# Grade 10 Optional Mathematics: Matrix Study Guide
## Unit: Algebra | Chapter: Matrix | Complete Solutions

---

## 🎯 Chapter Introduction

Welcome to the comprehensive study guide for the **Matrix** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered matrix operations, determinants, inverses, and solving systems of equations using Cramer's Rule. 📈

### Key Learning Areas
1.  **Matrix Fundamentals:** Definitions, order, and types of matrices.
2.  **Matrix Operations:** Multiplication of matrices and its conditions.
3.  **Determinants:** How to calculate the determinant of a 2x2 matrix.
4.  **Inverse of a Matrix:** Finding the adjoint and inverse of a 2x2 matrix.
5.  **Cramer's Rule:** A method for solving systems of linear equations.

---

## 🔑 Key Concepts & Formulas

Before we solve the problems, let's understand the main tools we'll be using.

### 1. Core Definitions

* **Matrix:** A rectangular array of numbers arranged in rows and columns, enclosed in brackets `[]` or parentheses `()`.
* **Order of a Matrix:** A matrix with `m` rows and `n` columns has an order of `m x n`.
* **Identity Matrix (I):** A square matrix with `1`s on the main diagonal and `0`s elsewhere. For a 2x2 matrix: $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. Multiplying any matrix by `I` leaves it unchanged (i.e., `AI = IA = A`).
* **Null Matrix (O):** A matrix where all elements are zero.
* **Transpose of a Matrix ($A^T$):** The matrix obtained by interchanging the rows and columns of matrix A.
* **Inverse Matrix ($A^{-1}$):** If A and B are two square matrices of the same order such that their product is the identity matrix ($AB = BA = I$), then B is the inverse of A, and A is the inverse of B.

---

### 2. Matrix Multiplication

* **Condition:** Two matrices A (order `m x n`) and B (order `p x q`) can be multiplied to get the product AB only if the **number of columns in A is equal to the number of rows in B** (i.e., `n = p`).
    * The resulting matrix AB will have an order of `m x q`.

---

### 3. Determinant of a 2x2 Matrix

The determinant of a matrix is a special number that can be calculated from its elements. For a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is denoted as `|A|` or `det(A)`.

* **Formula:**
    $$|A| = ad - bc$$

* **Singular Matrix:** A matrix whose determinant is zero (`|A| = 0`). A singular matrix does **not** have an inverse. This is the necessary condition for an inverse to be undefined.

---

### 4. Inverse of a 2x2 Matrix

* **Condition for Existence:** The inverse of a matrix A exists **only if its determinant is not zero** (i.e., $|A| \neq 0$).

* **Steps to Find the Inverse:**
    1.  **Find the Determinant:** Calculate $|A| = ad - bc$. If it's zero, stop. The inverse does not exist.
    2.  **Find the Adjoint:** For a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint of A, denoted `adj(A)`, is found by swapping the elements of the main diagonal and changing the signs of the other two elements:
        $$adj(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
    3.  **Calculate the Inverse:** Use the formula:
        $$A^{-1} = \frac{1}{|A|} \text{adj}(A)$$

---

### 5. Cramer's Rule

Cramer's Rule is a method for solving systems of linear equations using determinants.
For a system of two linear equations:
$$
\begin{aligned}
a_1x + b_1y &= c_1 \\
a_2x + b_2y &= c_2
\end{aligned}
$$
**Step 1: Define the Determinants.**
* **D (Coefficient Determinant):** The determinant of the coefficient matrix.
    $$D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$$
* **$D_x$:** Replace the x-column in D with the constant terms.
    $$D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}$$
* **$D_y$:** Replace the y-column in D with the constant terms.
    $$D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$$



**Step 2: Solve for x and y.**
$$
x = \frac{D_x}{D} \quad \text{and} \quad y = \frac{D_y}{D}
$$

* **Condition for a Unique Solution:** A unique solution exists only if the determinant of the coefficient matrix is non-zero ($D \neq 0$).

