y; substitute.
X = A⁻¹B if A⁻¹ is computed.
Check each algebraic step carefully.
For which k is [[k, 1], [2, 2]] invertible?
1) det = 2k − 2 = 2(k − 1).
2) Invertible when det ≠ 0 ⇒ k ≠ 1.
Answer: Invertible for all k ≠ 1
Non-zero determinant is the criterion.
When k=1 rows are proportional.
State the excluded value clearly.
Check each algebraic step carefully.
If A and B are invertible, write (AB)⁻¹ in terms of A⁻¹ and B⁻¹.
1) (AB)⁻¹ = B⁻¹ A⁻¹ (reverse order).
Answer: B⁻¹ A⁻¹
Inverse of a product reverses the order.
Like undoing operations: last first.
Requires both invertible.
Check each algebraic step carefully.