Question

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?

Answer 1

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 `2xx2` matrix `[(4,k),(k,1)]` is `0`; when the determinant is `0`, the two lines are guaranteed parallel (but not necessarily overlapping).

The determinant of `[(4,k),(k,1)]` is

`|(4,k),(k,1)|=(4xx1)-(kxxk)`
`color(white)(|(4,k),(k,1)|)=4-k^2`

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

`4-k^2=0`
`color(white)(4-)k^2=4`
`color(white)(4-)k^color(white)(2)=+-2`

Now, we test both `k="+"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.