Class 12 Applied Maths Matrices Practice Questions with Solutions | Boundless Maths
This page covers all topics in Unit 2 of CBSE Class 12 Applied Mathematics — Algebra, which carries 10 marks in the board exam. Sub-topics include Types of Matrices and Matrix Operations, Transpose with Symmetric and Skew-Symmetric Matrices, Determinants including minors and cofactors, Inverse of a Matrix using the adjoint method, and solving a System of Linear Equations using both the Matrix Method and Cramer's Rule. You'll find 10 practice MCQs with detailed answers, 5 short-answer questions with step-by-step solutions, and 2 case studies covering real-world applications of matrices and linear equations. All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026-27.
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Topics Covered in Unit 2 — Algebra

Key concepts and essential formulas for the CBSE Class 12 Applied Maths board exam

Topic 1

Matrices

Types of matrices — row, column, square, diagonal, identity, zero, symmetric, skew-symmetric — and the \(a_{ij}\) notation.

Key Formula
\(A = [a_{ij}]_{m\times n}\) — element in row \(i\), col \(j\)
Transpose: \((A^T)_{ij} = A_{ji}\)
Topic 2

Matrix Operations

Addition, subtraction, scalar multiplication, matrix multiplication — with commutativity and associativity properties.

Key Formula
\((AB)^T = B^T A^T\)
Note: \(AB \neq BA\) in general
Topic 3

Symmetric & Skew-Symmetric

Symmetric: \(A = A^T\). Skew-symmetric: \(A = -A^T\) (diagonal elements are always zero). Decomposition of any square matrix.

Key Formula
\(A = \dfrac{1}{2}(A+A^T) + \dfrac{1}{2}(A-A^T)\)
Sym. part + Skew-sym. part
Topic 4

Determinants

Evaluation by expansion, properties, minors, cofactors. Collinearity of three points via area formula.

Key Formula
\(|kA| = k^n\,|A|\) for an \(n\times n\) matrix
3 points collinear \(\Rightarrow\) area \(= 0\)
Topic 5

Inverse of a Matrix

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

Key Formula
\(A^{-1} = \dfrac{1}{|A|}\,\text{adj}(A)\)
Verify: \(A \cdot A^{-1} = A^{-1} \cdot A = I\)
Topic 6

System of Linear Equations

Matrix method and Cramer's Rule for solving systems of 2 or 3 simultaneous linear equations.

Key Formulas
\(AX = B \Rightarrow X = A^{-1}B\)
Cramer's: \(x = D_x/D,\quad y = D_y/D\)
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Practice MCQs with Answers — Unit 2 Algebra

