Q 268

Question

Use Cramer’s Rule to Solve Systems of Equations.

In the following exercises, solve each system of equations using Cramer’s Rule.

$\{\begin{cases} x + y - 3 z = - 1 \\ y - z = 0 \\ - x + 2 y = 1 \end{cases}$

Step-by-Step Solution

Verified

Answer

The system is consistent and dependent and it has infinitely many solutions.

1Step 1 Given system of equations are,

$\{\begin{cases} x + y - 3 z = - 1 \\ y - z = 0 \\ - x + 2 y = 1 \end{cases}$

The objective is , we need to solve the system of equations by using the Cramer's rule.

2Step 2 Find the determinant D by using the coefficients of the variables.

$D = |\begin{matrix} 1 & 1 & - 3 \\ 0 & 1 & - 1 \\ - 1 & 2 & 0 \end{matrix}| = 1 (0 + 2) - 1 (0 - 1) - 3 (0 + 1) = 1 (2) - 1 (- 1) - 3 (1) = 2 + 1 - 3 = 0$

Here we cannot use the Cramer's rule to solve this system. But by looking at the determinants $D_{x}, D_{y}, D_{z}$ , we can determine whether the given system is inconsistent or dependent.

3Step 3 Evaluate the determinant D x by using the constants to replace the coefficients of x .

$D_{x} = |\begin{matrix} - 1 & 1 & - 3 \\ 0 & 1 & - 1 \\ 1 & 2 & 0 \end{matrix}| = - 1 (0 + 2) - 1 (0 + 1) - 3 (0 - 1) = - 1 (2) - 1 (1) - 3 (- 1) = - 2 - 1 + 3 = 0$

4Step 4 Evaluate the determinant D y by using the constants to replace the coefficients of y .

$D_{y} = |\begin{matrix} 1 & - 1 & - 3 \\ 0 & 0 & - 1 \\ - 1 & 1 & 0 \end{matrix}| = 1 (0 + 1) + 1 (0 - 1) - 3 (0 - 0) = 1 (1) + 1 (- 1) - 3 (0) = 1 - 1 - 0 = 0$

5Step 5 Evaluate the determinant D z by using the constants to replace the coefficients of z .

$D_{z} = |\begin{matrix} 1 & 1 & - 1 \\ 0 & 1 & 0 \\ - 1 & 2 & 1 \end{matrix}| = 1 (1 - 0) - 1 (0 - 0) - 1 (0 + 1) = 1 (1) - 1 (0) - 1 (1) = 1 - 0 - 1 = 0$

Here all the determinants are equal to zero.

So, the system is consistent and dependent and it has infinitely many solutions.

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