For the matrix `A=[(3, 1), (7, 5)]` , find `x` and `y` so that `A^2+x I=y Adot`

Answer

`x=y=8, A^(-1)=[(5//8,-1//8),(-7//8,3//8)]`

Explanation

We have
`A=[(3,1),(7,5)]`
`A^(2)=A A=[(3,1),(7,5)][(3,1),(7,5)]`
`=[(9+7,3+5),(21+35,7+25)]=[(16,8),(56,32)]`
Now, `A^(2)+xI=yA`
`implies [(16,8),(56,32)]+x[(1,0),(0,1)]=y[(3,1),(7,5)]`
`[(16+x,8+0),(56+0,32+x)]=[(3y,y),(7y,5y)]`
`16+x=3y, y=8, 7y=56, 5y=32+x`
Putting `y=8` in `16+x=3y`, we get `x=24-16=8`. Clearly, `x=8` and `y=8` also satify `7y=56` and `5y=32+x`. Hence,`x=8` and `y=8`. We have
`|A|=|(3,1),(7,5)|=8 ne 0`
So, A is invertible.
Putting `x=8, y=8` in `A^(2)+ xI=yA`, we get
`A^(2)+8I=8A`
`implies A^(-1) (A^(2)+8I)=8A^(-1) A`
`A^(-1) A^(2)+8A^(-1)I=8A^(-1) A`
`A+8A^(-1)=8I`
`[ :' A^(-1) A^(2)=(A^(-1) A)A=IA=A, A^(-1)I=A^(-1) and A^(-1) A=I]`
`8A^(-1)=8I-A`
or `A^(-1)=1/8 (8I-A)=1/8 {[(8,0),(0,8)]-[(3,1),(7,5)]}`
`=1/8 [(8-3,0-1),(0-7,8-5)]=1/8 [(5,-1),(-7,3)]=[(5//8,-1//8),(-7//8,3//8)]`