Finding b12 entry of a 2×2 complex matrix inverse

[Razberryz]

Homework Statement

Consider the following matrix.

A =

2 + 4i...1 + 5i

2 − 3i...2 + 3i

Let B = A^-1. Find b₁₂ (i.e., find the entry in row 1, column 2 of A⁻¹)

Homework Equations

A^-1 = 1/(ad - cb)*
[ d -b ]
[ -c a ]
<--imagine as 2x2 matrix with first row (d,-b) and second row (-c,a)

The Attempt at a Solution

1/(ad - cb) =

1/((2+4i)(2+3i) - (2-3i)(1+5i))

1/((4+14i-12) - (2+7i+15))

1/(4+14i-12-2-7i-15)

1/(-25+7i)

[(d,-b)(-c,a)] =

[ d -b ]
[ -c a ] =

2 + 3i...-1 - 5i

-2 + 3i...2 + 4i

So for first row, second column, we should be multiplying 1/(-25+7i) by -1 - 5i. I'm getting -10/576 + 12i/576, but my answer is wrong. Where is my mistake?

 

[Razberryz]

Homework Statement

Consider the following matrix.

A =

2 + 4i...1 + 5i

2 − 3i...2 + 3i

Let B = A^-1. Find b₁₂ (i.e., find the entry in row 1, column 2 of A⁻¹)

Homework Equations

A^-1 = 1/(ad - cb)*
[ d -b ]
[ -c a ]
<--imagine as 2x2 matrix with first row (d,-b) and second row (-c,a)

The Attempt at a Solution

1/(ad - cb) =

1/((2+4i)(2+3i) - (2-3i)(1+5i))

1/((4+14i-12) - (2+7i+15))

1/(4+14i-12-2-7i-15)

1/(-25+7i)

[(d,-b)(-c,a)] =

[ d -b ]
[ -c a ] =

2 + 3i...-1 - 5i

-2 + 3i...2 + 4i

So for first row, second column, we should be multiplying 1/(-25+7i) by -1 - 5i. I'm getting -10/576 + 12i/576, but my answer is wrong. Where is my mistake?

I just realized my mistake, it was arithmetic.