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Δβα\Delta _{\beta \alpha}
01
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Linear algebra
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Δβα\Delta _{\beta \alpha}
04
Linear algebra
Starter
Attempt to solve this system using the Gaussian elimination method you learned from chapter 1:
{2x+3x+4z=15x+6y+7z=28x+9y+10z=4\left\{ \begin{array}{l} 2x+3x+4z=1 \\ 5x+6y+7z=2 \\ 8x+9y+10z=4 \end{array} \right.
What linear algebra is involved in AA HL?
Generally, you'll see matrix operations in AA HL when it comes to solving 3 equations in a system. Although this doesn't sound like the linear algebra you expected when you first clicked on this chapter, I promise you, Gaussian elimination becomes interesting when involving more than 2 equations!
A deeper dive into Gaussian elimination
From Chapter 1, you know that there's different forms of an augmented matrix that lets you find solutions to systems involving 2 equations, and those rules almost exactly apply to systems with 3, despite having that 3rd row and an additional variable:
Note N04.0a - Reduced row form "forms"
1 solution:
(1□□□01□□001□)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 1 & \square \\ \end{pmatrix}
0 solutions:
(1□□□01□□000□)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 0 & \square \\ \end{pmatrix}
Infinite solutions:
(1□□□01□□0000)\begin{pmatrix} 1 & \square & \square & \square \\ 0 & 1 & \square & \square \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}
Now, with three equations, we can find points of intersection in 3D space. When you solve these, the best way to achieve the zeros are as follows: get the zero in the center left first, then the bottom left, then the bottom middle.
Example E04.0a - Solving systems with Gaussian elimination
Consider the system
{x−2y+5z=12x−y+8z=2−3x−11z=−3\left\{ \begin{array}{l} x-2y+5z=1 \\ 2x-y+8z=2 \\ -3x-11z=-3 \end{array} \right.
.
As usual, transition the system into a matrix to get
(1−2512−182−30−11−3)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 2 & -1 & 8 & 2 \\ -3 & 0 & -11 & -3 \\ \end{pmatrix}
(note how, since there's no
yy
-variable for that last function, I just subbed in a 0)
Then perform the operation
R2=2R1−R2R_2=2R_1-R_2
to get that middle left zero:
(1−2510−320−30−11−3)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ -3 & 0 & -11 & -3 \\ \end{pmatrix}
Then perform the operation
R3=−3R1−R3R_3=-3R_1-R_3
to get that bottom left zero:
(1−2510−32006−40)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ 0 & 6 & -4 & 0 \\ \end{pmatrix}
Finally perform the operation
R3=2R2+R3R_3=2R_2+R_3
to get those bottom zeros:
(1−2510−3200000)\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & -3 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}
You can see however, we get infinite solutions (because the bottom row is all zeros). From here, we assign
z=tz=t
and sub that
tt
into the equations we extract from the matrices above. So,
−3y+2(t)=0-3y+2(t)=0
which re-arranges to produce
y=2t3y=\frac{2t}{3}
and then sub that and
tt
into
x−2(2t3)+5t=1x-2\left(\frac{2t}{3} \right)+5t=1
which re-arranges to get
x=1−11t3x=1-\frac{11t}{3}
.
These are therefore your solutions:
z=tz=t
,
y=2t3y=\frac{2t}{3}
, and
x=1−11t3x=1-\frac{11t}{3}
The example above, shows how you solve it when a) a variable is missing, and b) there are infinite solutions. All questions related to simultaneous equations will take any of these forms:
- Find the solutions of a system
- Find the value of a variable in a system (like
aa
) which makes the system have no, one, or infinite solutions
So, you should only really focus on the reduced row forms for this topic, and from there, you can find any answer you need! Below are some practice questions.
Coursework C04.0a - Solving systems using Gaussian elimination
1) Consider the system
{x−2y−z=−12x+y+3z=13x+8y+9z=a\left\{ \begin{array}{l} x-2y-z=-1 \\ 2x+y+3z=13 \\ x+8y+9z=a \end{array} \right.
. For what values of
aa
does the system have no solutions?
2) Consider the system
{x+2y+z=32x−y+4z=1x+7y−z=k\left\{ \begin{array}{l} x+2y+z=3 \\ 2x-y+4z=1 \\ x+7y-z=k \end{array} \right.
. For what values of
kk
does the system have no solutions? How about infinite solutions? Find the solutions in this case. Finally, explain why the system never has a unique solution.
3) Consider the system
{x+4y+mz=−m(m+1)x+4y+z=14x+4y+z=1\left\{ \begin{array}{l} x+4y+mz=-m \\ (m+1)x+4y+z=1 \\ 4x+4y+z=1 \end{array} \right.
. Show that for one value of
mm
, the system has no solutions. Show that for a different value of
mm
, the system has infinitely many solutions and find the solutions in this case.
4) Consider the system
{x+4y−z=k−2x+z=35x+4y+(k−6)z=−3\left\{ \begin{array}{l} x+4y-z=k \\ -2x+z=3 \\ 5x+4y+(k-6)z=-3 \end{array} \right.
. Show that for one value of
kk
, the system has infinitely many solutions and find the solutions in this case. Show that there is a unique solution for all other values of
kk
. Write this solution in terms of
kk
.
That is the extent to which the IB includes linear algebra in their course, hence why this page is also quite short!