Solving x+3y=4-kz & 4x-2y=5+10z: Find the Solution Here

[chwala]

Homework Statement

See attached

Relevant Equations

linear equations

Find the solution here;

I will need to check which approach they used...probably some row- reduced echelon somewhere... anyway my approach;

$x+3y=4-kz$
$4x-2y=5+10z$
.................................
$4x+12y=16-4kz$
$4x-2y=5+10z$
.................................
$14y=11-4kz-10z$
.................................
From third equation,
$x+y+2z=1$
$y=1-2z-x$
$y=1-2z-(4-kz-3y)$
$-2y=-3-2z+kz$ multiplying this by $7$
.................................
Therefore,
$14y=11-4kz-10z$
$-14y=-21-14z+7zk$
...................................
subtracting we get,
$10=3kz-24z$
$10=z(3k-24)$
.................................
For no unique solution, we set
$3k-24=0$
$k=8$

Probably you may inform me which approach they used...just the form...otherwise, i can check this up later...

Arrrrgh i have seen it...using determinant! cheers guys.

 

[chwala]

Using determinant, we shall have;
$1(-4+10)-3(8+10)+k(4+2)=0$
$6-54+6k=0$
$6k=48$
$k=8$ Bingo! learned this many years ago...going deep into my brain to remember...

 

[Orodruin]

Beware of the wording in this problem. Comparing the wording in your title ”has no solution” it has quite different meaning from the problem’s ”has no unique solution”.

In this case, there indeed are solutions. They are just not unique.

 

[chwala]

Beware of the wording in this problem. Comparing the wording in your title ”has no solution” it has quite different meaning from the problem’s ”has no unique solution”.

In this case, there indeed are solutions. They are just not unique.

Thanks...let me amend that...

 

[chwala]

We can also use the row-reduction method,

$\begin{bmatrix} 1 & 3 & k \\ 2 & -1 & -5 \\ 1 & 1 & 2 \end{bmatrix}$

$2R_1 -R_2$

$\begin{bmatrix} 1 & 3 & k \\ 0 & 7 & 2k+5 \\ 1 & 1 & 2 \end{bmatrix}$

$R_1-R_3$

$\begin{bmatrix} 1 & 3 & k \\ 0 & 7 & 2k+5 \\ 0 & 2 & k-2 \end{bmatrix}$

$2R_2-7R_3$

$\begin{bmatrix} 1 & 3 & k \\ 0 & 14 & 4k+10 \\ 0 & 0 & 7k-14 \end{bmatrix}$

$4k+10-(7k-14)=0$
$-3k+24=0$
$k=8$

Bingo!

 

[chwala]

Sorry allow me to ask...since ms are allowed to be followed 'strictly' to the latter...does it mean that any other approach/ method to solution wouldn't be accepted? like method shown in post $1$ and $5.$ Cheers.

 

[Mark44]

.since ms are allowed to be followed 'strictly' to the latter

"ms are allowed to be followed... " -- What are "ms"?
Also, the phrase is "to the letter."

.does it mean that any other approach/ method to solution wouldn't be accepted? like method shown in post 1 and 5. Cheers.

Method in post 1 is an approach that would be taken by someone with no knowledge of matrix reduction methods. Method in post 5 relies solely on the determinant to be able to discern whether the matrix of coefficients is invertible.

An alternate method that is better than that used in post 5 is to use row reduction on the augmented matrix. That is, the 3 x 4 matrix that contains a 4th column that has the constants on the right sides of the given three equations.

When I worked the problem, the third row was $\begin{bmatrix}0 & 0 & -24 + 3k & | &0 \end{bmatrix}$. This means that if k = 8, the bottom row consists solely of 0 elements, which further means that the three given equations are linearly dependent. Graphically, if k = 8, the three planes intersect at the least in a line in space.

On the other hand, if the bottom row had ended up with a nonzero constant in the 4th column, for example like this... $\begin{bmatrix}0 & 0 & -24 + 3k & | &1 \end{bmatrix}$, that would represent the equation 0x + 0y + 0z = 1 (again with k = 8), which has no solution. In this case the system of equations would have no solution at all; i.e., no single point lies on all three of the given planes.