Linear algebra: system of linear equations?

Question

determine the value(s) of k such that the given system of linear equation has exactly one solution:

4x + ky = 7
kx + y = 0

i know the answer is

all k is not equal to +/-2

but how do you reach that answer???

Answer 1

4x + ky = 7
kx + y = 0

Let's solve this by substitution. From the second equation, since kx + y = 0, then y = -kx, so plugging this into the first equation,

4x + k(-kx) = 7

Solving this,

4x - x(k^2) = 7
x(4 - k^2) = 7
x = 7/(4 - k^2)

Note that x is undefined if the denominator is equal to 0; that is,

4 - k^2 = 0

Solving this gives us
k^2 = 4
k = +/- 2

Which means k cannot equal +/- 2.

Answer 2

The matrix version of the problem is

AX = b

where A is a 2x2 matrix
[4, k; k, 1]

X is a vector variable
[x; y]

and b is a vector constant
[7; 0]

There will be a unique solution as long as A is non-singular.

There will not be a unique solution if A *is* singular.

For A to be singular
det(A) = 0

where det() is the "determinant" of a square matrix

for a 2x2 matrix [a, b; c, d]
det([a, b; c, d]) = ad - bc

Hence in your case

4*1 - k*k = 0
k^2 = 4
k = +/- 2

Answer 3

Since you're in linear algebra, I assume you want this in terms of matrices.

Set the matrix:

A = [4    k]
      [k     1]

and you want to have a consistent system of equations, meaning det(A) not 0. The only time that doesn't happen is when det(A)=0, and so you find:

det(A) = 4 - k²

set det(A)=0

4 - k² = 0

k = ± 2

Those are the only times det(A)=0, thus the only times your set of equations is inconsistent.

Answer 4

First, treat it like 2 equations you are trying to solve. I'm going to multiply the second equation by -k and then add to the first equation:

-k^2x + 4x = 7
from which I get
(4-k^2)*x = 7

The only case for which I can't solve this is for k = 2 or k = -2
because (4-k^2) would become zero, and 0*x cannot = 7.

Answer 5

eqn. 1: 4x+ky=7
eqn. 2: kx+y=0

from eqn. 1, solve for y: y=(7-4x)/k

plug this into eqn. 2: kx +(7-4x)/k=0

after some algebra: (k^2)x+7-4x=0 => (k^2-4)x+7=0, or x=-7/(k^2-4)

which is real if and only if (k^2-4) is not equal to zero. Thus, k^2 cannot be equal to 4.

=>k is not equal to +/-2.

Answer 6

4x + ky = 7
-k²x - ky = 0----ADD

(4 - k²).x = 7
x = 7 / (4 - k²)

28 / (4 - k²) + ky = 7
ky = 7 - 28 / (4 - k²)
y = (7 / k) - 28 / k.(4 - k²)
y = [7.(4 - k²) - 28 ] / [ k.(4 - k²) ]

x and y will have solutions for k ≠ 2

Answer 7

You can not have k = +/-2 since if this was so you will have

4x +2y = 7
2x +y = 0

or

4x -2y = 7
-2x + y = 0

These two sets of equations correspond to two parallel lines which of course do not intersect.

Answer 8

in (2) relation:
x=-y/k
replace x in (1) relation:
4(-y/k)+ky-7=0
-4y+k^2y-7k=0
yk^2-7k-4y=0
det=(-7)^2-4y(-4y)
det=49+16y^2
we got double square if:
det =0
16y^2+49=0 or 16y^2=-49 imp. → R
There's no solution for k.

Answer 9

(4x+ky)/(kx+y) <> any integer
4x+ky divided by kx +y is an integer only when k = +/- 2.