Unit 2 Algebra: Matrices Practice Questions with Solutions | Class 12 Applied Maths | Boundless Maths

🔢 Unit 2: Algebra — Matrices & Determinants

CBSE Class 12 Applied Maths — Free Study Resources

📌 Weightage: 10 Marks in Board Exams

Class 12 Applied Maths Unit 2 — Algebra (Matrices) Study Material

This page covers all topics in Unit 2 of CBSE Class 12 Applied Mathematics — Algebra, which carries 10 marks in the board exam. Topics include Types of Matrices and Matrix Operations, Symmetric and Skew-Symmetric Matrices, Determinants, Inverse of a Matrix, and System of Linear Equations using the Matrix Method and Cramer's Rule.

You'll find 10 interactive MCQs with detailed answers, 5 short-answer questions with step-by-step solutions, and 2 case studies. All content is aligned to the latest CBSE Applied Maths syllabus 2026-27.

MatricesMatrix OperationsSymmetric MatricesDeterminantsInverse of MatrixSystem of Linear EquationsCramer's RuleCBSE 2026-27
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Topics & Key Formulas — Unit 2

6 topics covered · essential formulas to memorise for the board exam

1. Matrices

Types — row, column, square, diagonal, identity, zero. Element notation and equality.

\(A = [a_{ij}]_{m\times n}\) — element at row \(i\), col \(j\)

Transpose: \((A^T)_{ij} = A_{ji}\)

2. Matrix Operations

Addition, subtraction, scalar multiplication, matrix multiplication and their properties.

\((AB)^T = B^T A^T\)

Note: \(AB \neq BA\) in general

3. Symmetric & Skew-Symmetric

Symmetric: \(A = A^T\). Skew-symmetric: \(A = -A^T\), diagonal entries always zero.

\(A = \tfrac{1}{2}(A+A^T) + \tfrac{1}{2}(A-A^T)\)

Sym. part + Skew-sym. part

4. Determinants

Expansion, minors, cofactors, properties. Area of triangle and collinearity.

\(|kA| = k^n\,|A|\) for \(n\times n\) matrix

3 points collinear \(\Rightarrow\) area \(= 0\)

5. Inverse of a Matrix

Singular (\(|A|=0\)) has no inverse. Non-singular: inverse exists via adjoint method.

\(A^{-1} = \dfrac{1}{|A|}\,\text{adj}(A)\)

Verify: \(A \cdot A^{-1} = I\)

6. System of Linear Equations

Matrix Method and Cramer's Rule for 2 and 3 variable simultaneous equations.

\(AX = B \;\Rightarrow\; X = A^{-1}B\)

Cramer's: \(x = D_x/D,\quad y = D_y/D\)

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Practice MCQs with Answers — Unit 2 Algebra: Matrices

Select your answer, then click Show Answer to check and reveal the explanation

Question 1CBSE 2022
If matrix \(A = [a_{ij}]_{2\times 2}\) where \(a_{ij} = i + j\), then \(A\) is equal to:
\(\begin{bmatrix}1&2\\3&4\end{bmatrix}\)
\(\begin{bmatrix}2&3\\3&4\end{bmatrix}\)
\(\begin{bmatrix}1&1\\2&2\end{bmatrix}\)
\(\begin{bmatrix}1&2\\1&2\end{bmatrix}\)
Step 1 — Understand the formula: \(a_{ij} = i + j\) means each element equals its row number plus column number.

