Unit 2: Matrices & Determinants — Class 12 Applied Maths
Unit 2 of 8 10 Marks CBSE 2026–27 Algebra

Algebra — Matrices & Determinants
Unit 2 — Free Study Resources

CBSE Class 12 Applied Mathematics · Free MCQs, Solved Examples & Case Studies

Unit 2 carries 10 marks in the CBSE board exam. Everything you need to master it: 10 interactive MCQs with instant feedback, 5 step-by-step solved examples, and 2 board-pattern case studies. Covering Matrices, Matrix Operations, Symmetric & Skew-Symmetric Matrices, Determinants, Inverse of a Matrix, and System of Linear Equations using the Matrix Method and Cramer's Rule. Aligned to the CBSE 2026–27 syllabus. Once you've attempted the questions and self-assessed your answers, tap ✨ My Report to see your personalised performance breakdown.

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Unit 2 · 10 Marks

Topics & Key Formulas

Six topics examined across MCQs, short answers, and case studies. Memorise the formulas below — they appear in multiple question types every year.

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 \(n\times n\) matrix3 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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Interactive Practice

Practice MCQs with Answers — Unit 2 Algebra: Matrices

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

Q1CBSE 2022Matrices
If matrix \(A = [a_{ij}]_{2\times 2}\) where \(a_{ij} = i + j\), what is \(A\) 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 \(a_{12} = a_{21} = 3\), this matrix is symmetric.
Q2CBSE 2022Matrices
If \(A^2 = A\), what is \((I + A)^2 - 3A\) equal to?
\(I\)
\(2A\)
\(3I\)
\(A\)
Step 1 — Expand \((I+A)^2\): \(I^2 + IA + AI + A^2\)

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

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

Step 4 — Subtract \(3A\): \(I + 3A - 3A = \mathbf{I}\)
Q3CBSE 2023Matrices
If \(A = \begin{bmatrix}1&0\\-2&1\end{bmatrix}\) and \(B = \begin{bmatrix}-5&10\\-10&-5\end{bmatrix}\), what is \(AB\)?
\(\begin{bmatrix}-5&10\\0&-5\end{bmatrix}\)
\(\begin{bmatrix}0&-5\\25&10\end{bmatrix}\)
\(\begin{bmatrix}10&-25\\-5&0\end{bmatrix}\)
\(\begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
Row 1 of A = \([1,0]\): \((AB)_{11}=(1)(-5)+(0)(-10)=-5\); \((AB)_{12}=(1)(10)+(0)(-5)=10\)

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

\(AB = \begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
Q4CBSE 2023Matrices
If \(\begin{bmatrix}x+y & x+2\\ 2x-y & 16\end{bmatrix} = \begin{bmatrix}8&5\\1&3y+1\end{bmatrix}\), what are the values of \(x\) and \(y\)?
\(x=3,\;y=5\)
\(x=5,\;y=3\)
\(x=2,\;y=7\)
\(x=7,\;y=2\)
Position (1,2): \(x + 2 = 5 \Rightarrow x = 3\)
Position (1,1): \(x + y = 8 \Rightarrow 3 + y = 8 \Rightarrow y = 5\)
Verify (2,1): \(2(3)-5=1\) ✓  (2,2): \(3(5)+1=16\) ✓
Q5CBSE 2023-CompttMatrices
If \(A\) and \(B\) are matrices such that \(AB=A\) and \(BA=B\), what does \(B^2\) equal?
\(B\)
\(A\)
\(I\)
\(O\)
\(B^2 = B\cdot B\). Substitute \(B = BA\): \(B^2 = (BA)\cdot B = B\cdot(AB)\). Substitute \(AB = A\): \(B^2 = B\cdot A = BA = B\).
Q6CBSE 2023-CompttMatrices
If \(A = \begin{bmatrix}5&x\\y&0\end{bmatrix}\) is symmetric, what must be true?
\(x=0,\;y=5\)
\(x=5,\;y=0\)
\(x=y\)
\(x+y=0\)
Definition: \(A = A^T\) means \(a_{ij} = a_{ji}\).

