You are given three matrices,
$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$
$ B = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$
$ C = \begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}$
A very quick inspection gives $C = A - 3B$ so that $1A - 3B - 1C = 0$
Thus the three matrices are not linearly independent.
The vector space of symmetric 2 x 2 matrices has dimension 3, ie three linearly independent matrices are needed to form a basis.
The standard basis is defined by$ M = \begin{bmatrix} x & y \\ y & z \end{bmatrix} = x\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + y\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + z\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$
Clearly the given $A,B,C$ cannot be equivalent, having only two independent matrices.
So, why not?
Look for: Copying error? Did you mis-write or mis-type something?
Sign error?
Misunderstood question?
Is one of the other answers correct so this one then could be wrong?
And then after all other routes are exhausted, yes texts sometimes contain typos.
You can just start with the identity matrix and apply transformations that don't change the determinant:
- Adding to column (row) another column (row) multiplied by an integer.
- Performing an even permutation of the columns (rows).
Hart to tell what is degrees of freedom for a discrete set. Its dimension is zero.
But you can think of it a cutting all the $n^2$ dimensions that you had by one equation. So, $n^2-1$.
Best Answer
First we count the non-singular $2\times 2$ matrices. The first row can be any of the $8$ non-zero vectors. Then the second row can be anything but a multiple of the first row. There are $3$ such multiples. Thus there are $(8)(6)$ non-singular $2\times 2$ matrices.
Mutiplying a row by $2$ multiplies the determinant by $2$, giving a bijection between matrices with determinant $1$ and those with determinant $2$. So there are $24$ with determinant $1$.
Remark: The idea generalizes to larger matrices, and other finite fields.