Step 2 — Calculate each element:
\(a_{11} = 1+1 = 2\),  \(a_{12} = 1+2 = 3\)
\(a_{21} = 2+1 = 3\),  \(a_{22} = 2+2 = 4\)

Step 3 — Write the matrix: \(A = \begin{bmatrix}2&3\\3&4\end{bmatrix}\)

Note: Since \(a_{12} = a_{21} = 3\), we have \(A = A^T\) — this matrix is symmetric.
Question 2CBSE 2022
If \(A^2 = A\), then \((I + A)^2 - 3A\) is equal to:
\(I\)
\(2A\)
\(3I\)
\(A\)
Step 1 — Expand \((I+A)^2\):
\((I+A)^2 = I^2 + IA + AI + A^2\)

Step 2 — Simplify each term:
\(I^2 = I\),  \(IA = A\),  \(AI = A\),  \(A^2 = A\) (given — idempotent matrix)

Step 3 — Substitute:
\((I+A)^2 = I + A + A + A = I + 3A\)

Step 4 — Subtract \(3A\):
\((I+A)^2 - 3A = I + 3A - 3A = \mathbf{I}\)
Question 3CBSE 2023
If \(A = \begin{bmatrix}1&0\\-2&1\end{bmatrix}\) and \(B = \begin{bmatrix}-5&10\\-10&-5\end{bmatrix}\), then \(AB\) is:
\(\begin{bmatrix}-5&10\\0&-5\end{bmatrix}\)
\(\begin{bmatrix}0&-5\\25&10\end{bmatrix}\)
\(\begin{bmatrix}10&-25\\-5&0\end{bmatrix}\)
\(\begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
Method: \((AB)_{ij}\) = (Row \(i\) of \(A\)) · (Column \(j\) of \(B\))

Row 1 of A = \([1,\ 0]\):
\((AB)_{11} = (1)(-5) + (0)(-10) = -5\)
\((AB)_{12} = (1)(10) + (0)(-5) = 10\)

Row 2 of A = \([-2,\ 1]\):
\((AB)_{21} = (-2)(-5) + (1)(-10) = 10 - 10 = 0\)
\((AB)_{22} = (-2)(10) + (1)(-5) = -20 - 5 = -25\)

\(AB = \begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
Question 4CBSE 2023
If \(\begin{bmatrix}x+y & x+2\\ 2x-y & 16\end{bmatrix} = \begin{bmatrix}8&5\\1&3y+1\end{bmatrix}\), then \(x\) and \(y\) are:
\(x=3,\;y=5\)
\(x=5,\;y=3\)
\(x=2,\;y=7\)
\(x=7,\;y=2\)
Rule: Two matrices are equal if every corresponding element is equal.

Step 1 — Position (1,2): \(x + 2 = 5 \Rightarrow x = 3\)
Step 2 — Position (1,1): \(x + y = 8 \Rightarrow 3 + y = 8 \Rightarrow y = 5\)

Step 3 — Verify remaining positions:
Position (2,1): \(2(3) - 5 = 1\) ✓
Position (2,2): \(3(5) + 1 = 16\) ✓
Question 5CBSE 2023-Comptt
If \(A\) and \(B\) are matrices such that \(AB=A\) and \(BA=B\), then \(B^2\) equals:
\(B\)
\(A\)
\(I\)
\(O\)
Step 1: Write \(B^2 = B \cdot B\)

Step 2 — Substitute \(B = BA\) (from \(BA = B\)):
\(B^2 = (BA) \cdot B\)

Step 3 — Regroup using associativity:
\(B^2 = B \cdot (AB)\)

Step 4 — Substitute \(AB = A\):
\(B^2 = B \cdot A = BA\)

Step 5 — Use \(BA = B\):
\(B^2 = B\)
Question 6CBSE 2023-Comptt
If \(A = \begin{bmatrix}5&x\\y&0\end{bmatrix}\) is symmetric, then:
\(x=0,\;y=5\)
\(x=5,\;y=0\)
\(x=y\)
\(x+y=0\)
Definition: A matrix is symmetric if \(A = A^T\), meaning \(a_{ij} = a_{ji}\) for all \(i, j\).