\(A^T = \begin{bmatrix}5&y\\x&0\end{bmatrix}\). Setting \(A = A^T\): diagonals match automatically; off-diagonal gives \(x = y\).
Q7CBSE 2024-CompttMatrices
If \(A = \begin{bmatrix}4&1\\3&2\end{bmatrix}\), what does \(A^2 - 6A\) equal?
\(3I\)
\(-5I\)
\(5I\)
\(-3I\)
\(A^2 = \begin{bmatrix}19&6\\18&7\end{bmatrix}\),  \(6A = \begin{bmatrix}24&6\\18&12\end{bmatrix}\)

\(A^2 - 6A = \begin{bmatrix}19-24&6-6\\18-18&7-12\end{bmatrix} = \begin{bmatrix}-5&0\\0&-5\end{bmatrix} = -5I\)
Q8CBSE 2022Determinants
If \(A\) is a \(3\times3\) matrix with \(|A|=4\), what does \(|3A|\) equal?
\(27\)
\(81\)
\(108\)
\(256\)
Property: \(|kA| = k^n \cdot |A|\) for an \(n \times n\) matrix.

\(|3A| = 3^3 \times 4 = 27 \times 4 = \mathbf{108}\)

Common mistake: Writing \(3 \times 4 = 12\) — incorrect. The exponent is the matrix order \(n\), not 1.
Q9CBSE 2022Determinants
If \((1,3)\), \((x,5)\) and \((2,7)\) are collinear, what does \(x\) equal?
\(2\)
\(\dfrac{3}{2}\)
\(1\)
\(\dfrac{3}{4}\)
Three points are collinear iff area of triangle = 0:

\(\dfrac{1}{2}\begin{vmatrix}1&3&1\\x&5&1\\2&7&1\end{vmatrix} = 0\)

Expanding: \((5-7) - 3(x-2) + (7x-10) = 0\)
\(-2 - 3x + 6 + 7x - 10 = 0\)
\(4x - 6 = 0 \Rightarrow x = \dfrac{3}{2}\)
Q10CBSE 2025Determinants
If \(\begin{vmatrix}2x&5\\4&x\end{vmatrix}=\begin{vmatrix}3&5\\4&6\end{vmatrix}\), what does \(x\) equal?
\(\dfrac{3}{2}\)
\(6\)
\(3\)
\(\pm3\)
LHS: \(2x^2 - 20\)   RHS: \(18-20 = -2\)

\(2x^2 - 20 = -2 \Rightarrow x^2 = 9 \Rightarrow x = \pm3\)

Verify: both \(x=3\) and \(x=-3\) give LHS = \(-2\) ✓

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Step-by-Step Solutions

Short Answer Questions

2-mark and 3-mark questions with complete working. Click Show Solution to reveal each answer.

