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## Main Question or Discussion Point

Can anyone explain to me how to count the total # of non-invertible 2x2 matrices?

I have the answer from the book, which is r^3+r^2-r provided r is a prime. But it doesn't explain how to get there, and I couldn't figure it out. I haven't been practicing linear algebra for quite a long time.

I know a 2x2 matrix is non-invertible iff the determinant is zero, and a matrix is non-invertible iff rows (columns) are linearly dependent, one row (column) must be a scalar multiple of the other.

If we have a 2x2 matrix A:

a b

c d

A is non-invertible iff det(A) = ad-bc = 0

If I have a set of number r that is prime to choose from for each entry in the matrix. It's easier to play with the zeros, but then I got very confused when it combines with the numbers.

Can anyone be so graciously helping me out and explain it in detail? Many thanks!

I have the answer from the book, which is r^3+r^2-r provided r is a prime. But it doesn't explain how to get there, and I couldn't figure it out. I haven't been practicing linear algebra for quite a long time.

I know a 2x2 matrix is non-invertible iff the determinant is zero, and a matrix is non-invertible iff rows (columns) are linearly dependent, one row (column) must be a scalar multiple of the other.

If we have a 2x2 matrix A:

a b

c d

A is non-invertible iff det(A) = ad-bc = 0

If I have a set of number r that is prime to choose from for each entry in the matrix. It's easier to play with the zeros, but then I got very confused when it combines with the numbers.

Can anyone be so graciously helping me out and explain it in detail? Many thanks!