---

## 📚 Exercise Solutions

### Short Questions

***

#### Question 1: Find an Unknown Matrix
**If $\begin{bmatrix} 4 \\ 1 \end{bmatrix} \cdot C = \begin{bmatrix} -8 & 12 \\ -2 & 3 \end{bmatrix}$, find the matrix C.**

**Solution:**
The first matrix is $2 \times 1$. The resulting matrix is $2 \times 2$.
For the multiplication to be possible, matrix C must have an order of $1 \times 2$.
Let $C = \begin{bmatrix} a & b \end{bmatrix}$.

$$
\begin{aligned}
\begin{bmatrix} 4 \\ 1 \end{bmatrix} \begin{bmatrix} a & b \end{bmatrix} &= \begin{bmatrix} -8 & 12 \\ -2 & 3 \end{bmatrix} \\
\begin{bmatrix} 4 \cdot a & 4 \cdot b \\ 1 \cdot a & 1 \cdot b \end{bmatrix} &= \begin{bmatrix} -8 & 12 \\ -2 & 3 \end{bmatrix} \\
\begin{bmatrix} 4a & 4b \\ a & b \end{bmatrix} &= \begin{bmatrix} -8 & 12 \\ -2 & 3 \end{bmatrix}
\end{aligned}
$$
Equating the corresponding elements:
From the first column: $a = -2$.
From the second column: $b = 3$.

Thus, **$C = \begin{bmatrix} -2 & 3 \end{bmatrix}$**.

***

#### Question 2: Find an Unknown Matrix
**If $\begin{bmatrix} 4 \\ 3 \end{bmatrix} \cdot p = \begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}$, find the matrix p.**

**Solution:**
Let the matrix $p$ be a $1 \times 2$ matrix, $p = \begin{bmatrix} a & b \end{bmatrix}$.

$$
\begin{aligned}
\begin{bmatrix} 4 \\ 3 \end{bmatrix} \begin{bmatrix} a & b \end{bmatrix} &= \begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix} \\
\begin{bmatrix} 4a & 4b \\ 3a & 3b \end{bmatrix} &= \begin{bmatrix} 8 & -4 \\ 6 & -3 \end{bmatrix}
\end{aligned}
$$
Equating the corresponding elements:
From the first column: $3a = 6 \implies a=2$.
From the second column: $3b = -3 \implies b=-1$.

Thus, **$p = \begin{bmatrix} 2 & -1 \end{bmatrix}$**.

***

#### Question 3: Matrix Multiplication
**If $A = \begin{bmatrix} 1 & 2 & -3 \\ -2 & 0 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 \\ -4 & 0 \\ 5 & -1 \end{bmatrix}$, find AB.**

**Solution:**
Matrix A is $2 \times 3$ and B is $3 \times 2$. The product AB will be a $2 \times 2$ matrix.
$$
\begin{aligned}
AB &= \begin{bmatrix} 1 & 2 & -3 \\ -2 & 0 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & 0 \\ 5 & -1 \end{bmatrix} \\
&= \begin{bmatrix} (1)(2) + (2)(-4) + (-3)(5) & (1)(3) + (2)(0) + (-3)(-1) \\ (-2)(2) + (0)(-4) + (5)(5) & (-2)(3) + (0)(0) + (5)(-1) \end{bmatrix} \\
&= \begin{bmatrix} 2 - 8 - 15 & 3 + 0 + 3 \\ -4 + 0 + 25 & -6 + 0 - 5 \end{bmatrix} \\
&= \begin{bmatrix} -21 & 6 \\ 21 & -11 \end{bmatrix}
\end{aligned}
$$
Thus, **$AB = \begin{bmatrix} -21 & 6 \\ 21 & -11 \end{bmatrix}$**.

***

#### Question 4: Singular Matrix
**If the inverse of the matrix $\begin{bmatrix} x & 3 \\ 4 & 2 \end{bmatrix}$ cannot be defined, find the value of x.**

**Solution:**
An inverse of a matrix cannot be defined if the matrix is singular, which means its determinant is 0.
Let $A = \begin{bmatrix} x & 3 \\ 4 & 2 \end{bmatrix}$.
$$
\begin{aligned}
|A| &= 0 \\
(x)(2) - (3)(4) &= 0 \\
2x - 12 &= 0 \\
2x &= 12 \\
x &= 6
\end{aligned}
$$
Thus, **$x=6$**.