📝Self-assess your answer after revealing each solution below — your responses build your free Unit 2 AI Performance Report. Tap ✨ My Report (bottom-right) once you're done.
Q1CBSE 2024Matrices
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 \(w\).
Equate corresponding elements: \(x-y=-1\)…(i), \(2x+z=5\)…(ii), \(2x-y=0\)…(iii), \(3z+w=13\)…(iv)
Subtract (i) from (iii): \(x=1\). Substitute into (i): \(y=2\).
From (ii): \(2(1)+z=5 \Rightarrow z=3\). From (iv): \(3(3)+w=13 \Rightarrow w=4\).
Verification: \(\begin{bmatrix}-1&5\\0&13\end{bmatrix}\) ✓
\(x=1,\quad y=2,\quad z=3,\quad w=4\)
Self-assess: ✓ Saved
Q2CBSE 2024Matrices
If \(A = \begin{bmatrix}1&0\\-1&7\end{bmatrix}\), find \(k\) such that \(A^2 - 8A + kI = O\).
Compute \(A^2\): \(A^2 = \begin{bmatrix}1&0\\-8&49\end{bmatrix}\)
Compute \(8A\): \(8A = \begin{bmatrix}8&0\\-8&56\end{bmatrix}\)
\(A^2 - 8A = \begin{bmatrix}-7&0\\0&-7\end{bmatrix} = -7I\)
Substitute: \(-7I + kI = O \Rightarrow k-7=0\)
\(k = 7\)
Self-assess: ✓ Saved
Q3CBSE 2025Matrices
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 \(BA\).
Row 1 of B = \([1,1,-5]\):
\((BA)_{11}=5,\;(BA)_{12}=-6,\;(BA)_{13}=-9\)
Row 2 of B = \([-5,1,-5]\):
\((BA)_{21}=-7,\;(BA)_{22}=-6,\;(BA)_{23}=-15\)
Row 3 of B = \([1,-2,4]\):
\((BA)_{31}=-4,\;(BA)_{32}=0,\;(BA)_{33}=3\)
\(BA=\begin{bmatrix}5&-6&-9\\-7&-6&-15\\-4&0&3\end{bmatrix}\)
Self-assess: ✓ Saved
Q4CBSE 2023Solving Linear Equations
Solve by Cramer's Rule: \(\;2x-y=17,\quad 3x+5y=6\)
Coefficient determinant \(D\):
\(D=\begin{vmatrix}2&-1\\3&5\end{vmatrix}=10+3=13 \neq 0\) — unique solution exists.
\(D_x\) (replace \(x\)-column with constants):
\(D_x=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91\)
\(D_y\) (replace \(y\)-column with constants):
\(D_y=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39\)
\(x=\dfrac{91}{13}=7\),  \(y=\dfrac{-39}{13}=-3\)   Verification: \(2(7)-(-3)=17\) ✓  \(3(7)+5(-3)=6\) ✓
\(x=7,\quad y=-3\)
Self-assess: ✓ Saved
Q5CBSE 2023-CompttMatrices
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: \(A = P + Q\) where \(P = \tfrac{1}{2}(A+A^T)\) is symmetric and \(Q = \tfrac{1}{2}(A-A^T)\) is skew-symmetric.
\(A^T\): \(\begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}\)
Symmetric part \(P = \tfrac{1}{2}(A+A^T)\):
\(P=\begin{bmatrix}7&-2&-2\\-2&1&0\\-2&0&1\end{bmatrix}\)  \(P^T=P\) ✓
Skew-symmetric part \(Q = \tfrac{1}{2}(A-A^T)\):
\(Q=\begin{bmatrix}0&-1&-1\\1&0&0\\1&0&0\end{bmatrix}\)  \(Q^T=-Q\) ✓
Verification: \(P+Q=\begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}=A\) ✓
\(A = P + Q\) 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)
Self-assess: ✓ Saved

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4–5 Mark Questions

Case Studies — Real-World Application Questions

Board-pattern case-based questions — click Show Solution under each part to reveal full working, then self-assess your answer.

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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.

Solving Linear Equations
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: \(x+y+z=10;\quad 2x+y=13;\quad x+y-4z=0\)
Self-assess: ✓ Saved
(b)

Write the system in matrix form \(AX=B\) and identify the coefficient matrix.

Solving Linear Equations
\(\underbrace{\begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}}_{A}\;\underbrace{\begin{bmatrix}x\\y\\z\end{bmatrix}}_{X}=\underbrace{\begin{bmatrix}10\\13\\0\end{bmatrix}}_{B}\)
Coefficient matrix: \(A=\begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}\)
Self-assess: ✓ Saved
(c)(i)

Find the cofactor matrix of \(A\).

Determinants
Formula: \(A_{ij} = (-1)^{i+j} M_{ij}\) (minor after deleting row \(i\), col \(j\)).
Row 1: \(A_{11}=-4,\;A_{12}=8,\;A_{13}=1\)
Row 2: \(A_{21}=5,\;A_{22}=-5,\;A_{23}=0\)
Row 3: \(A_{31}=-1,\;A_{32}=2,\;A_{33}=-1\)
Cofactor matrix \(=\begin{bmatrix}-4&8&1\\5&-5&0\\-1&2&-1\end{bmatrix}\)
Self-assess: ✓ Saved
(c)(ii) OR

Find the number of students in each group using the matrix method.