Step 1 — Find \(A^T\): \(A^T = \begin{bmatrix}5&y\\x&0\end{bmatrix}\)

Step 2 — Set \(A = A^T\) and compare element by element:
Diagonals: \(5 = 5\) ✓ and \(0 = 0\) ✓
Off-diagonal: position (1,2) gives \(x = y\)

Conclusion: The only condition needed for symmetry is \(x = y\).
Question 7CBSE 2024-Comptt
If \(A = \begin{bmatrix}4&1\\3&2\end{bmatrix}\), then \(A^2 - 6A\) equals:
\(3I\)
\(-5I\)
\(5I\)
\(-3I\)
Step 1 — Compute \(A^2 = A \times A\):
\((A^2)_{11} = (4)(4)+(1)(3) = 19\),  \((A^2)_{12} = (4)(1)+(1)(2) = 6\)
\((A^2)_{21} = (3)(4)+(2)(3) = 18\),  \((A^2)_{22} = (3)(1)+(2)(2) = 7\)
\(A^2 = \begin{bmatrix}19&6\\18&7\end{bmatrix}\)

Step 2 — Compute \(6A\):
\(6A = \begin{bmatrix}24&6\\18&12\end{bmatrix}\)

Step 3 — Subtract:
\(A^2 - 6A = \begin{bmatrix}19-24&6-6\\18-18&7-12\end{bmatrix} = \begin{bmatrix}-5&0\\0&-5\end{bmatrix} = -5I\)
Question 8CBSE 2022
If \(A\) is a \(3\times3\) matrix with \(|A|=4\), then \(|3A|\) equals:
\(27\)
\(81\)
\(108\)
\(256\)
Property: For an \(n \times n\) matrix, \(|kA| = k^n \cdot |A|\).

Why this works: Multiplying the whole matrix by \(k\) multiplies every row by \(k\). Since there are \(n\) rows and each multiplies the determinant by \(k\), the overall effect is \(k^n\).

Applying here: \(n = 3\), \(k = 3\), \(|A| = 4\)
\(|3A| = 3^3 \times 4 = 27 \times 4 = \mathbf{108}\)

Common mistake: Writing \(3 \times 4 = 12\) — this only holds when multiplying a single row, not the whole matrix.
Question 9CBSE 2022
If \((1,3)\), \((x,5)\) and \((2,7)\) are collinear, then \(x\) equals:
\(2\)
\(\dfrac{3}{2}\)
\(1\)
\(\dfrac{3}{4}\)
Key fact: Three points are collinear if and only if the area of the triangle they form equals zero.

Step 1 — Set up the determinant for area:
\(\dfrac{1}{2}\begin{vmatrix}1&3&1\\x&5&1\\2&7&1\end{vmatrix} = 0\)

Step 2 — Expand along Row 1:
\(1\begin{vmatrix}5&1\\7&1\end{vmatrix} - 3\begin{vmatrix}x&1\\2&1\end{vmatrix} + 1\begin{vmatrix}x&5\\2&7\end{vmatrix} = 0\)

Step 3 — Evaluate each 2×2 determinant:
\((5-7) - 3(x-2) + (7x-10) = 0\)
\(-2 - 3x + 6 + 7x - 10 = 0\)
\(4x - 6 = 0\)

Step 4 — Solve: \(x = \dfrac{6}{4} = \dfrac{3}{2}\)
Question 10CBSE 2025
If \(\begin{vmatrix}2x&5\\4&x\end{vmatrix}=\begin{vmatrix}3&5\\4&6\end{vmatrix}\), then \(x\) equals:
\(\dfrac{3}{2}\)
\(6\)
\(3\)
\(\pm3\)
Step 1 — Compute LHS determinant:
\(\begin{vmatrix}2x&5\\4&x\end{vmatrix} = (2x)(x) - (5)(4) = 2x^2 - 20\)

Step 2 — Compute RHS determinant:
\(\begin{vmatrix}3&5\\4&6\end{vmatrix} = (3)(6) - (5)(4) = 18 - 20 = -2\)

