(d) Explain Cramer's rule for solution of linear equation and solve the system of linear equations using Cramer's rule. $$ \begin{array}{c}x+2 y=4 \ 3 x-y=5\end{array} $$

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Cramer's Rule is a powerful method in linear algebra used to solve systems of linear equations with as many equations as unknowns, provided that the system has a unique solution. It leverages determinants of matrices constructed from the coefficients of the variables in the system.

### **Cramer's Rule Explained**

Consider a system of two linear equations:

$$
\begin{cases}
a_1x + b_1y = c_1 \quad \text{(1)} \
a_2x + b_2y = c_2 \quad \text{(2)}
\end{cases}
$$

**Cramer's Rule** states that the solution for the variables $$x$$ and $$y$$ can be found using determinants as follows:

$$
x = \frac{D_x}{D} \quad \text{and} \quad y = \frac{D_y}{D}
$$

where:

- $$D$$ is the **determinant** of the coefficient matrix:

$$
D =
\begin{vmatrix}
a_1 & b_1 \
a_2 & b_2
\end{vmatrix}
= a_1b_2 - a_2b_1
$$

- $$D_x$$ is the determinant obtained by replacing the $$x$$-coefficients with the constants from the right-hand side of the equations:

$$
D_x =
\begin{vmatrix}
c_1 & b_1 \
c_2 & b_2
\end{vmatrix}
= c_1b_2 - c_2b_1
$$

- $$D_y$$ is the determinant obtained by replacing the $$y$$-coefficients with the constants:

$$
D_y =
\begin{vmatrix}
a_1 & c_1 \
a_2 & c_2
\end{vmatrix}
= a_1c_2 - a_2c_1
$$

**Important Note:** Cramer's Rule can only be applied if $$D \neq 0$$, ensuring that the system has a unique solution.

### **Applying Cramer's Rule to the Given System**

Let's solve the provided system of equations:

$$
\begin{cases}
x + 2y = 4 \quad \text{(1)} \
3x - y = 5 \quad \text{(2)}
\end{cases}
$$

**Step 1: Identify Coefficients**

From the system, we can identify:

$$
a_1 = 1, \quad b_1 = 2, \quad c_1 = 4 \
a_2 = 3, \quad b_2 = -1, \quad c_2 = 5
$$

**Step 2: Compute Determinant $$D$$**

$$
D =
\begin{vmatrix}
1 & 2 \
3 & -1
\end{vmatrix}
= (1)(-1) - (3)(2) = -1 - 6 = -7
$$

Since $$D = -7 \neq 0$$, the system has a unique solution.

**Step 3: Compute Determinant $$D_x$$**

Replace the $$x$$-coefficients with the constants $$c_1$$ and $$c_2$$:

$$
D_x =
\begin{vmatrix}
4 & 2 \
5 & -1
\end{vmatrix}
= (4)(-1) - (5)(2) = -4 - 10 = -14
$$

**Step 4: Compute Determinant $$D_y$$**

Replace the $$y$$-coefficients with the constants $$c_1$$ and $$c_2$$:

$$
D_y =
\begin{vmatrix}
1 & 4 \
3 & 5
\end{vmatrix}
= (1)(5) - (3)(4) = 5 - 12 = -7
$$

**Step 5: Solve for $$x$$ and $$y$$**

$$
x = \frac{D_x}{D} = \frac{-14}{-7} = 2 \
y = \frac{D_y}{D} = \frac{-7}{-7} = 1
$$

### **Solution**

The system of equations has a unique solution:

$$
x = 2 \quad \text{and} \quad y = 1
$$

### **Verification**

To ensure the solution is correct, plug the values back into the original equations:

1. **First Equation:**
   $$
   x + 2y = 2 + 2(1) = 2 + 2 = 4 \quad \checkmark
   $$

2. **Second Equation:**
   $$
   3x - y = 3(2) - 1 = 6 - 1 = 5 \quad \checkmark
   $$

Both equations are satisfied, confirming that $$x = 2$$ and $$y = 1$$ is indeed the correct solution.
To solve the system of equations using Cramer's Rule: $$ \begin{cases} x + 2y = 4 \ 3x - y = 5 \end{cases} $$ 1. **Compute Determinant $$D$$:** $$ D = (1)(-1) - (3)(2) = -7 $$ Since $$D \neq 0$$, a unique solution exists. 2. **Compute $$D_x$$:** $$ D_x = (4)(-1) - (5)(2) = -14 $$ $$ x = \frac{D_x}{D} = \frac{-14}{-7} = 2 $$ 3. **Compute $$D_y$$:** $$ D_y = (1)(5) - (3)(4) = -7 $$ $$ y = \frac{D_y}{D} = \frac{-7}{-7} = 1 $$ **Solution:** $$ x = 2, \quad y = 1 $$ **Verification:** Plugging the values back into the original equations confirms the solution is correct.

(d) Explain Cramer’s rule for solution of linear
equation and solve the system of linear
equations using Cramer’s rule.

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(d) Explain Cramer’s rule for solution of linear equation and solve the system of linear equations using Cramer’s rule.