***

#### Question 5: Find an Inverse Matrix
**If matrix $A = \begin{bmatrix} 5 & 3 \\ 7 & 4 \end{bmatrix}$, find $A^{-1}$.**

**Solution:**
**Step 1: Find the determinant.**
$$
|A| = (5)(4) - (3)(7) = 20 - 21 = -1
$$
Since $|A| \neq 0$, the inverse exists.

**Step 2: Find the adjoint.**
$$
adj(A) = \begin{bmatrix} 4 & -3 \\ -7 & 5 \end{bmatrix}
$$
**Step 3: Calculate the inverse.**
$$
A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-1} \begin{bmatrix} 4 & -3 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 7 & -5 \end{bmatrix}
$$
Thus, **$A^{-1} = \begin{bmatrix} -4 & 3 \\ 7 & -5 \end{bmatrix}$**.

***

#### Question 6: Finding a Value from an Inverse
**If the inverse of the matrix $\begin{bmatrix} m & 2 \\ 7 & 3 \end{bmatrix}$ is the matrix $\begin{bmatrix} 3 & -2 \\ -7 & m \end{bmatrix}$, find the value of m.**

**Solution:**
Let $A = \begin{bmatrix} m & 2 \\ 7 & 3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -2 \\ -7 & m \end{bmatrix}$.
By the definition of an inverse, $A \cdot A^{-1} = I$.
$$
\begin{aligned}
\begin{bmatrix} m & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ -7 & m \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\
\begin{bmatrix} 3m - 14 & -2m + 2m \\ 21 - 21 & -14 + 3m \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\
\begin{bmatrix} 3m - 14 & 0 \\ 0 & 3m - 14 \end{bmatrix} &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\end{aligned}
$$
Equating the top-left elements:
$$
\begin{aligned}
3m - 14 &= 1 \\
3m &= 15 \\
m &= 5
\end{aligned}
$$
Thus, **$m=5$**.

***

#### Question 7: Matrix and its Transpose
**If matrix $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, find $A \cdot A^T$.**

**Solution:**
First, find the transpose of A, $A^T$.
$$
A^T = \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix}
$$
Now, multiply A by $A^T$.
$$
\begin{aligned}
A \cdot A^T &= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 1 & 2 \end{bmatrix} \\
&= \begin{bmatrix} (3)(3) + (1)(1) & (3)(-1) + (1)(2) \\ (-1)(3) + (2)(1) & (-1)(-1) + (2)(2) \end{bmatrix} \\
&= \begin{bmatrix} 9+1 & -3+2 \\ -3+2 & 1+4 \end{bmatrix} \\
&= \begin{bmatrix} 10 & -1 \\ -1 & 5 \end{bmatrix}
\end{aligned}
$$
Thus, **$A \cdot A^T = \begin{bmatrix} 10 & -1 \\ -1 & 5 \end{bmatrix}$**.

***

#### Question 8: Find an Unknown from a Matrix Equation
**If $A = \begin{bmatrix} 2 & 0 \\ -3 & x \end{bmatrix}$, $B = \begin{bmatrix} 4 & 0 \\ -9 & 1 \end{bmatrix}$ and $A^2 = B$, find the value of x.**

**Solution:**
First, calculate $A^2$.
$$
\begin{aligned}
A^2 &= A \cdot A = \begin{bmatrix} 2 & 0 \\ -3 & x \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -3 & x \end{bmatrix} \\
&= \begin{bmatrix} (2)(2) + (0)(-3) & (2)(0) + (0)(x) \\ (-3)(2) + (x)(-3) & (-3)(0) + (x)(x) \end{bmatrix} \\
&= \begin{bmatrix} 4 & 0 \\ -6 - 3x & x^2 \end{bmatrix}
\end{aligned}
$$
Now, set $A^2 = B$.
$$
\begin{bmatrix} 4 & 0 \\ -6 - 3x & x^2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 1 \end{bmatrix}
$$
Equating the bottom-right elements:
$$x^2 = 1 \implies x = \pm 1$$
Equating the bottom-left elements:
$$-6 - 3x = -9 \implies -3x = -3 \implies x = 1$$
Both conditions must be true, so the only valid solution is **$x=1$**.