Step 3 — Set LHS = RHS and solve:
\(2x^2 - 20 = -2\)
\(2x^2 = 18 \Rightarrow x^2 = 9 \Rightarrow x = \pm3\)

Verify: Both \(x=3\) and \(x=-3\) give LHS \(= 2(9)-20 = -2\) ✓

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Short Answer Questions — Step-by-Step Solutions

2-mark and 3-mark solved questions — click Show Solution to reveal full working

Question 1CBSE 2024
If \(\begin{bmatrix}x-y & 2x+z\\ 2x-y & 3z+w\end{bmatrix} = \begin{bmatrix}-1&5\\0&13\end{bmatrix}\), find \(x,\,y,\,z\) and \(w\).
Step 1 — Equate corresponding elements. Two matrices are equal if and only if every element at the same position is equal:
Position (1,1): \(x - y = -1\) …(i)
Position (1,2): \(2x + z = 5\) …(ii)
Position (2,1): \(2x - y = 0\) …(iii)
Position (2,2): \(3z + w = 13\) …(iv)
Step 2 — Solve for \(x\) and \(y\) using equations (i) and (iii).
Subtract (i) from (iii): \((2x-y)-(x-y) = 0-(-1) \Rightarrow x = 1\)
Substitute into (i): \(1 - y = -1 \Rightarrow y = 2\)
Step 3 — Solve for \(z\) using (ii): \(2(1) + z = 5 \Rightarrow z = 3\)
Step 4 — Solve for \(w\) using (iv): \(3(3) + w = 13 \Rightarrow w = 4\)
Verification: \(\begin{bmatrix}1-2&2(1)+3\\2(1)-2&3(3)+4\end{bmatrix} = \begin{bmatrix}-1&5\\0&13\end{bmatrix}\) ✓
\(x=1,\quad y=2,\quad z=3,\quad w=4\)
Question 2CBSE 2024
If \(A = \begin{bmatrix}1&0\\-1&7\end{bmatrix}\), find \(k\) such that \(A^2 - 8A + kI = O\).
Step 1 — Compute \(A^2 = A \times A\):
\((A^2)_{11} = (1)(1)+(0)(-1) = 1\)
\((A^2)_{12} = (1)(0)+(0)(7) = 0\)
\((A^2)_{21} = (-1)(1)+(7)(-1) = -1-7 = -8\)
\((A^2)_{22} = (-1)(0)+(7)(7) = 49\)
\(A^2 = \begin{bmatrix}1&0\\-8&49\end{bmatrix}\)
Step 2 — Compute \(8A\):
\(8A = \begin{bmatrix}8&0\\-8&56\end{bmatrix}\)
Step 3 — Compute \(A^2 - 8A\):
\(A^2 - 8A = \begin{bmatrix}1-8&0-0\\-8+8&49-56\end{bmatrix} = \begin{bmatrix}-7&0\\0&-7\end{bmatrix} = -7I\)
Step 4 — Find \(k\):
Substitute into \(A^2 - 8A + kI = O\):
\(-7I + kI = O \Rightarrow (k-7)I = O\)
Since \(I \neq O\), we need \(k - 7 = 0\), so \(k = 7\).
\(k = 7\)
Question 3CBSE 2025
Given \(A=\begin{bmatrix}2&0&1\\3&4&5\\0&2&3\end{bmatrix}\) and \(B=\begin{bmatrix}1&1&-5\\-5&1&-5\\1&-2&4\end{bmatrix}\), find \(BA\).
Method: Both are 3×3 matrices so \(BA\) is 3×3. Entry \((BA)_{ij}\) = (Row \(i\) of \(B\)) · (Column \(j\) of \(A\)).
Row 1 of B = \([1,\ 1,\ -5]\):
\((BA)_{11} = (1)(2)+(1)(3)+(-5)(0) = 5\)
\((BA)_{12} = (1)(0)+(1)(4)+(-5)(2) = -6\)