Solving Linear Equations
\(|A|\) = \(1(-4)+1(8)+1(1)=5 \neq 0\) — unique solution exists.
\(\text{adj}(A)\) = transpose of cofactor matrix = \(\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
\(A^{-1} = \dfrac{1}{5}\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
\(X = A^{-1}B\):
Row 1: \(\tfrac{1}{5}(-40+65+0)=5\)   Row 2: \(\tfrac{1}{5}(80-65+0)=3\)   Row 3: \(\tfrac{1}{5}(10+0+0)=2\)
Hard workers \(x=5\) · Honest \(y=3\) · Vigilant \(z=2\)  |  Verification: \(5+3+2=10\) ✓  \(2(5)+3=13\) ✓  \(5+3-4(2)=0\) ✓
Self-assess: ✓ Saved
🎁

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.

Solving Linear Equations
Condition 1: \((x-8)(y+10)=xy \Rightarrow 10x-8y=80 \Rightarrow 5x-4y=40\) …(i)
Condition 2: \((x+16)(y-10)=xy \Rightarrow -10x+16y=160 \Rightarrow 5x-8y=-80\) …(ii)
\(5x-4y=40\) and \(5x-8y=-80\)
Self-assess: ✓ Saved
(ii)

Write the system in matrix form \(AX=B\).

Solving Linear Equations
\(\underbrace{\begin{bmatrix}5&-4\\5&-8\end{bmatrix}}_{A}\;\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{X}=\underbrace{\begin{bmatrix}40\\-80\end{bmatrix}}_{B}\)
Matrix form: \(\begin{bmatrix}5&-4\\5&-8\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}40\\-80\end{bmatrix}\)
Self-assess: ✓ Saved
(iii)(a)

Find \(A^{-1}\).

Determinants
\(|A|\) \(=(5)(-8)-(-4)(5)=-40+20=-20 \neq 0\) — inverse exists.
\(\text{adj}(A)\) \(=\begin{bmatrix}-8&4\\-5&5\end{bmatrix}\)
\(A^{-1}=\dfrac{1}{-20}\begin{bmatrix}-8&4\\-5&5\end{bmatrix}=\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\)
Verification: \(A\cdot A^{-1}=I\) ✓
\(A^{-1}=\dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\)
Self-assess: ✓ Saved
(iii)(b) OR

Find \(x\) and \(y\) — the number of children and the amount per child.

Solving Linear Equations
\(X=A^{-1}B = \dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\begin{bmatrix}40\\-80\end{bmatrix}\)
Row 1: \(x=\tfrac{320+320}{20}=32\)   Row 2: \(y=\tfrac{200+400}{20}=30\)
Verification: \(5(32)-4(30)=40\) ✓  \(5(32)-8(30)=-80\) ✓
Number of children: \(x = 32\)  |  Amount per child: ₹30
Self-assess: ✓ Saved
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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.

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❌ Common Mistakes to Avoid in Unit 2

  • Using \(|kA|=k|A|\) instead of the correct \(|kA|=k^n|A|\) for an \(n\times n\) matrix
  • Forgetting to transpose the cofactor matrix to get adj(A)
  • Sign errors in cofactor calculation — always apply \((-1)^{i+j}\)
  • Not verifying \(A\cdot A^{-1}=I\) — this step earns marks in case studies
  • Writing \(AB=BA\) — matrix multiplication is NOT commutative in general
  • Missing the verification step when solving a system by matrix method
Common Questions

Frequently Asked Questions

Answers to questions students frequently 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: \((I+A)^2-3A=I+2A+A^2-3A=I\). 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 \(X=A^{-1}B\). Always verify your answer by substituting back into the original equations — this earns the verification mark in case studies.
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