***

#### Question 9: Finding a Value from Determinant
**If $M = \begin{bmatrix} 5 & 3 \\ 6 & K \end{bmatrix}$ and $|M| = 27$, find the value of K.**

**Solution:**
Calculate the determinant of M and set it equal to 27.
$$
\begin{aligned}
|M| &= (5)(K) - (3)(6) \\
27 &= 5K - 18 \\
45 &= 5K \\
K &= 9
\end{aligned}
$$
Thus, **$K=9$**.

***

#### Question 10: Prove AB is a Null Matrix
**If $A = \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -5 \\ 2 & -5 \end{bmatrix}$, show that AB is a null matrix.**

**Solution:**
Multiply the matrices A and B.
$$
\begin{aligned}
AB &= \begin{bmatrix} 1 & -1 \\ 3 & -3 \end{bmatrix} \begin{bmatrix} 2 & -5 \\ 2 & -5 \end{bmatrix} \\
&= \begin{bmatrix} (1)(2) + (-1)(2) & (1)(-5) + (-1)(-5) \\ (3)(2) + (-3)(2) & (3)(-5) + (-3)(-5) \end{bmatrix} \\
&= \begin{bmatrix} 2-2 & -5+5 \\ 6-6 & -15+15 \end{bmatrix} \\
&= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\end{aligned}
$$
Since the resulting matrix contains only zeros, **AB is a null matrix**.

***

#### Question 11: Finding a Matrix from its Inverse
**If the inverse of a matrix A is $A^{-1} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$, find the matrix A.**

**Solution:**
The inverse of an inverse matrix is the original matrix, i.e., $(A^{-1})^{-1} = A$. So, we find the inverse of the given $A^{-1}$.
Let $M = A^{-1} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$.

**Step 1: Find the determinant of M.**
$$
|M| = (1)(4) - (3)(2) = 4 - 6 = -2
$$
**Step 2: Find the adjoint of M.**
$$
adj(M) = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}
$$
**Step 3: Calculate the inverse of M.**
$$
A = M^{-1} = \frac{1}{|M|}adj(M) = \frac{1}{-2} \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3/2 \\ 1 & -1/2 \end{bmatrix}
$$
Thus, **$A = \begin{bmatrix} -2 & 3/2 \\ 1 & -1/2 \end{bmatrix}$**.

***

#### Question 12: Prove AB is an Identity Matrix
**If $A = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix}$, prove that AB is an identity matrix.**

**Solution:**
Multiply the matrices A and B.
$$
\begin{aligned}
AB &= \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix} \\
&= \begin{bmatrix} (3)(3) + (4)(-2) & (3)(-4) + (4)(3) \\ (2)(3) + (3)(-2) & (2)(-4) + (3)(3) \end{bmatrix} \\
&= \begin{bmatrix} 9-8 & -12+12 \\ 6-6 & -8+9 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\end{aligned}
$$
Since the result is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, **AB is an identity matrix**.

***

#### Question 13: Find Unknowns in a Matrix Equation
**If $\begin{pmatrix} -2 & 0 \\ 0 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -2 \\ -10 \end{pmatrix}$, find the values of x and y.**

**Solution:**
Perform the matrix multiplication on the left side:
$$
\begin{pmatrix} (-2)(x) + (0)(y) \\ (0)(x) + (-5)(y) \end{pmatrix} = \begin{pmatrix} -2 \\ -10 \end{pmatrix} \\
\begin{pmatrix} -2x \\ -5y \end{pmatrix} = \begin{pmatrix} -2 \\ -10 \end{pmatrix}
$$
Equating the corresponding elements:
$$-2x = -2 \implies x = 1$$
$$-5y = -10 \implies y = 2$$
Thus, **$x=1$ and $y=2$**.

