Answer to Question #173821 in Algebra for isaac

Question

For what value(s) of k, if any, will the system have no solution, a unique solution, and infinitely many solutions? (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

x − 2y + 8z = 6

x + y + z = k

2x − y + 9z = k²

Answer 1

Consider the system of linear equations

x-2y+8z=6

x +y +z=k

2x-y+9z=k²

The augmented matrix for the above linear system:

A = $\begin{bmatrix} 1 & -2 & 8 | & 6 \\ 1 & 1 & 1 | & k \\ 2 & -1& 9 | & k^2 \end{bmatrix}$

The row reduction process as follows:

$R_2\implies R_2-R_1$

A = $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 2 & -1& 9&| & k^2 \end{bmatrix}$

$R_3\implies R_3-2R_1$

A = $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 3& -7&| & k^2-12 \end{bmatrix}$

$R_3\implies R_3-R_2$

A = $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$ ........ (1)

The system has no solution if the the rank of the matrix $\begin{bmatrix} 1 & -2 & 8 \\ 0 & 3 & -7 \\ 0 & 0& 0 \end{bmatrix}$ is not equal to rank of matrix $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$

Let $k^2-k-6\ne 0$

$(k+2)(k-3)\ne0$

$k\ne-2,3$

Thus, the rank of matrix $\begin{bmatrix} 1 & -2 & 8 \\ 0 & 3 & -7 \\ 0 & 0& 0 \end{bmatrix}$ is 2.

Thus, the rank of matrix $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$ is 3.

Hence the system has no solution if $k\ne-2,3$

The system has unique solution if the rank of matrix $\begin{bmatrix} 1 & -2 & 8 \\ 0 & 3 & -7 \\ 0 & 0& 0 \end{bmatrix}$ is equal to rank of matrix $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$ $=3$

This is not possible.

So, the system has no unique solution.

The system has infinitely many solutions if the rank of matrix $\begin{bmatrix} 1 & -2 & 8 \\ 0 & 3 & -7 \\ 0 & 0& 0 \end{bmatrix}$ is equal to rank of matrix $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$ less than 3.

Let $k^2-k-6=0$

$(k+2)(k-3)=0$

$k=-2,3$

Thus, the rank of matrix $\begin{bmatrix} 1 & -2 & 8 \\ 0 & 3 & -7 \\ 0 & 0& 0 \end{bmatrix}$ is 2.

Thus, the rank of matrix $\begin{bmatrix} 1 & -2 & 8&| & 6 \\ 0 & 3 & -7&| & k-6 \\ 0 & 0& 0&| & k^2-k-6 \end{bmatrix}$ is 2.

Hence, the system has infinitely many solutions if $k=-2,3$ .