 # Cramer's Rule (Example Problem)

### Solve the following system of equations using Cramer's Rule:

$$V_1 - 5V_2 - V_3 = 0$$ $$3V_1 - 9V_2 + 8V_3 = 18$$ $$2V_1 - 7V_2 + 4V_3 = 0$$

### Step 1: Create the augmented matrix of coefficients:

$$\begin{pmatrix} 1&-5&-1&0\\ 3&-9&8&18\\ 2&-7&4&0\\ \end{pmatrix}$$

### Step 2: Create the matrix of coefficients:

$$\begin{pmatrix} 1&-5&-1\\ 3&-9&8\\ 2&-7&4\\ \end{pmatrix}$$

### Step 3: Calculate the determinant of the matrix of coefficients:

$$D = \begin{vmatrix} 1&-5&-1\\ 3&-9&8\\ 2&-7&4\\ \end{vmatrix}$$ $$= 1 \begin{vmatrix} -9&8\\ -7&4\\ \end{vmatrix} - (-5) \begin{vmatrix} 3&8\\ 2&4\\ \end{vmatrix} + (-1) \begin{vmatrix} 3&-9\\ 2&-7\\ \end{vmatrix}$$ $$D = 1\Big(4(-9)-8(-7)\Big) + 5\Big( 4(3)-8(2) \Big) - 1\Big( 3(-7)-2(-9) \Big)$$ $$= 1(20) + 5(-4) - 1(-3)$$

$$D = 3$$

### Step 4: Create a matrix that consists of the matrix of coefficients with its 1st column replaced by the 4th column of the augmented matrix of coefficients

...and proceed to calculate the determinant of this matrix: $$D_{V1} = \begin{vmatrix} 0&-5&-1\\ 18&-9&8\\ 0&-7&4\\ \end{vmatrix}$$ $$D_{V1} = 0\Big(4(-9)-8(-7)\Big) + 5\Big( 4(3)-8(2) \Big) - 1\Big( 3(-7)-2(-9) \Big)$$ $$= 0 - (-5)\Big[4(18)-0\Big] + (-1)\Big[ 18(-7)-0 \Big]$$

$$D_{V1} = 486$$

### Step 5: Create a matrix that consists of the matrix of coefficients with its 2nd column replaced by the 4th column of the augmented matrix of coefficients

...and proceed to calculate the determinant of this matrix: $$D_{V2} = \begin{vmatrix} 1&0&-1\\ 3&18&8\\ 2&0&4\\ \end{vmatrix}$$ $$D_{V2} = 1\Big(4(18)-0)\Big) - 0 + (-1)\Big( 0-18(2) \Big)$$

$$D_{V2} = 108$$

### Step 6: Create a matrix that consists of the matrix of coefficients with its 3rd column replaced by the 4th column of the augmented matrix of coefficients

...and proceed to calculate the determinant of this matrix: $$D_{V3} = \begin{vmatrix} 1&-5&0\\ 3&-9&18\\ 2&-7&0\\ \end{vmatrix}$$ $$D_{V3} = 1\Big( 0-18(-7) \Big) - (-5)\Big( 0-18(2) \Big) + 0$$

$$D_{V3} = -54$$

### Step 7: Use Cramer's Rule to solve the system of equations:

From the previous page, we know that:

$$V_1 = \frac{D_{V1}}{D} = \frac{486}{3}$$ $$V_1 = 162$$

$$V_2 = \frac{D_{V2}}{D} = \frac{108}{3}$$ $$V_2 = 36$$

$$V_3 = \frac{D_{V3}}{D} = \frac{-54}{3}$$ $$V_3 = -18$$