Home / Mathematics / Chapter 4 — Determinants
Class 12 Maths · Chapter 4 · Bihar Board

Determinants & Applications

Master determinants with visual diagrams, expansion techniques, area of triangle, adjoint & inverse, step-marked solutions, MCQs, Assertion-Reasoning, and PYQs.

3 Orders (1×1, 2×2, 3×3) Visual Expansion Step Marking
8Properties
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

What is a determinant? Let's visualize it.

4.1 Introduction to Determinants

A determinant is a scalar value associated with every square matrix. It is denoted by $\det(A)$ or $|A|$ or $\Delta$.

Key Idea

Determinants help us determine whether a system of linear equations has a unique solution. If $\det(A) \neq 0$, the system has a unique solution.

Important: Only square matrices have determinants.

4.2 Determinant of Different Orders

Order 1 ($1 \times 1$)

For $A = [a_{11}]$, $\det(A) = a_{11}$.

Order 2 ($2 \times 2$)

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$

Trick: "Main diagonal product minus anti-diagonal product"

Order 3 ($3 \times 3$)

We expand along a row or column using cofactors. For example, expanding along $R_1$:

$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1(b_2 c_3 - b_3 c_2) - b_1(a_2 c_3 - a_3 c_2) + c_1(a_2 b_3 - a_3 b_2)$$

4.3 Area of a Triangle

Area of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:

$$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$$
Key Points
  • Always take absolute value since area is positive.
  • If area = 0, points are collinear.
  • If area given, use both positive and negative values of determinant.
Section 02

🎨 Visual Diagrams

See how determinants work

V1 2×2 Determinant — Cross Multiplication

ad - bc a b c d + (main diagonal) = ad - (anti diagonal) = bc = ad - bc
Multiply main diagonal (a·d), subtract anti-diagonal product (b·c). That's your 2×2 determinant.

V2 3×3 Expansion Along First Row

Cofactor signs (+ - +) Sign Pattern (-1)^(i+j) + − + − + − + − + Expanding along Row 1: |A| = a₁₁ M₁₁ − a₁₂ M₁₂ + a₁₃ M₁₃ Where Mᵢⱼ = minor (determinant after removing row i and column j) Or in terms of cofactors: |A| = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃
Signs alternate +, −, + across the first row. The same pattern extends to all rows and columns.

V3 How to Find a Minor

Delete row i and column j Original 3×3 1 2 3 4 5 6 7 8 9 → Minor M₂₂ (2×2) 1 3 7 9 = 9 - 21 = -12 M₂₂
To find minor M₂₂, delete row 2 and column 2. The remaining 2×2 determinant is the minor.

V4 Area of Triangle Using Determinant

Geometric Interpretation X Y A(x₁,y₁) B(x₂,y₂) C(x₃,y₃) Area = ½ | determinant of vertices with third column all 1s |
The absolute value of the determinant (divided by 2) gives the area of the triangle. If the determinant equals zero, the three points are collinear.

V5 Adjoint of a 2×2 Matrix — Shortcut

Swap diagonal, negate off-diagonal A a₁₁ a₁₂ a₂₁ a₂₂ → adj A adj A a₂₂ −a₁₂ −a₂₁ a₁₁ Green (swap): a₁₁ ↔ a₂₂ | Red (negate): a₁₂, a₂₁ For 2×2 only — larger orders need cofactor method
For a 2×2 matrix, the adjoint is easy: swap the diagonal elements, negate the off-diagonal elements.

V6 Inverse via Adjoint

A⁻¹ = (1/|A|) · adj(A) |A| ≠ 0 Check first! → adj(A) Transpose of cofactor matrix [Aᵢⱼ]ᵀ = A⁻¹ = adj(A)/|A|
Golden rule: First check $|A| \neq 0$. If yes, inverse exists. Then compute adj(A) and divide each element by $|A|$.
Section 03

Area of Triangle

Application of determinants in coordinate geometry

A1 Formula & Conditions

$$\text{Area} = \frac{1}{2}\left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$$
Key Cases
  • Area = 0 → Points are collinear
  • Area given → Consider both + and − values
  • Always take absolute value for final area
  • Works for any orientation of the triangle
Section 04

Minors & Cofactors

Essential for expansion and adjoint

M1 Definitions

Minor (Mᵢⱼ)

The determinant obtained by deleting row i and column j from the original determinant.