10 MCQs from CBSE board papers — click "Show Answer" to reveal step-by-step explanations

Question 1
If matrix \(A = [a_{ij}]_{2\times 2}\) where \(a_{ij} = i + j\), then \(A\) is equal to:
  • A\(\begin{bmatrix}1&2\\3&4\end{bmatrix}\)
  • B\(\begin{bmatrix}2&3\\3&4\end{bmatrix}\)
  • C\(\begin{bmatrix}1&1\\2&2\end{bmatrix}\)
  • D\(\begin{bmatrix}1&2\\1&2\end{bmatrix}\)
✓ Correct Answer: (B)
Given: \(a_{ij} = i + j\)
Compute each element: \(a_{11}=1+1=2,\quad a_{12}=1+2=3,\quad a_{21}=2+1=3,\quad a_{22}=2+2=4\)
Therefore \(A = \begin{bmatrix}2&3\\3&4\end{bmatrix}\)
Question 2
If \(A^2 = A\), then \((I + A)^2 - 3A\) is equal to:
  • A\(I\)
  • B\(2A\)
  • C\(3I\)
  • D\(A\)
✓ Correct Answer: (A) \(I\)
Note: \(A^2 = A\) means \(A\) is an idempotent matrix.
Expand: \((I+A)^2 - 3A = I^2 + IA + AI + A^2 - 3A\)
Since \(IA = AI = A\) and \(A^2 = A\): \(= I + A + A + A - 3A = I + 3A - 3A\)
\(= I\)
Question 3
If \(A = \begin{bmatrix}1&0\\-2&1\end{bmatrix}\) and \(B = \begin{bmatrix}-5&10\\-10&-5\end{bmatrix}\), then \(AB\) is:
  • A\(\begin{bmatrix}-5&10\\0&-5\end{bmatrix}\)
  • B\(\begin{bmatrix}0&-5\\25&10\end{bmatrix}\)
  • C\(\begin{bmatrix}10&-25\\-5&0\end{bmatrix}\)
  • D\(\begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
✓ Correct Answer: (D)
Row 1 of \(A\) × columns of \(B\): \((-5, 10)\) — element \((1,1)=1(-5)+0(-10)=-5\); element \((1,2)=1(10)+0(-5)=10\)
Row 2 of \(A\) × columns of \(B\): element \((2,1)=(-2)(-5)+1(-10)=10-10=0\); element \((2,2)=(-2)(10)+1(-5)=-25\)
\(AB = \begin{bmatrix}-5&10\\0&-25\end{bmatrix}\)
Question 4
If \(\begin{bmatrix}x+y & x+2\\ 2x-y & 16\end{bmatrix} = \begin{bmatrix}8&5\\1&3y+1\end{bmatrix}\), then the values of \(x\) and \(y\) are:
  • A\(x = 3,\ y = 5\)
  • B\(x = 5,\ y = 3\)
  • C\(x = 2,\ y = 7\)
  • D\(x = 7,\ y = 2\)
✓ Correct Answer: (A) \(x=3,\ y=5\)
Equate position \((1,2)\): \(x+2 = 5 \Rightarrow x = 3\)
Equate position \((1,1)\): \(x+y = 8 \Rightarrow 3+y = 8 \Rightarrow y = 5\)
Verify: \(2x-y = 6-5 = 1\,\checkmark\) and \(3y+1 = 16\,\checkmark\)
\(x = 3,\quad y = 5\)
Question 5
If \(A\) and \(B\) are two matrices such that \(AB = A\) and \(BA = B\), then \(B^2\) is equal to:
  • A\(B\)
  • B\(A\)
  • C\(I\)
  • D\(O\)
✓ Correct Answer: (A) \(B\)
Write \(B^2 = B\cdot B\). Replace one \(B\) using \(B = BA\): \(B^2 = (BA)\cdot B = B\cdot(AB)\)
