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?

1 Answer
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 #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.