Question

How do you solve `x+3y=11` and `x+4y=14` using matrices?

Answer 1

x=2; y=3

Explanation:

you must solve by applying the basic propositions:

`x=D_x/D; y=D_y/D`

where D is the determinant of the matrix

`|quad1quad3|`
`|quad1 quad 4|`

Its determinant is obtained by:

`D=1*4-3*1=1`

`D_x and D_y` are respectively the determinants of the matrices

`|quad11quad3|`
`|quad14quad4|`

`|quad1quad11|`
`|quad1quad14|`

`D_x=11*4-3*14=44-42=2`

`D_y=1*14-11*1=3`

So

`x=D_x/D=2/1=2`

`y=D_y/D=3/1=3`

Answer 2

`x=2" ; "y=3`

Two methods given

Explanation:

`color(blue)("Using Gauss Jordon elimination")`

`" "color(white)(.)xcolor(white)(.)ycolor(white)(.)"answer"`
`" "[[1,3,"|",11],[1,4,"|",14]]`
`" "Row2-Row1`
`" "darr`

`" "[[1,3,"|",11],[0,1,"|",3]]`
`" "Row1-3(Row2)`
`" "darr`

`" "[[1,0,"|",2],[0,1,"|",3]]" "larr [[x],[y]]`
`color(white)(.)`

`=>[[1,0],[0,1]] color(white)(.)[[x],[y]] = [[2],[3]]`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Using Linear Algebra")`

`A=[[1,3],[1,4]]`
`color(white)(.)`

`X=[[x],[y]]`

`B=[[11],[14]]`

`=>AX=B`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Consider the Algebra method of determining "x)`
Given:`" "3x=4`

Divide both side by 3 `-> 1/3xx3x=1/3xx4`

We multiply the 3 by 3 inverse to turn the coefficient of `x` into 1

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Using the Algebra method shown above")`

Multiply both sides by `A" inverse. written as "color(blue)(A^(-))`

`color(brown)(=>color(blue)(A^(-))AX" "=" "color(blue)(A^(-))B)`

`=>X=A^(-)B` ...................................Equation(1)

Checked in Maple:

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine "A^(-))`

`color(brown)("Method 1")`

Write the original matrix as:

`" "[[1,3,"|",1,0],[1,4,"|",0,1]]`

And conduct a Gauss Jordon elimination on the LHS

`" "Row2-Row1`
`" "darr`

`" "[[1,3,"|",1,0],[0,1,"|",-1,1]]`
`" "Row1-3(Row2)`
`" "darr`

`" "[[1,0,"|",4,-3],[0,1,"|",-1,1]] => A^(-)=[[4,-3],[-1,1]]`
,.....................................................................................
`color(brown)("Method 2 - Shortcut approach")`

Multiply a modified version of `[[1,3],[1,4]]` by its inverted determinant.

For `[[a,b],[c,d]]` the determinant is `ab-cd`
and the modified matrix is `[[d,-b],[-c,a]]`

`D=(1xx4)-(1xx3)=1`

Thus `A^(-)= 1xx[[4,-3],[-1,1]]`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Finish the calculation for Equation(1)")`
`=>X=A^(-)B` ...................................Equation(1)

`[[4,-3],[-1,1]] [[11],[14]] = [[2],[3]]`