\((BA)_{13} = (1)(1)+(1)(5)+(-5)(3) = -9\)
Row 2 of B = \([-5,\ 1,\ -5]\):
\((BA)_{21} = (-5)(2)+(1)(3)+(-5)(0) = -7\)
\((BA)_{22} = (-5)(0)+(1)(4)+(-5)(2) = -6\)
\((BA)_{23} = (-5)(1)+(1)(5)+(-5)(3) = -15\)
Row 3 of B = \([1,\ -2,\ 4]\):
\((BA)_{31} = (1)(2)+(-2)(3)+(4)(0) = -4\)
\((BA)_{32} = (1)(0)+(-2)(4)+(4)(2) = 0\)
\((BA)_{33} = (1)(1)+(-2)(5)+(4)(3) = 3\)
\(BA=\begin{bmatrix}5&-6&-9\\-7&-6&-15\\-4&0&3\end{bmatrix}\)
Question 4CBSE 2023
Solve by Cramer's Rule: \(\;2x-y=17,\quad 3x+5y=6\)
Step 1 — Compute the coefficient determinant \(D\):
\(D=\begin{vmatrix}2&-1\\3&5\end{vmatrix}=(2)(5)-(-1)(3)=10+3=13\)
Since \(D = 13 \neq 0\), a unique solution exists.
Step 2 — Compute \(D_x\) (replace the \(x\)-column with constants):
\(D_x=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=(17)(5)-(-1)(6)=85+6=91\)
Step 3 — Compute \(D_y\) (replace the \(y\)-column with constants):
\(D_y=\begin{vmatrix}2&17\\3&6\end{vmatrix}=(2)(6)-(17)(3)=12-51=-39\)
Step 4 — Apply Cramer's Rule:
\(x=\dfrac{D_x}{D}=\dfrac{91}{13}=7\)    \(y=\dfrac{D_y}{D}=\dfrac{-39}{13}=-3\)
Verification:
\(2(7)-(-3)=17\) ✓    \(3(7)+5(-3)=6\) ✓
\(x=7,\quad y=-3\)
Question 5CBSE 2023-Comptt
Express \(A=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}\) as the sum of a symmetric and a skew-symmetric matrix.
Key result: Every square matrix can be written as \(A = P + Q\), where \(P = \tfrac{1}{2}(A+A^T)\) is symmetric and \(Q = \tfrac{1}{2}(A-A^T)\) is skew-symmetric.
Step 1 — Find \(A^T\) (swap rows and columns):
\(A^T=\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}\)
Step 2 — Compute \(A+A^T\):
\(A+A^T=\begin{bmatrix}14&-4&-4\\-4&2&0\\-4&0&2\end{bmatrix}\)
Step 3 — Symmetric part \(P = \tfrac{1}{2}(A+A^T)\):
\(P=\begin{bmatrix}7&-2&-2\\-2&1&0\\-2&0&1\end{bmatrix}\)
Verify: \(P^T = P\) ✓
Step 4 — Compute \(A-A^T\):
\(A-A^T=\begin{bmatrix}0&-2&-2\\2&0&0\\2&0&0\end{bmatrix}\)
Step 5 — Skew-symmetric part \(Q = \tfrac{1}{2}(A-A^T)\):
\(Q=\begin{bmatrix}0&-1&-1\\1&0&0\\1&0&0\end{bmatrix}\)
Verify: \(Q^T = -Q\) ✓ and all diagonal elements = 0 ✓
Verification — \(P + Q = A\):
\(\begin{bmatrix}7&-2&-2\\-2&1&0\\-2&0&1\end{bmatrix}+\begin{bmatrix}0&-1&-1\\1&0&0\\1&0&0\end{bmatrix}=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}=A\) ✓
\(A = P + Q\) where \(P=\begin{bmatrix}7&-2&-2\\-2&1&0\\-2&0&1\end{bmatrix}\) (symmetric) and \(Q=\begin{bmatrix}0&-1&-1\\1&0&0\\1&0&0\end{bmatrix}\) (skew-symmetric)

