Question

Apply the Gaussian elimination process to determine values of λ for which the
following linear system is consistent:
x − 3y + 4 = 0, 3x − 2y = λ, y = 6 − 2x .

Answer 1

Answer on Question #78495 – Math – Linear Algebra

Question

Apply the Gaussian elimination process to determine values of $\lambda$ for which the following linear system is consistent:

$$\begin{array}{l} x - 3y + 4 = 0 \\ 3x - 2y = \lambda \\ y = 6 - 2x \\ \end{array}$$

Solution

$$\begin{array}{l} x - 3y + 4 = 0 \\ 3x - 2y = \lambda \\ y = 6 - 2x \\ \end{array}$$ $$\begin{array}{l} x - 3y = -4 \\ 3x - 2y = \lambda \\ 2x + y = 6 \\ \end{array}$$

Augmented matrix

$$\begin{pmatrix} 1 & -3 & -4 \\ 3 & -2 & \lambda \\ 2 & 1 & 6 \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 & -3 & -4 \\ 3 & -2 & \lambda \\ 2 & 1 & 6 \\ \end{pmatrix} \xrightarrow{R_2 - (3)R_1} \begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & \lambda + 12 \\ 2 & 1 & 6 \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & \lambda + 12 \\ 2 & 1 & 6 \\ \end{pmatrix} \xrightarrow{R_3 - (2)R_1} \begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & \lambda + 12 \\ 0 & 7 & 14 \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & \lambda + 12 \\ 0 & 7 & 14 \\ \end{pmatrix} \xrightarrow{R_3 - R_2} \begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & \lambda + 12 \\ 0 & 0 & 2 - \lambda \\ \end{pmatrix}$$

The following linear system is consistent if

$$2 - \lambda = 0 \Rightarrow \lambda = 2$$

Then

$$\begin{pmatrix} 1 & -3 & -4 \\ 0 & 7 & 14 \\ 0 & 0 & 0 \\ \end{pmatrix} \xrightarrow{R_2/7} \begin{pmatrix} 1 & -3 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ $$\begin{pmatrix} 1 & -3 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{pmatrix} \xrightarrow{R_1 + (3)R_2} \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

The solution is $x = 2, y = 2$.

Answer: the system is consistent if $\lambda = 2$, then $x = y = 2$.

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