Now replace \(AB = A\): \(B^2 = B\cdot A = BA\)
And \(BA = B\) is given.
\(B^2 = B\)
Question 6
If \(A = \begin{bmatrix}5&x\\y&0\end{bmatrix}\) is symmetric, then:
  • A\(x=0,\ y=5\)
  • B\(x=5,\ y=0\)
  • C\(x=y\)
  • D\(x+y=0\)
✓ Correct Answer: (C) \(x = y\)
A matrix is symmetric if \(A = A^T\), meaning \(a_{ij} = a_{ji}\) for all \(i,j\).
\(A^T = \begin{bmatrix}5&y\\x&0\end{bmatrix}\). For \(A = A^T\), compare position \((1,2)\) and \((2,1)\): we need \(x = y\).
\(x = y\)
Question 7
If \(A = \begin{bmatrix}4&1\\3&2\end{bmatrix}\) and \(I = \begin{bmatrix}1&0\\0&1\end{bmatrix}\), then \((A^2 - 6A)\) equals:
  • A\(3I\)
  • B\(-5I\)
  • C\(5I\)
  • D\(-3I\)
✓ Correct Answer: (B) \(-5I\)
Compute \(A^2 = \begin{bmatrix}4&1\\3&2\end{bmatrix}\begin{bmatrix}4&1\\3&2\end{bmatrix} = \begin{bmatrix}4(4)+1(3)&4(1)+1(2)\\3(4)+2(3)&3(1)+2(2)\end{bmatrix} = \begin{bmatrix}19&6\\18&7\end{bmatrix}\)
Compute \(6A = \begin{bmatrix}24&6\\18&12\end{bmatrix}\)
Subtract: \(A^2 - 6A = \begin{bmatrix}19-24&6-6\\18-18&7-12\end{bmatrix} = \begin{bmatrix}-5&0\\0&-5\end{bmatrix}\)
\(A^2 - 6A = -5I\)
Question 8
If \(A\) is a square matrix of order \(3\times 3\) such that \(|A| = 4\), then \(|3A|\) is equal to:
  • A\(27\)
  • B\(81\)
  • C\(108\)
  • D\(256\)
✓ Correct Answer: (C) 108
Property: for an \(n\times n\) matrix, \(|kA| = k^n \cdot |A|\).
Here \(n=3\) and \(k=3\), so \(|3A| = 3^3 \times |A| = 27 \times 4\)
\(|3A| = 108\)
Question 9
If the points \((1,3)\), \((x,5)\) and \((2,7)\) are collinear, then the value of \(x\) is:
  • A\(2\)
  • B\(\dfrac{3}{2}\)
  • C\(1\)
  • D\(\dfrac{3}{4}\)
✓ Correct Answer: (B) \(\dfrac{3}{2}\)
Three points are collinear if the area of the triangle they form equals zero.
Area \(= \dfrac{1}{2}\begin{vmatrix}1&3&1\\x&5&1\\2&7&1\end{vmatrix} = 0\)
Expanding: \(1(5-7) - 3(x-2) + 1(7x-10) = 0\)
\(-2 - 3x + 6 + 7x - 10 = 0 \Rightarrow 4x = 6\)
\(x = \dfrac{3}{2}\)
Question 10
If \(\begin{vmatrix}2x&5\\4&x\end{vmatrix} = \begin{vmatrix}3&5\\4&6\end{vmatrix}\), then the value of \(x\) is:
  • A\(\dfrac{3}{2}\)
  • B\(6\)
  • C\(3\)
  • D\(\pm 3\)
✓ Correct Answer: (D) \(\pm 3\)
LHS: \(\begin{vmatrix}2x&5\\4&x\end{vmatrix} = 2x\cdot x - 5\cdot 4 = 2x^2 - 20\)
RHS: \(\begin{vmatrix}3&5\\4&6\end{vmatrix} = 18 - 20 = -2\)
Set equal: \(2x^2 - 20 = -2 \Rightarrow 2x^2 = 18 \Rightarrow x^2 = 9\)
\(x = \pm 3\)

