Matrices Part3

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System of Linear Equations

System of linear equations can be solved using matrix method.

Note that :

If the question IS a matrix question, we MUST use matrix method.
If it ISN'T a matrix question, using the normal simultaneous equations is much easier most of the time.

Let's see how it is done. First, we see that

$\begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} k \\ l \end{pmatrix}$

will mean

$ax+by=k\,$ and
$cx+dy=l\,$

In other words, the matrix equation represent the two linear equations. Conversely, any system of linear equations can be transformed into a matrix equation. We will then solve the matrix equation to obtain the solution to the system of linear equations.

Writing the Matrix Equation

If needed, rewrite the equations to the "standard form" (refer below) before writing the matrix equation to minimize careless mistakes.

$\begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{pmatrix} {\color{Red}x} \\ {\color{Blue}y} \end{pmatrix}=\begin{pmatrix} k \\ l \end{pmatrix}$ $\iff$ $\begin{align} a{\color{Red}x}+b{\color{Blue}y} &= k \\ c{\color{Red}x}+d{\color{Blue}y} &= l \end{align}$
$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\begin{pmatrix} {\color{Red}x} \\ {\color{Blue}y} \\ {\color{Green}z} \end{pmatrix}=\begin{pmatrix} j \\ k \\ l \end{pmatrix}$ $\iff$ $\begin{align} a{\color{Red}x}+b{\color{Blue}y}+c{\color{Green}z} &= j \\ d{\color{Red}x}+e{\color{Blue}y}+f{\color{Green}z} &= k \\ g{\color{Red}x}+h{\color{Blue}y}+i{\color{Green}z} &= l \\ \end{align}$

Examples

Rewrite the following system of linear equations into a matrix equation

$\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 6 \\ 1 \end{pmatrix}$
- Analyze :
  - $\begin{pmatrix} \quad & \quad \\ \quad & \quad \end{pmatrix}\begin{pmatrix} {\color{Red}x} \\ {\color{Red}y} \end{pmatrix}=\begin{pmatrix} \quad \\ \quad \end{pmatrix}$
  - $\begin{pmatrix} {\color{Red}2} & {\color{Red}3} \\ {\color{Red}1} & {\color{Red}1} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} \quad \\ \quad \end{pmatrix}$
  - $\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} {\color{Red}6} \\ {\color{Red}1} \end{pmatrix}$

$\begin{pmatrix} 1 & -3 \\ -2 & -1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} -7 \\ 5 \end{pmatrix}$

$\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -1 \\ 2 \end{pmatrix}$ or $\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -1 \\ -2 \end{pmatrix}$
- Analyze
  - $x+y+1 = 0\,$ needs to be changed to $x+y = -1 \,$
  - $\begin{align} x+y &=-1 \\ x-y & =2 \end{align}$ or $\begin{align} x+y &=-1 \\ -x+y &=-2 \end{align}$

$\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 0 \\ 7 \end{pmatrix}$
- $\begin{align} 2x-y &=0 \\ -x+y &=7 \end{align}$

$\begin{pmatrix} 1 & 2 & -1 \\ 3 & -2 & 4 \\ 2 & 5 & -7\end{pmatrix}\begin{pmatrix} x \\ y \\z \end{pmatrix}=\begin{pmatrix} 7 \\ -8 \\ 1 \end{pmatrix}$

$\begin{pmatrix} 1 & -2 & 0 \\ 2 & 0 & 3 \\ 0 & 5 & 7\end{pmatrix}\begin{pmatrix} x \\ y \\z \end{pmatrix}=\begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$
- Analyze
  - $\begin{array}{llll} x & -2y & & = -1 \\ 2x& & +3z & = 2 \\ & 5y & -7z & = 3 \\ \end{array}$ or $\begin{array}{llll} x & -2y & +0z & = -1 \\ 2x& +0y & +3z & = 2 \\ 0x& +5y & -7z & = 3 \\ \end{array}$

Solving

It's important to understand the steps

Check your answers!

Take note of the last two steps!

Example

Solve the following using matrix method

Important Note :

$\begin{align} 2x -3y &=-13 \\ x +5y &= 13 \end{align}$
$x=-2, y=3\,$
Rewrite as matrix equation. Don't need rearranging.
- $\begin{pmatrix} 2 & -3 \\ 1 & 5 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -13 \\ 13 \end{pmatrix}$
- Let $\mathbf{A}=\begin{pmatrix} 2 & -3 \\ 1 & 5 \end{pmatrix}$
- $\therefore \mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -13 \\ 13 \end{pmatrix}$
- $\mathbf{A}^{-1}\mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix}=\mathbf{A}^{-1}\begin{pmatrix} -13 \\ 13 \end{pmatrix}$
- $\mathbf{I}\begin{pmatrix} x \\ y \end{pmatrix}=\frac{1}{10-\left(-3\right)}\begin{pmatrix} 5 & 3 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} -13 \\ 13 \end{pmatrix}$
- $\begin{pmatrix} x \\ y \end{pmatrix}=\frac{1}{13}\begin{pmatrix} -26 \\ 39 \end{pmatrix}$
- $=\begin{pmatrix} -2 \\ 3 \end{pmatrix}$
- Done! Nope
- $\therefore x=-2, y=3$
- Done! Nope
- Check. $2\left(-2\right)-3\left(3\right)=-13, \left(-2\right)+5\left(3\right)=13$ . No problem here.