Cofactor (Aᵢⱼ)

$A_{ij} = (-1)^{i+j} M_{ij}$

Sign Pattern of Cofactors

$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$

Key Properties

Important
  • $\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ (expansion along R₁)
  • $a_{11}A_{21} + a_{12}A_{22} + a_{13}A_{23} = 0$ (different row × cofactors)
  • Minor of element in $n \times n$ determinant is a determinant of order $(n-1) \times (n-1)$
Section 05

Adjoint & Inverse

Finding the inverse of a matrix

Ad1 Adjoint of a Matrix

Adjoint (adj A)

The transpose of the cofactor matrix of A. If A = [aᵢⱼ] and Aᵢⱼ is the cofactor of aᵢⱼ, then adj(A) = [Aᵢⱼ]ᵀ.

For 2×2 Matrix

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then:

$$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

Trick: Swap diagonal, negate off-diagonal.

Key Theorem

$$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A| I$$

Ad2 Inverse of a Matrix

Invertible Condition

A square matrix A is invertible if and only if $|A| \neq 0$ (i.e., A is non-singular).

$$A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$$

Important Properties

Memorize
  • $(AB)^{-1} = B^{-1}A^{-1}$ (reverse order)
  • $(A^T)^{-1} = (A^{-1})^T$
  • $(A^{-1})^{-1} = A$
  • $|A^{-1}| = \frac{1}{|A|}$
  • $|\text{adj}(A)| = |A|^{n-1}$ for $n \times n$ matrix

Ad3 System of Linear Equations

For the system $a_1 x + b_1 y + c_1 z = d_1$, etc., we write:

$$AX = B$$

where A is the coefficient matrix, X = [x, y, z]ᵀ, B = [d₁, d₂, d₃]ᵀ.

Cases
  • If $|A| \neq 0$: Unique solution — $X = A^{-1}B$
  • If $|A| = 0$ and $(adj A)B \neq 0$: No solution (inconsistent)
  • If $|A| = 0$ and $(adj A)B = 0$: Infinite solutions or no solution
Quick Reference

📖 All Formulas at a Glance

Everything you need from Chapter 4

F1 Basic Determinants

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$
$$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1(b_2 c_3 - b_3 c_2) - b_1(a_2 c_3 - a_3 c_2) + c_1(a_2 b_3 - a_3 b_2)$$

F2 Properties of Determinants

  • $|A^T| = |A|$
  • If any two rows/columns are identical → $|A| = 0$
  • If any row/column is all zeros → $|A| = 0$
  • Interchanging two rows/columns → sign changes
  • $|kA| = k^n |A|$ for $n \times n$ matrix
  • $|AB| = |A| \cdot |B|$
  • $|adj(A)| = |A|^{n-1}$

F3 Area of Triangle

$$\Delta = \frac{1}{2}\left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$$

F4 Minors & Cofactors

$M_{ij}$ = Minor (delete row i, column j)
$A_{ij} = (-1)^{i+j} M_{ij}$
$\Delta = \sum_{j} a_{ij} A_{ij}$ (for any row i)

F5 Adjoint & Inverse

$$\text{adj}(A) = [A_{ij}]^T$$
$$A \cdot \text{adj}(A) = |A| \cdot I$$
$$A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A), \quad |A| \neq 0$$
For 2×2: $A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

F6 System of Equations

$AX = B \Rightarrow X = A^{-1}B$ (if $|A| \neq 0$)
Shortcuts

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — 2×2 Determinant Fast

"Down minus Up"

  • Main diagonal (top-left to bottom-right): $a \times d$
  • Anti-diagonal (top-right to bottom-left): $b \times c$
  • Result = $ad - bc$
Trick 2 — Choose Smart Row/Column

Expand along row/column with most zeros!