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Short Answer Questions — Step-by-Step Solutions

2-mark and 3-mark solved questions — click "Show Solution" to reveal step-by-step workings

Question 1
If \(\begin{bmatrix}x-y & 2x+z\\ 2x-y & 3z+w\end{bmatrix} = \begin{bmatrix}-1&5\\0&13\end{bmatrix}\), find the values of \(x,\ y,\ z\) and \(w\).
✓ Solution
Equate corresponding elements to form four equations: \(x - y = -1\) …(i), \(2x - y = 0\) …(ii), \(2x + z = 5\) …(iii), \(3z + w = 13\) …(iv)
Subtract (i) from (ii): \((2x-y)-(x-y)=0-(-1) \Rightarrow x=1\). Substitute into (i): \(1-y=-1 \Rightarrow y=2\).
Substitute \(x=1\) into (iii): \(2+z=5 \Rightarrow z=3\). Substitute \(z=3\) into (iv): \(9+w=13 \Rightarrow w=4\).
\(x = 1,\quad y = 2,\quad z = 3,\quad w = 4\)
Question 2
If \(A = \begin{bmatrix}1&0\\-1&7\end{bmatrix}\), find the value of \(k\) such that \(A^2 - 8A + kI = O\).
✓ Solution
Compute \(A^2 = \begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix} = \begin{bmatrix}1\cdot1+0\cdot(-1)&1\cdot0+0\cdot7\\(-1)\cdot1+7\cdot(-1)&(-1)\cdot0+7\cdot7\end{bmatrix} = \begin{bmatrix}1&0\\-8&49\end{bmatrix}\)
Compute \(8A = \begin{bmatrix}8&0\\-8&56\end{bmatrix}\)
\(A^2 - 8A = \begin{bmatrix}1-8&0-0\\-8-(-8)&49-56\end{bmatrix} = \begin{bmatrix}-7&0\\0&-7\end{bmatrix} = -7I\)
For \(A^2 - 8A + kI = O\): \(-7I + kI = O \Rightarrow (k-7)I = O \Rightarrow k = 7\)
\(k = 7\)
Question 3
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 \(BA\).
✓ Solution
Multiply each row of \(B\) with each column of \(A\). Row 1 of \(B\) \(\times\) columns of \(A\):
\(r_1c_1 = 1(2)+1(3)+(-5)(0)=5\), \(r_1c_2 = 1(0)+1(4)+(-5)(2)=-6\), \(r_1c_3 = 1(1)+1(5)+(-5)(3)=-9\)
Row 2 of \(B\) \(\times\) columns of \(A\):
\(r_2c_1 = (-5)(2)+1(3)+(-5)(0)=-7\), \(r_2c_2 = (-5)(0)+1(4)+(-5)(2)=-6\), \(r_2c_3 = (-5)(1)+1(5)+(-5)(3)=-15\)
Row 3 of \(B\) \(\times\) columns of \(A\):
\(r_3c_1 = 1(2)+(-2)(3)+4(0)=-4\), \(r_3c_2 = 1(0)+(-2)(4)+4(2)=0\), \(r_3c_3 = 1(1)+(-2)(5)+4(3)=3\)
\(BA = \begin{bmatrix}5&-6&-9\\-7&-6&-15\\-4&0&3\end{bmatrix}\)
Question 4
Solve using Cramer's Rule: \(2x - y = 17,\quad 3x + 5y = 6\)
✓ Solution
Form the coefficient determinant: \(D = \begin{vmatrix}2&-1\\3&5\end{vmatrix} = (2)(5)-(-1)(3) = 10+3 = 13\)
Replace the \(x\)-column with constants: \(D_x = \begin{vmatrix}17&-1\\6&5\end{vmatrix} = (17)(5)-(-1)(6) = 85+6 = 91\)
Replace the \(y\)-column with constants: \(D_y = \begin{vmatrix}2&17\\3&6\end{vmatrix} = (2)(6)-(17)(3) = 12-51 = -39\)
Apply Cramer's Rule: \(x = \dfrac{D_x}{D} = \dfrac{91}{13} = 7\) and \(y = \dfrac{D_y}{D} = \dfrac{-39}{13} = -3\)
\(x = 7,\quad y = -3\)
Question 5
Write \(A = \begin{bmatrix}7&-3&-3\\-1&1&0\\-1&0&1\end{bmatrix}\) as a sum of a symmetric and a skew-symmetric matrix.
✓ Solution
Find the transpose: \(A^T = \begin{bmatrix}7&-1&-1\\-3&1&0\\-3&0&1\end{bmatrix}\)
Symmetric part: \(P = \dfrac{1}{2}(A+A^T) = \dfrac{1}{2}\begin{bmatrix}14&-4&-4\\-4&2&0\\-4&0&2\end{bmatrix} = \begin{bmatrix}7&-2&-2\\-2&1&0\\-2&0&1\end{bmatrix}\)
Verify \(P = P^T\) ✓ (symmetric)
Skew-symmetric part: \(Q = \dfrac{1}{2}(A-A^T) = \dfrac{1}{2}\begin{bmatrix}0&-2&-2\\2&0&0\\2&0&0\end{bmatrix} = \begin{bmatrix}0&-1&-1\\1&0&0\\1&0&0\end{bmatrix}\)
Verify \(Q = -Q^T\) ✓ and diagonal elements are all zero ✓ (skew-symmetric)
\(A = P + Q\) 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 as per the latest CBSE Class 12 Applied Maths pattern

