# How do you determine the value(s) of k such that the system of linear equations has the indicated number of solutions: An infinite number of solutionsfor 4x + ky = 6 and kx +y = -3?

Sep 12, 2017

The system is dependent (has infinite solutions) when k=–2.

#### Explanation:

Express the equations in extended matrix form:

[(4,k,|,6),(k,1,|,–3)]

This system of linear equations can have infinite solutions only if the determinant of the $2 \times 2$ matrix $\left[\begin{matrix}4 & k \\ k & 1\end{matrix}\right]$ is $0$; when the determinant is $0$, the two lines are guaranteed parallel (but not necessarily overlapping).

The determinant of $\left[\begin{matrix}4 & k \\ k & 1\end{matrix}\right]$ is

$| \left(4 , k\right) , \left(k , 1\right) | = \left(4 \times 1\right) - \left(k \times k\right)$
$\textcolor{w h i t e}{| \left(4 , k\right) , \left(k , 1\right) |} = 4 - {k}^{2}$

We then set this equal to zero, since that's what we want:

$4 - {k}^{2} = 0$
$\textcolor{w h i t e}{4 -} {k}^{2} = 4$
$\textcolor{w h i t e}{4 -} {k}^{\textcolor{w h i t e}{2}} = \pm 2$

Now, we test both $k = \text{+} 2$ and k=–2 to see if each makes the original system inconsistent (no solution) or dependent (infinite solutions).

Try $k = 2$:

[(4,color(red)2,|,6),(color(red)2,1,|,–3)]=>"       "color(white)[(color(black)(1/2R_1)),(3)][(2,1,|,3),(2,1,|,–3)]

"                            "=>color(white)[(),(color(black)(R_2-R_1))][(2,1,|,3),(0,0,|,–6)]

The last row in the matrix says 0x+0y=–6, which simplifies to 0=–6. Since this is a contradiction, $k = 2$ makes the system inconsistent.

Try k=–2:

[(4,color(red)(–2),|,6),(color(red)(–2),1,|,–3)]=>"    "color(white)[(color(black)(1/2R_1)),(3)][(2,–1,|,3),(–2,1,|,–3)]

"                                  "=>color(white)[(),(color(black)(R_2+R_1))][(2,–1,|,3),(0,0,|,0)]

Since the last line of the matrix is all zeroes, k=–2 makes the system dependent, and so this is the value of $k$ we seek.