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Case Studies — Real-World Application Questions

4–5 mark case-based questions — click Show Solution under each part to reveal full working

Case Study 1: School Awards Distribution
10 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 13, while the combined strength of the first and second group is four times that of the third group. Let \(x\), \(y\) and \(z\) denote the number of students in the first, second and third group respectively.
(a) Write the system of linear equations
Translate each condition:
Total students = 10: \(x + y + z = 10\) …(i)
Double first + second = 13: \(2x + y = 13\) …(ii)
First + second = 4 × third: \(x + y - 4z = 0\) …(iii)
System: \(x+y+z=10;\quad 2x+y=13;\quad x+y-4z=0\)
(b) Write the system in matrix form \(AX=B\) and identify the coefficient matrix
Write as \(AX = B\) where \(A\) contains the coefficients of \(x, y, z\) from each equation:
\(\underbrace{\begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}}_{A}\;\underbrace{\begin{bmatrix}x\\y\\z\end{bmatrix}}_{X}=\underbrace{\begin{bmatrix}10\\13\\0\end{bmatrix}}_{B}\)
Coefficient matrix: \(A=\begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}\)
(c)(i) Find the cofactor matrix of \(A\)
Formula: \(A_{ij} = (-1)^{i+j} M_{ij}\) where \(M_{ij}\) is the 2×2 minor (determinant after deleting row \(i\) and column \(j\)).
Row 1 cofactors:
\(A_{11} = (+1)\begin{vmatrix}1&0\\1&-4\end{vmatrix} = (1)(-4)-(0)(1) = -4\)
\(A_{12} = (-1)\begin{vmatrix}2&0\\1&-4\end{vmatrix} = -[(2)(-4)-(0)(1)] = 8\)
\(A_{13} = (+1)\begin{vmatrix}2&1\\1&1\end{vmatrix} = (2)(1)-(1)(1) = 1\)
Row 2 cofactors:
\(A_{21} = (-1)\begin{vmatrix}1&1\\1&-4\end{vmatrix} = -[(1)(-4)-(1)(1)] = 5\)
\(A_{22} = (+1)\begin{vmatrix}1&1\\1&-4\end{vmatrix} = (1)(-4)-(1)(1) = -5\)
\(A_{23} = (-1)\begin{vmatrix}1&1\\1&1\end{vmatrix} = -[(1)(1)-(1)(1)] = 0\)
Row 3 cofactors:
\(A_{31} = (+1)\begin{vmatrix}1&1\\1&0\end{vmatrix} = (1)(0)-(1)(1) = -1\)
\(A_{32} = (-1)\begin{vmatrix}1&1\\2&0\end{vmatrix} = -[(1)(0)-(1)(2)] = 2\)
\(A_{33} = (+1)\begin{vmatrix}1&1\\2&1\end{vmatrix} = (1)(1)-(1)(2) = -1\)
Cofactor matrix \(=\begin{bmatrix}-4&8&1\\5&-5&0\\-1&2&-1\end{bmatrix}\)
(c)(ii) OR: Find the number of students in each group using the matrix method
Step 1 — Find \(|A|\) using Row 1 expansion:
\(|A| = 1(-4) + 1(8) + 1(1) = -4+8+1 = 5 \neq 0\) — unique solution exists.
Step 2 — Find \(\text{adj}(A)\) = transpose of cofactor matrix:
\(\text{adj}(A) = \begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
Step 3 — Find \(A^{-1}\):
\(A^{-1} = \dfrac{1}{5}\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
Step 4 — Compute \(X = A^{-1}B\):
\(X = \dfrac{1}{5}\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\begin{bmatrix}10\\13\\0\end{bmatrix}\)

