# 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

The system is dependent (has infinite solutions) when

#### 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

The determinant of

#|(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 **inconsistent** (no solution) or **dependent** (infinite solutions).

Try

#[(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

Try

#[(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,