$\begin{align} x +y +z &=1 \\ 2x -y +3z &=-1 \\ -x+2y-z &=-1 \end{align}$
$x=4,y=0,z=-3\,$
Similar to above. Just that need more steps to find inverse.
- $\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 2 & -1 \end{pmatrix}\begin{pmatrix}x \\ y \\ x\end{pmatrix}=\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}$
- Let $\mathbf{A}=\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 2 & -1 \end{pmatrix}$
- $\begin{align} &.\\ &.\\ &.\\ \mathbf{A}^{-1}&=-\frac{1}{3}\begin{pmatrix} -5 & 3 & 4 \\ -1 & 0 & -1 \\ 3 & -3 & -3 \end{pmatrix}\end{align}$
- $\mathbf{A}\begin{pmatrix} x \\ y \\z \end{pmatrix}=\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$
- $\mathbf{A}^{-1}\mathbf{A}\begin{pmatrix} x \\ y \\z \end{pmatrix}=\mathbf{A}^{-1}\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$
- $\mathbf{I}\begin{pmatrix} x \\ y \\z \end{pmatrix}=-\frac{1}{3}\begin{pmatrix} -5 & 3 & 4 \\ -1 & 0 & -1 \\ 3 & -3 & -3 \end{pmatrix}\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$
- $\begin{pmatrix} x \\ y \\z \end{pmatrix}=-\frac{1}{3}\begin{pmatrix} -12 \\ 0 \\ 9 \end{pmatrix}$
- $=\begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix}$
- $x=4,y=0,z=-3\,$

Using Inverses Already Found

Often, we are required to use the inverse found from earlier parts of the question to solve the system of linear equations (Or we will be asked to find/prove a matrix equation, from which we can find the inverse).

Sometimes though, it seems the matrix we get from the system of linear equations is not the same as the earlier given matrix. In this case, we must rearrange/rewrite the equations until it can give us the same matrix.

To do that, we can

Important Note :

Example

$\mathbf{P}=\begin{pmatrix} 1 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 2 & -3 \end{pmatrix} \mathbf{Q}=\begin{pmatrix} 2 & 3 & -1 \\ 6 & -3 & 1 \\ 4 & -2 & -2 \end{pmatrix}$ . Find $\mathbf{PQ}$ . Hence, find $\mathbf{P}^{-1}$ & $\mathbf{Q}^{-1}$ . Solve the following system of linear equations

$\begin{align} 2x+3y-z &=8 \\ 6x-3y+z &=0 \\ 4x-2y-2z &=2 \\ \end{align}$

$\begin{array}{l} x+y=5 \\ 2x=z \\ 2y-3z =-12 \\ \end{array}$

$\begin{array}{l} 2x+2y =8 \\ 2x-z =0 \\ 4y =6z-8 \\ \end{array}$

$\begin{array}{l} 2x=y+z \\ 2x+3y=z \\ 6x+z=3y\\ \end{array}$

Solving Problems

Exercise 4

1) Using matrix method, solve the following

a) $\begin{align} x + 2y &= -7 \\ 3x &= y \end{align}$ $x=-1,y=-3\,$
- $\begin{align} x + 2y &= -7 \\ 3x - y & = 0 \end{align}$
- $\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} -7 \\ 0 \end{pmatrix}$
- Let $\mathbf{A}=\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$
- $\begin{align} \mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix}&=\begin{pmatrix} -7 \\ 0 \end{pmatrix}\\ \mathbf{A}^{-1}\mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix}&=\mathbf{A}^{-1}\begin{pmatrix} -7 \\ 0 \end{pmatrix}\\ \mathbf{I}\begin{pmatrix} x \\ y \end{pmatrix}&=\frac{1}{-1-6}\begin{pmatrix} -1 & -2 \\ -3 & 1 \end{pmatrix}\begin{pmatrix} -7 \\ 0 \end{pmatrix}\\ \begin{pmatrix} x \\ y \end{pmatrix}&=-\frac{1}{7}\begin{pmatrix} 7 \\ 21 \end{pmatrix}\\ &=\begin{pmatrix} -1 \\ -3 \end{pmatrix} \end{align}$
- $\therefore x=-1, y=-3$