  • Every zero saves you a 2×2 determinant calculation
  • Less computation = fewer mistakes
Trick 3 — Sign Pattern Memory

Checkerboard rule for cofactors:

  • Top-left is always +
  • Signs alternate like chessboard
  • $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$
Trick 4 — Row Operations Shortcut

Make zeros smartly:

  • $R_i \to R_i + k R_j$ — determinant unchanged
  • $R_i \leftrightarrow R_j$ — sign flips
  • $R_i \to k R_i$ — determinant multiplies by k
Trick 5 — Collinearity Check

"If area = 0, points are collinear"

  • Just compute determinant with all 1s in column 3
  • If result = 0 → collinear
  • Faster than slope method
Trick 6 — Adjoint of 2×2 Memory

"Swap diagonal, flip signs of off-diagonal"

  • $a_{11} \leftrightarrow a_{22}$
  • $a_{12} \to -a_{12}$, $a_{21} \to -a_{21}$
Trick 7 — Inverse Existence Test

Compute |A| first.

  • If $|A| = 0$ → Inverse does NOT exist (singular)
  • If $|A| \neq 0$ → Inverse exists (non-singular)
  • Don't waste time finding adjoint if |A| = 0
Trick 8 — Special Determinant Values

Memorize these common ones:

  • $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} = 1$
  • Skew-symmetric odd order determinant = 0
  • Triangular matrix determinant = product of diagonal
Trick 9 — System Consistency Quick Check

For $AX = B$:

  • $|A| \neq 0$ → Unique solution exists
  • $|A| = 0$ AND $(adj A)B = 0$ → Infinite solutions
  • $|A| = 0$ AND $(adj A)B \neq 0$ → No solution
Trick 10 — Product Rule Reversal

When dealing with product of matrices:

  • $(AB)^{-1} = B^{-1}A^{-1}$ (reverse order!)
  • $|AB| = |A| \cdot |B|$ (order doesn't matter here)
  • $(AB)^T = B^T A^T$ (reverse order)
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. Evaluate $\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}$.

Step 1 — Apply 2×2 Formula1 Mark

$\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = (2)(2) - (4)(-1)$

Step 2 — Simplify1 Mark

$= 4 + 4 = 8$

Determinant = 8
📋 Total: 3 Marks

Q2. Evaluate $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.

Step 1 — Choose column with most zeros0.5 Mark

Column 3 has two zeros → expand along C₃.

Step 2 — Apply expansion1.5 Marks

$\Delta = 4 \cdot (-1)^{1+3} \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix}$

Only the first term survives because other elements in C₃ are 0.

$= 4 \cdot [(-1)(1) - (3)(4)] = 4(-1 - 12)$

Step 3 — Simplify1 Mark

$= 4 \cdot (-13) = -52$

$\Delta = -52$
📋 Total: 3 Marks

Q3. Find the area of the triangle whose vertices are (3, 8), (−4, 2) and (5, 1).

Step 1 — Set up determinant1 Mark

$\Delta = \frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \end{vmatrix}$

Step 2 — Expand1.5 Marks

$= \frac{1}{2}[3(2-1) - 8(-4-5) + 1(-4-10)]$

$= \frac{1}{2}[3(1) - 8(-9) + 1(-14)]$

$= \frac{1}{2}[3 + 72 - 14] = \frac{1}{2}(61)$

Step 3 — Take absolute value0.5 Mark

Area = $\frac{61}{2}$ sq units

Area = $\frac{61}{2}$ sq units
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

1. The value of $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$ is:

Correct: B — $(2)(5) - (3)(4) = 10 - 12 = -2$.

2. If $|A| = 0$, then A is called:

Correct: A — $|A| = 0$ means A is singular (inverse doesn't exist).

3. $|A^T|$ equals:

Correct: C — Transpose doesn't change the determinant value.

4. If two rows of a determinant are identical, then its value is:

Correct: D — Determinant with two identical rows is always 0.

5. Area of triangle with vertices (0,0), (1,0), (0,1) is:

Correct: B — Right triangle with legs 1, 1 → Area = 1/2.