Row 1: \(\dfrac{1}{5}[(-4)(10)+(5)(13)+(-1)(0)] = \dfrac{-40+65}{5} = \dfrac{25}{5} = 5\)
Row 2: \(\dfrac{1}{5}[(8)(10)+(-5)(13)+(2)(0)] = \dfrac{80-65}{5} = \dfrac{15}{5} = 3\)
Row 3: \(\dfrac{1}{5}[(1)(10)+(0)(13)+(-1)(0)] = \dfrac{10}{5} = 2\)
Hard workers \(x=5\) · Honest students \(y=3\) · Vigilant students \(z=2\)
Verification: \(5+3+2=10\) ✓  \(2(5)+3=13\) ✓  \(5+3-4(2)=0\) ✓
Case Study 2: Orphanage Donation
On her birthday, Prema decides to donate some money to children of an orphanage home. If there are 8 children less, everyone gets ₹10 more. However, if there are 16 children more, everyone gets ₹10 less. Let the number of children be \(x\) and the amount per child be ₹\(y\).
(i) Write the system of linear equations
Total donation = \(xy\) (number of children × amount per child)
Condition 1: 8 fewer children, each gets ₹10 more, total same:
\((x-8)(y+10) = xy\)
Expand: \(xy + 10x - 8y - 80 = xy\)
Simplify: \(10x - 8y = 80 \Rightarrow 5x - 4y = 40\) …(i)
Condition 2: 16 more children, each gets ₹10 less, total same:
\((x+16)(y-10) = xy\)
Expand: \(xy - 10x + 16y - 160 = xy\)
Simplify: \(-10x + 16y = 160 \Rightarrow 5x - 8y = -80\) …(ii)
\(5x-4y=40\)   and   \(5x-8y=-80\)
(ii) Write in matrix form \(AX=B\)
Rewrite both equations with coefficients of \(x\) and \(y\):
\(\underbrace{\begin{bmatrix}5&-4\\5&-8\end{bmatrix}}_{A}\;\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{X}=\underbrace{\begin{bmatrix}40\\-80\end{bmatrix}}_{B}\)
Matrix form: \(\begin{bmatrix}5&-4\\5&-8\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}\)
(iii)(a) Find \(A^{-1}\)
Step 1 — Compute \(|A|\):
\(|A| = (5)(-8) - (-4)(5) = -40 + 20 = -20\)
Since \(|A| = -20 \neq 0\), the inverse exists.
Step 2 — Find \(\text{adj}(A)\) for a 2×2 matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\): \(\text{adj}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
\(\text{adj}(A)=\begin{bmatrix}-8&4\\-5&5\end{bmatrix}\)
Step 3 — Compute \(A^{-1}\):
\(A^{-1}=\dfrac{1}{-20}\begin{bmatrix}-8&4\\-5&5\end{bmatrix}=\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\)
Verification \(A \cdot A^{-1} = I\):
\(\begin{bmatrix}5&-4\\5&-8\end{bmatrix}\cdot\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}=\dfrac{1}{20}\begin{bmatrix}40-20&-20+20\\40-40&-20+40\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I\) ✓
\(A^{-1}=\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\)
(iii)(b) OR: Find \(x\) and \(y\) — number of children and amount per child
Use \(X = A^{-1}B\):
\(X = \dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\begin{bmatrix}40\\-80\end{bmatrix}\)
Row 1 (gives \(x\)):
\(x = \dfrac{1}{20}[(8)(40)+(-4)(-80)] = \dfrac{320+320}{20} = \dfrac{640}{20} = 32\)
Row 2 (gives \(y\)):
\(y = \dfrac{1}{20}[(5)(40)+(-5)(-80)] = \dfrac{200+400}{20} = \dfrac{600}{20} = 30\)
Verification:
\(5(32)-4(30) = 160-120 = 40\) ✓
\(5(32)-8(30) = 160-240 = -80\) ✓
Number of children: \(x = 32\)  |  Amount per child: ₹30
🎯

How to Score Full Marks in Unit 2 — Exam Tips

Common mistakes examiners flag every year in CBSE Class 12 Applied Maths

✅ Tip 1

Always verify your matrix inverse after finding it. After computing \(A^{-1}\), multiply \(A \times A^{-1}\) and confirm you get the identity matrix \(I\). This 30-second check earns method marks in 3-mark questions and catches cofactor sign errors before they cost you the full answer. In case studies, examiners specifically look for this verification step.

