Linear Algebra - REF with another variable

[Schaus]

Homework Statement

Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values.
x₁−2x₂+2x₃ = −1
−3x₁+ax₂−10x₃ = 5
x₁+x₂−4x₃ = 3

Homework Equations

Augmented Matrix

The Attempt at a Solution

1 -2 2 -1
-3 a -10 5
1 1 -4 3
I switched the row with the variable to the bottom.
1 -2 2 -1
1 1 -4 3
-3 a -10 5
I minused Row 1 from Row 2
1 -2 2 -1
0 3 -6 5
-3 a -10 5
3Row 1 + Row 3
1 -2 2 -1
0 3 -6 5
0 (a-6) -4 2
Here is where I get confused. I don't know how to get the 3 in row 2 to equal 1 and get the (a-6) to equal 0.
Any advice as to where I went wrong or what I need to do next would be greatly appreciated!

 

[MarkFL]

Let's begin with the augmented matrix:

$\left[\begin{array}{ccc|c}1 & -2 & 2 & -1 \\ -3 & a & -10 & 5 \\ 1 & 1 & -4 & 3 \end{array}\right]$

You switched the 2nd and 3rd rows:

$\left[\begin{array}{ccc|c}1 & -2 & 2 & -1 \\1 & 1 & -4 & 3 \\ -3 & a & -10 & 5\end{array}\right]$

Next, you performed $R_2-R_1$:

$\left[\begin{array}{ccc|c}1 & -2 & 2 & -1 \\0 & 3 & -6 & 4 \\ -3 & a & -10 & 5\end{array}\right]$

Note: I have corrected an arithmetic error. Next, you performed $3R_1+R_3$:

$\left[\begin{array}{ccc|c}1 & -2 & 2 & -1 \\0 & 3 & -6 & 4 \\ 0 & a-6 & -4 & 2\end{array}\right]$

What can we multiply the second row by to get the 3 to be a 1?

 

[FactChecker]

Here is where I get confused. I don't know how to get the 3 in row 2 to equal 1

Divide that row by 3.

 

[NFuller]

Use the determinant! When $\text{det}[\mathbf{A}]=0$ there will be no unique solution.

 

[MarkFL]

Let's take FactChecker's advice and perform $\dfrac{1}{3}R_2$:

$\left[\begin{array}{ccc|c}1 & -2 & 2 & -1 \\0 & 1 & -2 & \frac{4}{3} \\ 0 & a-6 & -4 & 2\end{array}\right]$

Now, obviously, we need to perform $2R_2+R_1$:

$\left[\begin{array}{ccc|c}1 & 0 & -2 & \frac{5}{3} \\0 & 1 & -2 & \frac{4}{3} \\ 0 & a-6 & -4 & 2\end{array}\right]$

Now, we need to perform $x\cdot R_2+R_3$ such that:

$x\cdot1+(a-6)=0$

What is $x$, and what do you get after performing the operation?

 

[Schaus]

So I take (a+6)R₂-R₃?

1 0 -2 ----------- 5/3
0 1 -2 ------------ 4/3
0 0 (-2a-16) -------- ((4/3)a+10)

Sorry for the slow reply, I've been very busy with school.

 

[Mark44]

So I take (a+6)R₂-R₃?

1 0 -2 ----------- 5/3
0 1 -2 ------------ 4/3
0 0 (-2a-16) -------- ((4/3)a+10)

No, that won't work. In column 2, (a + 6) * 1 - (a - 6) $\neq$ 0.

Sorry for the slow reply, I've been very busy with school.

 

[Schaus]

So I should have multiplied by -(a-6)?
-(a-6)R2+R3

1 0 -2 ----------- 5/3
0 1 -2 ------------ 4/3
0 0 (2a-16) -------- ((4/3)a-6)

 

[Mark44]

So I should have multiplied by -(a-6)?
-(a-6)R2+R3

1 0 -2 ----------- 5/3
0 1 -2 ------------ 4/3
0 0 (2a-16) -------- ((4/3)a-6)

Yes, assuming that your work leading up to this last matrix is correct (I didn't check).
Now, for what value(s) of a will there be a unique solution, no solution, or an infinite number of solutions?

 

[Schaus]

2a-16 = ((4/3)a-6)
a=15
No solution = Never
Unique solution ---- a≠15
Infinitely many solutions--- a=15?

 

[Mark44]

2a-16 = ((4/3)a-6)
a=15
No solution = Never
Unique solution ---- a≠15
Infinitely many solutions--- a=15?

No. Where did 15 come from?

 

[Mark44]

A bit more of a hint. Suppose that each of the following was the bottom row of your augmented matrix:
1) 0 0 0 | 5
2) 0 0 0 | 0
3) 0 0 1 | 2
and that the 2nd and 3rd rows had leading entries of 1.

For which of the above 3rd rows would we expect a) a unique solution, b) no solution, c) an infinite number of solutions?

 

[Schaus]

1) No solution
2) Infinitely many solutions
3) One unique solution

I got the a=15 by making 2a-16 = (4/3a-6)

 

[Mark44]

1) No solution
2) Infinitely many solutions
3) One unique solution

Yes to all of the above.

I got the a=15 by making 2a-16 = (4/3a-6)

That seems like a reasonable thing to do, but it's not. You're not solving the equation 2a - 16 = (4/3)a - 6. Instead, you're trying to make the coefficient of z (what I'm calling the 3rd variable) equal to 1.

In other words, the bottom row of your matrix corresponds to the equation (2a - 16)z = (4/3)a - 6. What do you need to do to turn this into 1z = <something>? In particular, are there any values of a that will cause problems?