Solving a System of Equations using matrix algebra

[Genius]

Hi, I'm a student in eighth grade and have been studying matrices for the past few days. Unfortunately, I was caught off guard when while studying matrix algebra I came upon this.
This is an example of the instructions I got to solving a system of equations using matrix algebra

Step 1 Write the system as a matrix equation AX=B. The matrix A is the coefficient matrix, X is the matrix of variables and B is the matrix of constants.
2x-3y=19
x+4y=-7
[2 -3 [x [19
1 4] * y]= -7]

Step 2 Find the inverse of matrix A.
A^-1=1/8-(-3)[4 3=[(4/11) (3/11)
-1 2] (-1/11 (2/11)
Here's where I get confused
Step 3 Multiply each side of AX=B by A^-1 on the left to find the solution X=(A^-1)*B

My Solution
I've guessed that whoever wrote the book had a slip up and messed up and meant to write "Delete A from the equation and then multiply the left side of the equation by the inverse of A or A^-1 to get the solution x=(A^-1)*B.

I'm only the student so I don't know but I hope I was right because I've gambled an entire test on whether a syntax in the book was a mistake.

 

[skeeter]

A*X = B
A[sup:1zu27765]-1[/sup:1zu27765]*A*X = A[sup:1zu27765]-1[/sup:1zu27765]*B
I*X = A[sup:1zu27765]-1[/sup:1zu27765]*B
X = A[sup:1zu27765]-1[/sup:1zu27765]*B

matrix A is not deleted ... A[sup:1zu27765]-1[/sup:1zu27765]*A = I , the identity matrix.

go learn more about it ...

http://en.wikipedia.org/wiki/Identity_matrix

 

[soroban]

Hello, Genius!

Step 1
Write the system as a matrix equation: \(\displaystyle AX\:=\:B\)
\(\displaystyle A\) is the coefficient matrix, \(\displaystyle X\) is the matrix of variables, and \(\displaystyle B\) is the matrix of constants.

\(\displaystyle \begin{array}{ccc}2x-3y&=&19\\x+4y&=&\text{-}7 \end{array} \quad\Rightarrow\quad \begin{bmatrix}2& \text{-}3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} \;=\;\begin{bmatrix}19 \\ \text{-}7 \end{bmatrix}\)


Step 2
\(\displaystyle \text{Find the inverse of matrix A: }\;A^{-1} \:=\:\frac{1}{8-(\text{-}3)}\begin{bmatrix}4 & 3 \\ \text{-}1 & 2\end{bmatrix} \;=\;\begin{bmatrix}\frac{4}{11} & \frac{3}{11} \\ \\[-3mm] \text{-}\frac{1}{11} & \frac{2}{11} \end{bmatrix}\)


Here's where I get confused
Step 3
\(\displaystyle \text{Multiply each side of }AX\:=\:B\text{ by }A^{-1}\text{ (left-multiply) to find the solution: }X\:=\:A^{-1}\cdot B\)
. . This is correct . . .

The matrix \(\displaystyle A\) is not just "deleted" . . . There is a reason for its vanishing.

\(\displaystyle \text{We have: }\;AX \;=\;B\)

\(\displaystyle \text{Multiply both sides by }A^{-1}\!:\;\;A^{-1}\!\cdot\!(AX) \;=\;A^{-1}\!\cdot\!B\)

. . . . . . . . . . . . . . . . . . . .\(\displaystyle \underbrace{\left(A^{-1}\!\cdot\!A\right)}_{\text{This is }I}\!\cdot\!X \;=\;A^{-1}\!\cdot\!B\)
. . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \underbrace{I\!\cdot\!X} \;=\;A^{-1}\!\cdot\!B\)
. . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle X\;=\;\underbrace{A^{-1}\!\cdot\!B}_{\text{Solution}}\)


\(\displaystyle \text{Our solution is: }\;X \;=\;\begin{bmatrix}\frac{4}{11} & \frac{3}{11} \\ \\[-3mm] \text{-}\frac{1}{11} & \frac{2}{11} \end{bmatrix}\begin{bmatrix}19 \\ \text{-}7\end{bmatrix} \;= \;\begin{bmatrix}\frac{76}{11} - \frac{21}{11} \\ \\[-3mm] \text{-}\frac{19}{11} - \frac{14}{11}\end{bmatrix} \;=\; \begin{bmatrix}\frac{55}{11} \\ \\[-3mm] \text{-}\frac{33}{11}\end{bmatrix} \;=\;\begin{bmatrix}5 \\ \text{-}3 \end{bmatrix}\)