Class 12 Mathematics Chapter 4 Determinants – Extra Questions with Answers

Determinants underpin whether a matrix can be inverted, and these questions work through calculating |A|, recognising singular matrices, and finding inverses via the adjoint.

Last Updated: September 23, 2026

How to Approach the HOTS Questions in This Chapter

Very Short Answer Questions (1 mark)

Q1. Find |A| for A=[[2,3],[1,4]].
Ans: |A|=2(4)−3(1)=8−3=5.

Q2. What does it mean for a matrix to be singular?
Ans: Its determinant equals zero, so it has no inverse.

Q3. What is the formula for the inverse of a matrix using adjoint?
Ans: A⁻¹=adj(A)/|A|, provided |A|≠0.

Q4. If |A|=0 for a 3×3 matrix in AX=B, can we conclude there is no solution?
Ans: No, it could have no solution or infinitely many; further check via (adj A)B is needed.

Q5. State one property relating |A| and |A’|.
Ans: |A’|=|A| (determinant is unchanged by transpose).

Short Answer Questions (2–3 marks)

Q6. Find the cofactor A₁₂ of the matrix [[1,2,3],[4,5,6],[7,8,9]].
Ans: Delete row1,col2: [[4,6],[7,9]], minor=4(9)−6(7)=36−42=−6. Cofactor A₁₂=(−1)¹⁺²(−6)=6.

Q7. Using determinants, find the area of a triangle with vertices (0,0), (4,0), (0,3).
Ans: Area=½|0(0−3)+4(3−0)+0(0−0)|=½|12|=6 sq units.

Q8. If |A|=5 and A is 3×3, find |adj(A)|.
Ans: |adj(A)|=|A|ⁿ⁻¹=5²=25.

Higher-Order Thinking / Application Questions

Q9. Solve the system 2x+3y=5 and x−2y=−3 using the matrix method, showing all steps including the inverse calculation.
Ans: Write as AX=B where A=[[2,3],[1,−2]], X=[[x],[y]], B=[[5],[−3]]. First, |A|=2(−2)−3(1)=−4−3=−7, which is non-zero, so A is invertible and a unique solution exists. Find adj(A): cofactors are A₁₁=−2, A₁₂=−1, A₂₁=−3, A₂₂=2, so the cofactor matrix is [[−2,−1],[−3,2]], and adj(A) (transpose of cofactor matrix)=[[−2,−3],[−1,2]]. Then A⁻¹=adj(A)/|A|=(1/−7)[[−2,−3],[−1,2]]=[[2/7,3/7],[1/7,−2/7]]. Now X=A⁻¹B=[[2/7,3/7],[1/7,−2/7]]×[[5],[−3]]=[[(2/7)(5)+(3/7)(−3)],[(1/7)(5)+(−2/7)(−3)]]=[[(10−9)/7],[(5+6)/7]]=[[1/7],[11/7]]. So x=1/7, y=11/7. Verification: 2(1/7)+3(11/7)=2/7+33/7=35/7=5 ✓; (1/7)−2(11/7)=1/7−22/7=−21/7=−3 ✓.

Q10. Explain, with reasoning, why a system of linear equations AX=B has a unique solution when |A|≠0, but may have no solution or infinitely many solutions when |A|=0.
Ans: When |A|≠0, matrix A is invertible, meaning A⁻¹ exists uniquely (since A⁻¹=adj(A)/|A| is a well-defined single matrix). We can then multiply both sides of AX=B on the left by A⁻¹ to get A⁻¹AX=A⁻¹B, i.e., IX=A⁻¹B, i.e., X=A⁻¹B. Since A⁻¹ is unique, this gives exactly one value of X — a unique solution. However, when |A|=0, A has no inverse, so we cannot isolate X this way at all. Geometrically, |A|=0 means the equations represented by the rows of A are linearly dependent (e.g., two planes/lines that are parallel or coincide rather than intersecting at a single point). If the equations are consistent (compatible), i.e. (adj A)B=0 (a zero matrix), the system reduces to fewer independent equations than unknowns, giving infinitely many solutions (like two coincident lines). If the equations are inconsistent, i.e. (adj A)B≠0, the equations contradict each other (like two parallel, non-intersecting lines), giving no solution at all. This is why checking (adj A)B is the standard test to distinguish between these two |A|=0 cases.

Frequently Asked Questions

How would you use determinants to check whether a system of two linear equations has a unique solution?
Calculate the determinant of the coefficient matrix; if it is non zero, the system has a unique solution by Cramer rule.

How would you find the area of a triangle with vertices (1,2), (3,4) and (5,0) using determinants?
Using the determinant formula for area, Area = half times the absolute value of the determinant formed by the coordinates, which gives 6 square units.

Written by Satish

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