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For what value(s) of k, \begin{cases}x-2y+5z=2 \\ x+y+z=k \\ 2x-y+6z=k^2 \end{cases} if any, will...

Question:

For what value(s) of {eq}k, {/eq} {eq}\begin{cases}x-2y+5z=2 \\ x+y+z=k \\ 2x-y+6z=k^2 \end{cases} {/eq}, if any, will the system have:

a) Infinitely many solutions,

b) A unique solution, and

c) No solution.

Explain

System of Equations


A system of linear equations can have no solution, a unique solution or infinitely many solutions with different number of free variables.

To solve a system of equation, we perform Gaussian elimination to eliminate the unknowns from the equations in such a manner that the matrix of the system is a staircase type matrix or the form is a row echelon form.

A system is consistent if the last column of the augmented matrix is a non- pivot column.

A row echelon form with at least-one non-pivot column in the coefficient matrix, has infinitely many solutions, if the system is consistent.

The number of non-pivot columns gives the number of free parameters in the (infinite) solution.

Answer and Explanation:


To determine the values of {eq}\displaystyle k {/eq} such that the system of equations {eq}\displaystyle \begin{cases} x-2y+5z=2 \\ x+y+z=k \\ 2x-y+6z=k^2 \end{cases} {/eq}

has no solution, unique solution or infinitely many solutions, we will obtain a row echelon form of the augmented matrix and analyze the pivot and non-pivot columns.

The augmented form of the matrix, {eq}\displaystyle \left[\begin{array}{ccc|c} 1&-2&5&2\\ 1&1&1&k\\ 2&-1&6&k^2 \end{array}\right] {/eq}

has a row echelon form, calculated as below.

{eq}\displaystyle \begin{align} &\left[\begin{array}{ccc|c} 1&-2&5&2\\ 1&1&1&k\\ 2&-1&6&k^2 \end{array}\right]\overset{-R_1+R_2}{\implies}\left[\begin{array}{ccc|c} 1&-2&5&2\\ 0&3&-4&k-2\\ 2&-1&6&k^2 \end{array}\right]\overset{-2R_1+R_3}{\implies}\left[\begin{array}{ccc|c} 1&-2&5&2\\ 0&3&-4&k-2\\ 0&3&-4&k^2-4 \end{array}\right]\overset{-R_2+R_3}{\implies}\left[\begin{array}{ccc|c} \boxed{1}&-2&5&2\\ 0&\boxed{3}&-4&k-2\\ 0&0&0&k^2-k-2 \end{array}\right]. \end{align} {/eq}


a) For the system above to have infinitely many solutions, we need to have a consistent system (the last column of the augmented matrix to be a non-pivot column)

and to have at least one non-pivot column in the coefficient matrix.

The above row echelon form of the augmented matrix is consistent if the last column is a non-pivot column, so {eq}\displaystyle k^2-k-2=0. {/eq}

Because there is a non-pivot column in the coefficient matrix, the third column, then there is a free parameter so, there are infinitely many solutions,

therefore {eq}\displaystyle \boxed{\text{ the system has infinitely many solutions if }k=2 \text{ or } k=-1}. {/eq}


b) To have a unique solution, we need to have all three columns of the coefficient matrix, pivot columns so, because we have only two pivot columns, {eq}\displaystyle \boxed{\text{ there is no } k \text{ such that the system to has a unique solution }}. {/eq}


c) The system has no solution, when the augmented matrix is inconsistent, or the last column is a pivot column, so when {eq}\displaystyle k^2-k-2\neq 0, {/eq}

therefore, {eq}\displaystyle \boxed{\text{ the system has no solution when } k\in\mathbb{R}\setminus \{2,-1\}}. {/eq}


Learn more about this topic:

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Consistent System of Equations: Definition & Examples

from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 8
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