6. $\text{adj}(A)$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ is:

Correct: C — Swap diagonal, negate off-diagonal.

7. If $|A| = 5$, then $|A^{-1}|$ equals:

Correct: A — $|A^{-1}| = 1/|A| = 1/5$.

8. $A \cdot \text{adj}(A)$ equals:

Correct: D — Fundamental identity: $A \cdot \text{adj}(A) = |A| \cdot I$.

9. If $A$ is $3 \times 3$ with $|A| = 2$, then $|kA|$ equals:

Correct: B — $|kA| = k^n |A| = k^3 \cdot 2$ for $n=3$.

10. If $A$ is non-singular, then $(A^T)^{-1}$ equals:

Correct: C — $(A^T)^{-1} = (A^{-1})^T$.

11. If three points are collinear, area of triangle formed is:

Correct: A — Collinear points → area = 0 → determinant = 0.

12. $|AB|$ equals:

Correct: D — Determinant of product = product of determinants.

13. If A is a $3 \times 3$ matrix, then $|\text{adj}(A)|$ equals:

Correct: B — $|\text{adj}(A)| = |A|^{n-1} = |A|^2$ for $n=3$.

14. Value of $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$ is:

Correct: A — $\cos^2\theta + \sin^2\theta = 1$.

15. For a non-singular matrix A, $(A^{-1})^{-1}$ equals:

Correct: C — Inverse of inverse gives back the original matrix.

16. If $|A| \neq 0$ and $AB = I$, then $B$ equals:

Correct: B — By definition, if $AB = I$ then $B = A^{-1}$.

17. The number of ways to expand a 3×3 determinant is:

Correct: D — 3 rows + 3 columns = 6 ways. All give same value.

18. If $A$ is a square matrix of order 2 with $|A| = -3$, then $|\text{adj}(A)|$ equals:

Correct: A — $|\text{adj}(A)| = |A|^{n-1} = |A|^1 = -3$ for $n=2$.

19. $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$ equals:

Correct: C — Identity matrix determinant = product of diagonals = 1.

20. If A is a $3 \times 3$ non-singular matrix, $A^{-1}$ equals:

Correct: D — Fundamental formula: $A^{-1} = \text{adj}(A)/|A|$.
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) If any two rows of a determinant are identical, its value is zero.
Reason (R) Interchanging two identical rows doesn't change the determinant but its sign changes.
Correct: A — Both true. Since swap gives same determinant but sign flips, so $D = -D \Rightarrow 2D = 0 \Rightarrow D = 0$.
2
Assertion (A) $|A^T| = |A|$ for any square matrix A.
Reason (R) Transposing swaps rows and columns; expansion along either gives the same value.
Correct: A — Both true and R correctly explains A.
3
Assertion (A) $(AB)^{-1} = A^{-1}B^{-1}$.
Reason (R) $(AB)^{-1} = B^{-1}A^{-1}$.
Correct: D — A is false (order must be reversed: $B^{-1}A^{-1}$, not $A^{-1}B^{-1}$). R is true.
4
Assertion (A) Area of a triangle whose vertices are collinear is zero.
Reason (R) Determinant of the vertex matrix equals zero for collinear points.
Correct: A — Both true. R explains why A is true.
5
Assertion (A) Only square matrices have determinants.
Reason (R) Determinant is defined as a scalar value associated only with square matrices.
Correct: A — Both true, R correctly explains A.
6
Assertion (A) Every square matrix has an inverse.
Reason (R) $A^{-1} = \text{adj}(A)/|A|$ and requires $|A| \neq 0$.
Correct: D — A is false (only NON-SINGULAR matrices have inverse). R is true.
7
Assertion (A) $|kA| = k^n|A|$ for an $n \times n$ matrix A.
Reason (R) Each of the n rows gets multiplied by k, giving n factors of k.
Correct: A — Both true, R correctly explains A.
8
Assertion (A) $|AB| = |A| + |B|$ for square matrices A, B.
Reason (R) $|AB| = |A| \cdot |B|$.
Correct: D — A is false (product, not sum). R is true.
9
Assertion (A) If $AX = B$ and $|A| \neq 0$, then $X = A^{-1}B$ is unique.
Reason (R) Inverse of a non-singular matrix is unique.
Correct: A — Both true, R explains uniqueness of the solution.
10
Assertion (A) $|\text{adj}(A)| = |A|^{n-1}$ for an $n \times n$ non-singular matrix A.
Reason (R) $A \cdot \text{adj}(A) = |A| \cdot I$ and taking determinants on both sides gives the result.
Correct: A — Both true. $|A \cdot \text{adj}(A)| = |A|^n \cdot |I_n| \Rightarrow |A| \cdot |\text{adj}(A)| = |A|^n \Rightarrow |\text{adj}(A)| = |A|^{n-1}$.
Score: 0 / 10
Exam Focus

