Question

Consider the equations

$\left|\begin{array}{c}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{z}\mathbf{=}\mathbf{4}\\ \mathbf{x}\mathbf{+}\mathbf{ky}\mathbf{+}\mathbf{4}\mathbf{z}\mathbf{=}\mathbf{6}\\ \mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{z}\mathbf{=}\mathbf{6}\end{array}\right|$

where$\mathit{k}$is an arbitrary constant.

a. For which values of the constant does this system have a unique solution?

b. When is there no solution?

c. When are there infinitely many solutions?

Short answer

(a) The system of equations will have a unique solution for the value, $k\ne 1,2$.

(b)The system of equations will have no solution for the value,$k=1$ .

(c)The system of equations will have infinitely many solutions for the value, .$k=2$

Step-by-step solution

Transform the equations into a matrix form

In the system of equations, the variables can be eliminated by performing arithmetic operations on the equations.

Represent the system of equations in the form of matrix.

$\left|\begin{array}{ccccc}1& 2& 3& |& 4\\ 1& k& 4& |& 6\\ 1& 2& k+2& |& 6\end{array}\right|$

Perform the row operations

Subtract second row from the first row and third row from the first row.

$\left|\begin{array}{ccccc}1& 2& 3& |& 4\\ 0& k-2& 1& |& 2\\ 0& 0& k-1& |& 2\end{array}\right|$

Solve the equations

From second and third rows, it’s clear that, for,$k\ne 1,2$ the system will have unique solution, as, if $k=1$in the third row, then ,$0=2$ which is not true.

If,$k=2$ then, this implies system will have infinitely many solutions, as two of three rows become identical.

Final answer

(a) The system of equations will have a unique solution for the value, $k\ne 1,2$.

(b)The system of equations will have no solution for the value, $k=1$.

(c)The system of equations will have infinitely many solutions for the value, $k=2$.