***

#### Question 14: Find Unknowns in a Matrix Equation
**If $\begin{pmatrix} x + 1 \\ y - 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, find the values of x and y.**

**Solution:**
First, perform the matrix multiplication on the right side:
$$
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} (1)(2) + (0)(3) \\ (0)(2) + (-1)(3) \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}
$$
Now, equate the matrices:
$$
\begin{pmatrix} x + 1 \\ y - 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}
$$
Equating corresponding elements:
$$x+1 = 2 \implies x=1$$
$$y-2 = -3 \implies y = -1$$
Thus, **$x=1$ and $y=-1$**.

***

#### Question 15: Find a Value from Determinant
**If the determinant of the matrix $M = \begin{bmatrix} -5 & -8 \\ p & 12 \end{bmatrix}$ is 20, find the value of p.**

**Solution:**
Given $|M|=20$.
$$
\begin{aligned}
|M| &= (-5)(12) - (-8)(p) \\
20 &= -60 + 8p \\
80 &= 8p \\
p &= 10
\end{aligned}
$$
Thus, **$p=10$**.

---

### Solving Systems of Linear Equations

***

#### Question 16: Solve using Cramer's Rule
**Solve: $2x - y = 3$ and $y + 3x = 7$.**

**Solution:**
First, arrange the equations in standard form:
$$
\begin{aligned}
2x - y &= 3 \\
3x + y &= 7
\end{aligned}
$$
Now, find the determinants D, $D_x$, and $D_y$.
$$
\begin{aligned}
D &= \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} = (2)(1) - (-1)(3) = 2 + 3 = 5 \\
D_x &= \begin{vmatrix} 3 & -1 \\ 7 & 1 \end{vmatrix} = (3)(1) - (-1)(7) = 3 + 7 = 10 \\
D_y &= \begin{vmatrix} 2 & 3 \\ 3 & 7 \end{vmatrix} = (2)(7) - (3)(3) = 14 - 9 = 5
\end{aligned}
$$
Now, solve for x and y:
$$
x = \frac{D_x}{D} = \frac{10}{5} = 2
$$
$$
y = \frac{D_y}{D} = \frac{5}{5} = 1
$$
The solution is **$x=2, y=1$**.

***

#### Question 17: Solve using Cramer's Rule
**Solve: $y = 2x$ and $x - \frac{3}{2}y + 1 = 0$.**

**Solution:**
Arrange the equations in standard form:
$$
\begin{aligned}
2x - y &= 0 \\
x - \frac{3}{2}y &= -1 \implies 2x - 3y = -2
\end{aligned}
$$
Find the determinants:
$$
\begin{aligned}
D &= \begin{vmatrix} 2 & -1 \\ 2 & -3 \end{vmatrix} = (2)(-3) - (-1)(2) = -6 + 2 = -4 \\
D_x &= \begin{vmatrix} 0 & -1 \\ -2 & -3 \end{vmatrix} = (0)(-3) - (-1)(-2) = 0 - 2 = -2 \\
D_y &= \begin{vmatrix} 2 & 0 \\ 2 & -2 \end{vmatrix} = (2)(-2) - (0)(2) = -4 - 0 = -4
\end{aligned}
$$
Solve for x and y:
$$
x = \frac{D_x}{D} = \frac{-2}{-4} = \frac{1}{2}
$$
$$
y = \frac{D_y}{D} = \frac{-4}{-4} = 1
$$
The solution is **$x = 1/2, y = 1$**.