🔒  4 more exam tips — including cofactor sign patterns, what examiners penalise in Cramer's Rule, how to present the matrix method for full step-marks, and how to avoid the \(|kA|\) mistake — are in the Question Bank.

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Common Questions — Unit 2 Algebra: Matrices

Answers to questions students commonly ask about Class 12 Applied Maths Unit 2
Unit 2 Algebra covers five areas: Types of Matrices and Matrix Operations (addition, multiplication, scalar multiplication), Transpose with Symmetric and Skew-Symmetric Matrices including the decomposition result, Determinants including minors, cofactors and key properties, Inverse of a Matrix using the adjoint method, and solving a System of Linear Equations by the Matrix Method and Cramer's Rule.
Unit 2: Algebra carries 10 marks in the CBSE Class 12 Applied Maths board exam. Questions appear as 1-mark MCQs, 2–3 mark short answers and a 4–5 mark case study. It is one of the higher-weightage units in the syllabus.
A matrix A is symmetric if \(A=A^T\), meaning \(a_{ij}=a_{ji}\) for all \(i,j\). It is skew-symmetric if \(A=-A^T\), forcing every diagonal entry to zero. Key exam result: every square matrix equals the sum of a symmetric part \(\tfrac{1}{2}(A+A^T)\) and a skew-symmetric part \(\tfrac{1}{2}(A-A^T)\).
For \(A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\): compute \(|A|=ad-bc\). If \(|A|=0\), no inverse exists. Otherwise \(\text{adj}(A)=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\) and \(A^{-1}=\tfrac{1}{|A|}\,\text{adj}(A)\). Always verify by checking \(A\cdot A^{-1}=I\) — this earns a mark and catches sign errors.
Cramer's Rule solves a system using determinants. For \(ax+by=e\) and \(cx+dy=f\): form D (coefficient determinant), \(D_x\) (replace \(x\)-column with constants) and \(D_y\) (replace \(y\)-column with constants). Then \(x=D_x/D\) and \(y=D_y/D\), provided \(D\neq0\). Always write and evaluate all three determinants explicitly — examiners look for this.
For an \(n\times n\) matrix: \(|kA|=k^n\times|A|\). If \(A\) is \(3\times3\) and \(|A|=4\), then \(|3A|=27\times4=108\). A common mistake is writing \(3\times4=12\) — that would only be correct if you multiplied a single row by 3, not the whole matrix.
A matrix \(A\) is idempotent if \(A^2=A\). This lets you simplify expressions quickly: \((I+A)^2-3A=I+2A+A^2-3A=I+2A+A-3A=I\). CBSE Applied Maths MCQs frequently test this — recognising an idempotent condition saves calculation time under exam pressure.
Write the system as \(AX=B\). If \(|A|\neq0\), find \(A^{-1}=\tfrac{1}{|A|}\,\text{adj}(A)\). The unique solution is \(X=A^{-1}B\). Multiply row by row to get the variable values. Always verify your answer by substituting back into the original equations — this earns the verification mark in case studies.

Explore All Units — Class 12 Applied Maths

Free study material for all 8 units of the CBSE Applied Maths syllabus
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Unit 1
Numbers & Quantification
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Unit 2 · Current
Algebra — Matrices & Determinants
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Unit 3
Calculus
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Unit 4
Probability Distributions
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Unit 5
Inferential Statistics
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Unit 6
Index Numbers & Time-based Data
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Unit 7
Financial Mathematics
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Unit 8
Linear Programming

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