I don't even know where to start with this problem, I really don't understand it at all.
Could anyone help me please?
Problem:
Find, with explanation, all real numbers s such that the Matrix A - sI_{2} is NOT invertible
Thanks
An invertible 2x2 matrix has no inverse ie determinant is 0.
sI(subscript 2) is s times the 2x2 identity matrix, ie leading diagonal has elements s and others are 0.
So find A - sI, find an expression for the determinant, let it equal 0 and solve. (You should end up with a quadratic equation which will yield 2 solutions.) Give it a go.
It is sad that you would be working with matrices and not know that "a matrix is invertible if and only if its determinant is non-zero". So you are looking for s such that $\displaystyle \left|\begin{bmatrix} 1 & 2 \\ 3 & 2\end{bmatrix}- s\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\right|= \left |\begin{array}{cc}1- s & 2 \\ 3 & 2- a\end{array}\right|= (1- s)(2- s)- 6\ne 0$.
The matrix will NOT be invertible when $\displaystyle (1- s)(2- s)- 6= s^2- 3s- 4= 0$. Solve that quadratic equation.