Matrices & Determinants

Matrices and Determinants is a high-yield, largely mechanical block in CUET Mathematics. Master the operations, the determinant, and the inverse, and you gain reliable marks plus a tool for solving linear systems.

Matrix basics

• Add/subtract only matrices of the same order; multiply A(m x n) by B(n x p) to get (m x p) - matrix multiplication is not commutative (AB is generally not BA).
• Transpose A^T swaps rows and columns. A is symmetric if A^T = A, skew-symmetric if A^T = -A (diagonal entries zero).

Determinants

• For a 2x2 matrix [[a, b],[c, d]], det = ad - bc.
• Properties: swapping two rows changes the sign; a common factor of a row comes out; det(AB) = det(A) x det(B).
• A matrix is singular if det = 0 (then no inverse).

Inverse and systems

• Inverse: A-inverse = adj(A) / det(A), valid only when det(A) is not 0.
• Solving AX = B: X = A-inverse B (matrix method), or use Cramer's rule with determinants.

Exam Tricks & Tips

• 🎯 Check conformability before multiplying: columns of A must equal rows of B.
• 🎯 A matrix is invertible iff det is nonzero - compute the determinant first.
• 🎯 Use det(AB) = det A x det B and det(kA) = k^n det A (n = order) to save work.
• 🎯 For symmetric/skew-symmetric checks, compare A with A^T; skew-symmetric has zero diagonal.
• 🎯 Solve linear systems with X = A-inverse B when det is nonzero (unique solution).
• ❌ Common mistake: assuming AB = BA - matrix multiplication is not commutative, so keep the order.

Expected exam pattern

Multiply/transpose matrices, compute a determinant, find an inverse, and solve a 2- or 3-variable system. Moderate; formula-driven.

Quick recap

Same order to add; conformable to multiply; det 2x2 = ad - bc; invertible iff det is nonzero with A-inverse = adj/det; solve AX = B by X = A-inverse B. AB is generally not BA.

Cover the answer, recall, then check. 12 cards on matrices and determinants for CUET Mathematics.

Q1. When can two matrices be multiplied?
A1. When the number of columns of the first equals the number of rows of the second.

Q2. Is matrix multiplication commutative?
A2. No - in general AB is not equal to BA.

Q3. Compute the determinant of [[a, b],[c, d]].
A3. ad - bc.

Q4. When is a matrix called singular?
A4. When its determinant is 0 (then it has no inverse).

Q5. State the formula for the inverse of a matrix.
A5. A-inverse = adj(A) / det(A), provided det(A) is not 0.

Q6. What is a symmetric matrix?
A6. One where A transpose equals A (A^T = A).

Q7. Define a skew-symmetric matrix.
A7. One where A^T = -A; its diagonal entries are all zero.

Q8. How is the system AX = B solved by the matrix method?
A8. X = A-inverse times B (when det A is not 0).

Q9. State the multiplicative property of determinants.
A9. det(AB) = det(A) x det(B).

Q10. What happens to a determinant when two rows are swapped?
A10. Its sign changes.

Q11. For an n x n matrix, det(kA) equals what?
A11. k^n times det(A).

Q12. What is the most common matrix-algebra mistake?
A12. Assuming AB = BA - always preserve the multiplication order.

Matrices and determinants are the algebra of grids of numbers, and they power the solving of linear equations. CUET tests the mechanics — multiplication, determinant, inverse — and their application to systems of equations, so accuracy with the operations is everything.

What this tests / core idea: matrix operations (addition, multiplication, transpose), computing determinants, finding inverses, and solving linear systems.

Deep explanation

Beginner — matrix operations

A matrix is a rectangular array of order m × n. Addition is element-wise (same order needed). Multiplication AB is defined only when the columns of A equal the rows of B; the (i, j) entry is the dot product of row i of A with column j of B. Matrix multiplication is not commutative: AB ≠ BA in general.