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
✓ Solution
Total students: \(x + y + z = 10\)
Double first + second = 13: \(2x + y = 13\), i.e. \(2x + y + 0\cdot z = 13\)
First + second = four times third: \(x + y = 4z\), i.e. \(x + y - 4z = 0\)
System: \(x+y+z=10;\quad 2x+y=13;\quad x+y-4z=0\)
(b) Write the coefficient matrix \(A\)
✓ Solution
The coefficient matrix is formed by the coefficients of \(x\), \(y\) and \(z\) in each equation.
\(A = \begin{bmatrix}1&1&1\\2&1&0\\1&1&-4\end{bmatrix}\)
(c)(i) Write the cofactor matrix of \(A\)
✓ Solution
Compute each cofactor \(A_{ij} = (-1)^{i+j} M_{ij}\) where \(M_{ij}\) is the minor (determinant of the \(2\times2\) submatrix).
\(A_{11} = +\begin{vmatrix}1&0\\1&-4\end{vmatrix} = -4\), \(\quad A_{12} = -\begin{vmatrix}2&0\\1&-4\end{vmatrix} = 8\), \(\quad A_{13} = +\begin{vmatrix}2&1\\1&1\end{vmatrix} = 1\)
\(A_{21} = -\begin{vmatrix}1&1\\1&-4\end{vmatrix} = 5\), \(\quad A_{22} = +\begin{vmatrix}1&1\\1&-4\end{vmatrix} = -5\), \(\quad A_{23} = -\begin{vmatrix}1&1\\1&1\end{vmatrix} = 0\)
\(A_{31} = +\begin{vmatrix}1&1\\1&0\end{vmatrix} = -1\), \(\quad A_{32} = -\begin{vmatrix}1&1\\2&0\end{vmatrix} = 2\), \(\quad A_{33} = +\begin{vmatrix}1&1\\2&1\end{vmatrix} = -1\)
Cofactor Matrix \(= \begin{bmatrix}-4&8&1\\5&-5&0\\-1&2&-1\end{bmatrix}\)
(c)(ii) OR: Determine the number of students in each group
✓ Solution
Check: \(|A| = 1(-4)+1(8)+1(1) = 5 \neq 0\), so a unique solution exists.
\(\text{adj}(A)\) = transpose of cofactor matrix \(= \begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
\(A^{-1} = \dfrac{1}{5}\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\)
Multiply \(X = A^{-1}B\): \(\dfrac{1}{5}\begin{bmatrix}-4&5&-1\\8&-5&2\\1&0&-1\end{bmatrix}\begin{bmatrix}10\\13\\0\end{bmatrix} = \dfrac{1}{5}\begin{bmatrix}-40+65+0\\80-65+0\\10+0+0\end{bmatrix} = \dfrac{1}{5}\begin{bmatrix}25\\15\\10\end{bmatrix} = \begin{bmatrix}5\\3\\2\end{bmatrix}\)
Hard workers: \(x=5\) · Honest students: \(y=3\) · Vigilant students: \(z=2\)
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
✓ Solution
Total money is constant: \(xy\). Condition 1 — 8 fewer children, ₹10 more each: \((x-8)(y+10) = xy\)
Expand: \(xy + 10x - 8y - 80 = xy \Rightarrow 10x - 8y = 80 \Rightarrow 5x - 4y = 40\) …(i)
Condition 2 — 16 more children, ₹10 less each: \((x+16)(y-10) = xy\)
Expand: \(xy - 10x + 16y - 160 = xy \Rightarrow -10x + 16y = 160 \Rightarrow 5x - 8y = -80\) …(ii)
\(5x - 4y = 40\quad\) and \(\quad 5x - 8y = -80\)
(ii) Write the system in matrix form \(AX = B\)
✓ Solution
\(\underbrace{\begin{bmatrix}5&-4\\5&-8\end{bmatrix}}_{A}\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{X} = \underbrace{\begin{bmatrix}40\\-80\end{bmatrix}}_{B}\)
(iii)(a) Find the inverse of matrix \(A\)
✓ Solution
Compute the determinant: \(|A| = 5\times(-8) - (-4)\times 5 = -40 + 20 = -20\)
Since \(|A| \neq 0\), the inverse exists. For \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\), \(\text{adj}(A) = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
So \(\text{adj}(A) = \begin{bmatrix}-8&4\\-5&5\end{bmatrix}\)
\(A^{-1} = \dfrac{1}{-20}\begin{bmatrix}-8&4\\-5&5\end{bmatrix} = \dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\)
(iii)(b) OR: Determine the number of children \(x\) and amount per child \(y\)
✓ Solution
Use the matrix method: \(X = A^{-1}B = \dfrac{1}{20}\begin{bmatrix}8&-4\\5&-5\end{bmatrix}\begin{bmatrix}40\\-80\end{bmatrix}\)
Row 1: \(\dfrac{1}{20}\bigl[8\times40 + (-4)\times(-80)\bigr] = \dfrac{1}{20}(320+320) = \dfrac{640}{20} = 32\)
Row 2: \(\dfrac{1}{20}\bigl[5\times40 + (-5)\times(-80)\bigr] = \dfrac{1}{20}(200+400) = \dfrac{600}{20} = 30\)
\(\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}32\\30\end{bmatrix}\)
∴ Number of children: \(x = 32\)  |  Amount per child: \(y =\) ₹30

How to Score Full Marks in Unit 2 — Exam Tips

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

Always verify your matrix inverse

After finding \(A^{-1}\), compute \(A \times A^{-1}\) and confirm you get the identity matrix \(I\). This 30-second check earns method marks in 3-mark questions — and catches calculation errors before they cost you the full answer.