📚 Previous Year Questions

Frequently asked questions in board exams

2024

Q. Evaluate: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

Step 1 — Formula0.5 Mark

$= (2)(-1) - (4)(-5)$

Step 2 — Simplify0.5 Mark

$= -2 + 20 = 18$

= 18
2024

Q. Find the value of $x$ if $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$.

Step 1 — Expand both1 Mark

$x^2 - 36 = 36 - 36$

$x^2 - 36 = 0$

Step 2 — Solve1 Mark

$x^2 = 36 \Rightarrow x = \pm 6$

$x = \pm 6$
2023

Q. Find the area of the triangle with vertices (1, 0), (6, 0), (4, 3).

Step 1 — Set up1 Mark

$\Delta = \frac{1}{2}\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$

Step 2 — Expand along C₂1 Mark

$\Delta = \frac{1}{2}\left[0 - 0 + 3 \cdot (-1)^{3+2}\begin{vmatrix} 1 & 1 \\ 6 & 1 \end{vmatrix}\right]$

$= \frac{1}{2}(-3)(1 - 6) = \frac{1}{2}(-3)(-5) = \frac{15}{2}$

Area = $\frac{15}{2}$ sq units
2023

Q. Find adjoint of $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Step 1 — Apply formula1 Mark

Swap diagonal, negate off-diagonal:

$\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$

adj(A) = $\begin{bmatrix} 4 & -3 \\ -1 & 2 \end{bmatrix}$
2022

Q. If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, show that $A \cdot \text{adj}(A) = |A| I$.

Step 1 — Compute |A|0.5 Mark

$|A| = 4 - 6 = -2$

Step 2 — Compute adj(A)1 Mark

$\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Step 3 — Verify1.5 Marks

$A \cdot \text{adj}(A) = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}$

$= -2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = |A| I$ ✓

Verified
2022

Q. Using determinants, find the equation of the line joining A(1, 3) and B(0, 0).

Step 1 — Determinant form1 Mark

For any point (x, y) on AB, $\begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0$

Step 2 — Expand1 Mark

$0(3-y) - 0(1-x) + 1(1 \cdot y - 3 \cdot x) = 0$

$y - 3x = 0 \Rightarrow y = 3x$

Equation: $y = 3x$
2021

Q. If $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$, find $A^{-1}$ using the equation $A^2 - 4A + I = O$.

Step 1 — Rearrange1 Mark

$A^2 - 4A + I = O \Rightarrow A^2 - 4A = -I$

Multiply by $A^{-1}$: $A - 4I = -A^{-1}$

Step 2 — Solve for A⁻¹1 Mark

$A^{-1} = 4I - A = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
2020

Q. Solve the system using matrix method: $x + 2y = 5$, $3x + 4y = 11$.

Step 1 — Matrix form1 Mark

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 5 \\ 11 \end{bmatrix}$

Step 2 — Find A⁻¹1.5 Marks

$|A| = 4 - 6 = -2$, $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

$A^{-1} = \frac{-1}{2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$

Step 3 — Solve X = A⁻¹B1.5 Marks

$X = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}\begin{bmatrix} 5 \\ 11 \end{bmatrix} = \begin{bmatrix} -10 + 11 \\ 15/2 - 11/2 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

$x = 1, y = 2$