***

#### Question 18: Solve using Cramer's Rule
**Solve: $\frac{4}{x} + \frac{5}{y} = 28$ and $\frac{7}{x} + \frac{3}{y} = 67$.**

**Solution:**
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$$
\begin{aligned}
4u + 5v &= 28 \\
7u + 3v &= 67
\end{aligned}
$$
Find the determinants D, $D_u$, and $D_v$.
$$
\begin{aligned}
D &= \begin{vmatrix} 4 & 5 \\ 7 & 3 \end{vmatrix} = (4)(3) - (5)(7) = 12 - 35 = -23 \\
D_u &= \begin{vmatrix} 28 & 5 \\ 67 & 3 \end{vmatrix} = (28)(3) - (5)(67) = 84 - 335 = -251 \\
D_v &= \begin{vmatrix} 4 & 28 \\ 7 & 67 \end{vmatrix} = (4)(67) - (28)(7) = 268 - 196 = 72
\end{aligned}
$$
Solve for u and v:
$$
u = \frac{D_u}{D} = \frac{-251}{-23} = 11 \quad \implies \frac{1}{x} = 11 \implies x = \frac{1}{11}
$$
$$
v = \frac{D_v}{D} = \frac{72}{-23} \quad \implies \frac{1}{y} = -\frac{72}{23} \implies y = -\frac{23}{72}
$$
The solution is **$x=\frac{1}{11}, y=-\frac{23}{72}$**.

***

#### Question 19: Solve using Cramer's Rule
**Solve: $3y + 4x = 2xy$ and $18y - 4x = 5xy$.**

**Solution:**
Divide both equations by $xy$ to get them in linear form.
Equation 1:
$$
\frac{3y}{xy} + \frac{4x}{xy} = 2 \implies \frac{3}{x} + \frac{4}{y} = 2
$$
Equation 2:
$$
\frac{18y}{xy} - \frac{4x}{xy} = 5 \implies \frac{18}{x} - \frac{4}{y} = 5
$$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The system is:
$$
\begin{aligned}
3u + 4v &= 2 \\
18u - 4v &= 5
\end{aligned}
$$
Find the determinants:
$$
\begin{aligned}
D &= \begin{vmatrix} 3 & 4 \\ 18 & -4 \end{vmatrix} = (3)(-4) - (4)(18) = -12 - 72 = -84 \\
D_u &= \begin{vmatrix} 2 & 4 \\ 5 & -4 \end{vmatrix} = (2)(-4) - (4)(5) = -8 - 20 = -28 \\
D_v &= \begin{vmatrix} 3 & 2 \\ 18 & 5 \end{vmatrix} = (3)(5) - (2)(18) = 15 - 36 = -21
\end{aligned}
$$
Solve for u and v:
$$
u = \frac{D_u}{D} = \frac{-28}{-84} = \frac{1}{3} \quad \implies \frac{1}{x} = \frac{1}{3} \implies x = 3
$$
$$
v = \frac{D_v}{D} = \frac{-21}{-84} = \frac{1}{4} \quad \implies \frac{1}{y} = \frac{1}{4} \implies y = 4
$$
The solution is **$x=3, y=4$**.

***

#### Question 20: Solve using Cramer's Rule
**Solve: $4(x - 1) + 5(y + 2) = 10$ and $5(x - 1) - 3(y + 2) + 6 = 0$.**

**Solution:**
First, simplify both equations.
Equation 1:
$$
\begin{aligned}
4x - 4 + 5y + 10 &= 10 \\
4x + 5y + 6 &= 10 \\
4x + 5y &= 4
\end{aligned}
$$
Equation 2:
$$
\begin{aligned}
5x - 5 - 3y - 6 + 6 &= 0 \\
5x - 3y - 5 &= 0 \\
5x - 3y &= 5
\end{aligned}
$$
Now find the determinants for the system: $4x+5y=4$ and $5x-3y=5$.
$$
\begin{aligned}
D &= \begin{vmatrix} 4 & 5 \\ 5 & -3 \end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\
D_x &= \begin{vmatrix} 4 & 5 \\ 5 & -3 \end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\
D_y &= \begin{vmatrix} 4 & 4 \\ 5 & 5 \end{vmatrix} = (4)(5) - (4)(5) = 20 - 20 = 0
\end{aligned}
$$
Solve for x and y:
$$
x = \frac{D_x}{D} = \frac{-37}{-37} = 1
$$
$$
y = \frac{D_y}{D} = \frac{0}{-37} = 0
$$
The solution is **$x=1, y=0$**.