Answer to Question #326553 in Linear Algebra for Arrdly

Question

Use the Gauss-Jordan process to determine for which value (s) of λ will the following system have no solutions?

$\begin{bmatrix} 1 & 2 & -3 & 4\\ 3 & -1 & 5 & 2\\ 4 & 1 & λ2 -14 & λ +2 \end{bmatrix}$

Answer 1

In the process of solving the system by the Gauss method, we must obtain at least one equation, where there will be zeros in front of all the x, and on the right the number is not equal to 0:

$\begin{bmatrix} 1&2&-3&4 \\ 3& -1&5&2\\ 4&1&2\lambda-14&\lambda+2 \end{bmatrix}$

Subtract from the second line the first multiplied by 3, and from the third - the first multiplied by 4:

$\begin{bmatrix} 1&2&-3&4 \\ 0& -7&14&-10\\ 0&-7&2\lambda-2&\lambda-14 \end{bmatrix}$

Subtract the second from the third line and divide the second by -7:

$\begin{bmatrix} 1&2&-3&4 \\ 0& 1&-2&\frac{10}7\\ 0&0&2\lambda-16&\lambda-4 \end{bmatrix}$

The system will not have solutions if:

$\begin{cases} 2\lambda-16=0\\ \lambda-4\not =0 \end{cases} \implies\begin{cases} \lambda=8\\ \lambda \not =4 \end{cases}$

Checking:

$\begin{bmatrix} 1&2&-3&4 \\ 0& 1&-2&\frac{10}7\\ 0&0&0&4 \end{bmatrix}$

We get: $0x_1 + 0x_2 + 0x_3 = 4 \to$ the system has no solutions.

Answer: $\lambda=8$