b) $\begin{align} 2x -y +z &= 8 \\ x +3y -4z &= 13 \\ 3x + y &=5 \end{align}$ $x=2,y=-1,z=3 \,$
- $\begin{array}{llll} 2x &-y &+z &= 8 \\ x &+3y &-4z &= 13 \\ 3x &+y & &=5 \end{array}$
- $\begin{pmatrix} 2 & -1 & 1 \\ 1 & 3 & -4 \\3 & 1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\z \end{pmatrix}=\begin{pmatrix} 8 \\ -13 \\ 5 \end{pmatrix}$
- Let $\mathbf{A}=\begin{pmatrix} 2 & -1 & 1 \\ 1 & 3 & -4 \\3 & 1 & 0 \end{pmatrix}$
- $\begin{align}\left|\mathbf{A}\right| & =2\begin{vmatrix} 3 & -4 \\ 1 & 0 \\ \end{vmatrix}-\left(-1\right)\begin{vmatrix} 1 & -4 \\ 3 & 0 \\ \end{vmatrix}+1\begin{vmatrix} 1 & 3 \\ 3 & 1 \\ \end{vmatrix}\\ & = 2\left(4\right)+12-8\\ & = 12 \end{align}$
- $\mathbf{A}^T= \begin{pmatrix} 2 & 1 & 3\\ -1 & 3 & 1 \\ 1 & -4 & 0 \end{pmatrix}$
- $\begin{align} adj \mathbf{A}&=\begin{pmatrix} \begin{vmatrix} 3 & 1 \\ -4 & 0 \\ \end{vmatrix} & -\begin{vmatrix} -1 & 1 \\ 1 & 0 \\ \end{vmatrix} & \begin{vmatrix} -1 & 3 \\ 1 & -4 \\ \end{vmatrix} \\ -\begin{vmatrix} 1 & 3 \\ -4 & 0 \\ \end{vmatrix} & \begin{vmatrix} 2 & 3 \\ 1 & 0 \\ \end{vmatrix} & -\begin{vmatrix} 2 & 1 \\ 1 & -4 \\ \end{vmatrix} \\ \begin{vmatrix} 1 & 3 \\ 3 & 1 \\ \end{vmatrix} & -\begin{vmatrix} 2 & 3 \\ -1 & 1 \\ \end{vmatrix} & \begin{vmatrix} 2 & 1 \\ -1 & 3 \\ \end{vmatrix} \end{pmatrix}\\ &=\begin{pmatrix} 4 & 1 & 1\\ -12 & -3 & 9 \\ -8 & -5 & 7 \end{pmatrix} \end{align}$
- $\begin{align} \mathbf{A}^{-1} & = \frac{1}{\left|\mathbf{A}\right|}adj \mathbf{A}\\ & = \frac{1}{12}\begin{pmatrix} 4 & 1 & 1\\ -12 & -3 & 9 \\ -8 & -5 & 7 \end{pmatrix} \end{align}$
- $\begin{align} \therefore \mathbf{A}\begin{pmatrix} x \\ y \\z \end{pmatrix}&=\begin{pmatrix} 8 \\ -13 \\ 5 \end{pmatrix}\\ \mathbf{A}^{-1}\mathbf{A}\begin{pmatrix} x \\ y \\z \end{pmatrix}&= \mathbf{A}^{-1}\begin{pmatrix} 8 \\ -13 \\ 5 \end{pmatrix}\\ \mathbf{I}\begin{pmatrix} x \\ y \\z \end{pmatrix}&=\frac{1}{\left|\mathbf{A}\right|}adj \mathbf{A}\\ & = \frac{1}{12}\begin{pmatrix} 4 & 1 & 1\\ -12 & -3 & 9 \\ -8 & -5 & 7 \end{pmatrix} \begin{pmatrix} 8 \\ -13 \\ 5 \end{pmatrix}\\ & = \frac{1}{12}\begin{pmatrix} 24 \\ -12 \\ 36 \end{pmatrix}\\ & = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}\\ \end{align}$
- $\therefore x=2,y=-1,z=3$

2) $\mathbf{P}=\begin{pmatrix} 2 & 1 & -1 \\ 0 & -2 & 3 \\ -2 & 2 & -2 \end{pmatrix} \mathbf{Q}=\begin{pmatrix} 2 & 0 & -1 \\ 6 & 6 & 6 \\ 4 & 6 & 4 \end{pmatrix}$ . Find $\mathbf{PQ}$ . Hence, find $\mathbf{P}^{-1}$ & $\mathbf{Q}^{-1}$ . Solve the following system of linear equations