Question

How do you solve `x+3y=5` and `x+4y=6` using matrices?

Answer 1

From `vecR = A * vecx` multiplying both sides by `A^-1`
`A^-1*vecR = A^-1A*vecx = I * vecx`
where `A =((a_11,a_12), (a_21,a_22 )) =((1,3), (1,4 ))`
`vecx = ((x), (y)), vecR = ((5), (6))`
`A^-1 = ((4,-3 ), (-1, 1 ))`

Answere `=>((x), (y)) = ((2), (1))`

Explanation:

Write,
`x+3y=5`
`x+4y=6`
`((1,3), (1,4 )) ((x), (y)) = ((5), (6))`
Let `A =((a_11,a_12), (a_21,a_22 )) =((1,3), (1,4 )), vecx = ((x), (y)), vecR = ((5), (6))`

"we need to find "
`A^-1` such that `I = A^-1 A " where " I = ((1,0 ), (0,1 ))` and we can write in vector matrix form:

`vecR = A * vecx` multiplying both sides by `A^-1`
`A^-1*vecR = A^-1A*vecx = I * vecx`
in our case:
`A^-1((5), (6)) = ((1,0 ), (0,1 )) ((x), (y))`
and the solution to `x, y`
`x = 5*a'_(11) + 6*a'_(12)`
`y = 5*a'_(21) + 6*a'_(22)`

but how do you find:
`A^-1 = ((a'_11, a'_12), (a'_21, a'_22))`?
`A^-1 = 1/det(A) * ((a_22, -a_12 ), (-a_21, a_11 ))` where
`det(A) = (a_11a_22 - a_12a_21) = 4-3= 1`
Thus `A^-1 = ((4,-3 ), (-1, 1 ))`
`((x), (y)) =((4,-3 ), (-1, 1 )) ((5), (6)) =((2), (1))`