Intermediate — determinants

For a 2 × 2 matrix [[a, b], [c, d]], the determinant is ad − bc. For 3 × 3, expand along a row/column with cofactors. A matrix is invertible (non-singular) exactly when its determinant is non-zero. Properties: det(AB) = det(A)·det(B); swapping two rows flips the sign.

Advanced — inverse and solving systems

The inverse is A⁻¹ = (1/det A)·adj(A), where adj(A) is the transpose of the cofactor matrix. To solve AX = B, use X = A⁻¹B (matrix method) or Cramer's Rule (xᵢ = det(Aᵢ)/det(A)). If det(A) = 0, the system has no unique solution.

Worked example

Q. Find the determinant of A = [[2, 3], [1, 4]] and state whether A is invertible.
det(A) = (2)(4) − (3)(1) = 8 − 3 = 5. Since det(A) = 5 ≠ 0, A is invertible, and A⁻¹ = (1/5)·[[4, −3], [−1, 2]]. Answer: det = 5; A is invertible.

CUET relevance

A core Mathematics topic: compute a determinant, decide invertibility, multiply matrices, or solve a 2×2/3×3 system. Frequent and formula-driven (3–5 marks).

Speed tricks & shortcuts

• 2×2 determinant = ad − bc (main diagonal minus anti-diagonal product).
• Invertible ⇔ determinant ≠ 0.
• For AX = B, X = A⁻¹B; use Cramer's Rule for quick 2-variable systems.

Key formulas

• (AB){ij} = Σ a{ik}b_{kj}; AB ≠ BA; (AB)ᵀ = BᵀAᵀ; (AB)⁻¹ = B⁻¹A⁻¹.
• Symmetric Aᵀ=A, skew-symmetric Aᵀ=−A.
• det(AB)=det A·det B; det(kA)=kⁿ det A; det(Aᵀ)=det A.
• A⁻¹ = adj A/det A (det A ≠ 0); A·adj A = det A·I.
• Solving AX=B: X = A⁻¹B; Cramer xᵢ = Dᵢ/D.
• Area of triangle = ½|det of vertex coordinates|.

Usage: check det A ≠ 0 before inverting or claiming a unique solution.

Worked Example

Problem: For A = [[2, 1], [1, 3]]: (a) find det(A), (b) find A⁻¹, and (c) use A⁻¹ to solve the system 2x + y = 5, x + 3y = 10.

Solution:
(a) For a 2×2 matrix [[a, b], [c, d]], the determinant is ad − bc.
det(A) = (2)(3) − (1)(1) = 6 − 1 = 5.
Since det(A) = 5 ≠ 0, A is invertible (non-singular).

(b) Inverse of a 2×2 matrix: A⁻¹ = (1/det A) × [[d, −b], [−c, a]] (swap the diagonal entries, negate the off-diagonal, divide by the determinant).
A⁻¹ = (1/5) × [[3, −1], [−1, 2]] = [[3/5, −1/5], [−1/5, 2/5]].
Check: A·A⁻¹ = (1/5)[[2,1],[1,3]]·[[3,−1],[−1,2]] = (1/5)[[6−1, −2+2],[3−3, −1+6]] = (1/5)[[5,0],[0,5]] = [[1,0],[0,1]] ✓ (identity).

(c) Write the system as AX = B with X = [x, y]ᵀ and B = [5, 10]ᵀ. Then X = A⁻¹B.
X = (1/5)[[3, −1], [−1, 2]] · [5, 10]ᵀ
= (1/5) [ (3·5 + (−1)·10), ((−1)·5 + 2·10) ]ᵀ
= (1/5) [ (15 − 10), (−5 + 20) ]ᵀ
= (1/5) [ 5, 15 ]ᵀ
= [1, 3]ᵀ.
So x = 1, y = 3. Verify in the original equations: 2(1) + 3 = 5 ✓; 1 + 3(3) = 10 ✓.

Answer: (a) det(A) = 6 − 1 = 5. (b) A⁻¹ = (1/5)[[3, −1], [−1, 2]]. (c) X = A⁻¹B = [1, 3]ᵀ, i.e. x = 1, y = 3.