🔒
4 more exam tips for Unit 2 — plus tips for all 8 units — are in the Question Bank Including: common cofactor sign errors, what examiners penalise in Cramer's Rule, and how to present the matrix method solution for full marks. Get the Question Bank — Exam Tips for All Units →

Common Questions — Unit 2 Algebra: Matrices

Frequently asked questions about Class 12 Applied Maths Unit 2

Unit 2 Algebra covers five main 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 properties, Inverse of a Matrix using the adjoint method, and solving a System of Linear Equations using 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 answer questions, and a 4–5 mark case study. It is one of the higher-weightage units, making determinants, the inverse method and system of equations priority topics for practice.
A matrix A is symmetric if \(A = A^T\), meaning every element mirrors across the main diagonal (\(a_{ij} = a_{ji}\)). A matrix is skew-symmetric if \(A = -A^T\), which forces every diagonal entry to be zero and \(a_{ij} = -a_{ji}\). An important exam result: every square matrix can be expressed as the sum of a symmetric part \(\tfrac{1}{2}(A+A^T)\) and a skew-symmetric part \(\tfrac{1}{2}(A-A^T)\).
For \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\): Step 1 — compute \(|A| = ad - bc\). If \(|A| = 0\), the matrix is singular (no inverse exists). Step 2 — write \(\text{adj}(A) = \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\). Step 3 — \(A^{-1} = \dfrac{1}{|A|}\,\text{adj}(A)\). Always verify by checking \(A \cdot A^{-1} = I\).
Cramer's Rule solves a system of linear equations using determinants. For \(ax + by = e\) and \(cx + dy = f\): form D = determinant of the coefficient matrix, \(D_x\) = determinant after replacing the \(x\)-column with the constants, and \(D_y\) = determinant after replacing the \(y\)-column with constants. Then \(x = D_x/D\) and \(y = D_y/D\), provided \(D \neq 0\). The same logic extends to three-variable systems.
For an \(n\times n\) matrix: \(|kA| = k^n \times |A|\). This is one of the most frequently tested MCQ properties. For example, if \(A\) is a \(3\times3\) matrix and \(|A|=4\), then \(|3A| = 3^3 \times 4 = 108\). A common mistake is to write \(3 \times 4 = 12\) — confusing multiplying a single row (which gives \(k \cdot |A|\)) with multiplying the whole matrix.
A matrix \(A\) is idempotent if \(A^2 = A\). This property lets you simplify expressions quickly: for example, \((I+A)^2 - 3A = I + 2A + A^2 - 3A = I + 2A + A - 3A = I\). CBSE Applied Maths MCQs frequently test this — recognising idempotent matrices saves time under exam pressure.
Write the system as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column vector of unknowns, and \(B\) is the constants column. If \(|A| \neq 0\), find \(A^{-1} = \tfrac{1}{|A|}\,\text{adj}(A)\). The unique solution is \(X = A^{-1}B\). Multiply \(A^{-1}\) by \(B\) row by row to get the values. This method is the standard approach for 3-mark and case-study questions.

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  • ✅ Concise & comprehensive as per latest CBSE syllabus
  • ✅ Perfect for last-minute revision before exams
  • ✅ Instant PDF — print & pin up for quick revision anytime
Buy Formula Deck — ₹199

📚 Complete Question Bank — ₹699

  • ✅ All 8 Units covered — Key Concepts, Exam Tips & Common Mistakes
  • ✅ MCQs, Assertion-Reason, unit-wise CBSE Sample Paper & Previous Year Questions with Solutions
  • ✅ 2025–26 CBSE Sample Paper with Solution
  • ✅ Last Year's Board Paper — fully solved
  • ✅ 2 Full Mock Papers — solved
⭐ Best Value — Limited Period
Combo: Formula Deck + Question Bank
₹898
₹599
Save ₹299

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