I have got the following problem where I is idenitity matrix, S is variable and A is a matrix
[math]I= \left[ \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \right] ,A= \left[ \begin{matrix} -5 & -1\\ 3 & -1 \end{matrix} \right][/math]For SI-A, I got:
[math]\left[ \begin{matrix} S & 0\\ 0 & S \end{matrix} \right] - \left[ \begin{matrix} -5 & -1\\ 3 & -1 \end{matrix} \right] = \left[ \begin{matrix} S+5 & 1\\ -3 & S+1 \end{matrix} \right][/math]However, in the sheet my professor gave me, her solution for SI-A was different, it was:
[math]\left[ \begin{matrix} S+1 & 1\\ -3 & S+1 \end{matrix} \right][/math]
I will attach the sheet (sorry for the bad handwriting)
Note this is not asking for solving a homework (as you can see, it's a solved example), I am just asking if she used a different way or something.
[math]I= \left[ \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \right] ,A= \left[ \begin{matrix} -5 & -1\\ 3 & -1 \end{matrix} \right][/math]For SI-A, I got:
[math]\left[ \begin{matrix} S & 0\\ 0 & S \end{matrix} \right] - \left[ \begin{matrix} -5 & -1\\ 3 & -1 \end{matrix} \right] = \left[ \begin{matrix} S+5 & 1\\ -3 & S+1 \end{matrix} \right][/math]However, in the sheet my professor gave me, her solution for SI-A was different, it was:
[math]\left[ \begin{matrix} S+1 & 1\\ -3 & S+1 \end{matrix} \right][/math]
I will attach the sheet (sorry for the bad handwriting)
Note this is not asking for solving a homework (as you can see, it's a solved example), I am just asking if she used a different way or something.
