I have an exam in two days and I'm practicing some problems from past exam papers, hence, no solutions to refer to. So could someone please please check if this solution is accurate? Thank you for your time in advance!
- Given is the matrix M
\begin{matrix} 2 & 4 & 2 \\ 3 & 1 & -7 \\ 2 & 2 & -2 \\ \end{matrix}
(a) Is M is regular or singular?
So essentially, we need to find out the determinant of M and check if it's equal to 0.
$$2 \begin{bmatrix} 1 & -7 \\ 2 & -2 \\ \end{bmatrix} + (-4)\begin{bmatrix} 3 & -7 \\ 2 & -2 \\ \end{bmatrix} + 2 \begin{bmatrix} 3 & 1 \\ 2 & 2 \\ \end{bmatrix} = 0 $$
In conclusion, M is singular since the determinant of M is 0, i.e. the matrix is not invertable.
(b) Determine matrix X satisfying the matrix equation $X* N=M{^2}+2X$ , where matrix M is given as above and N is
\begin{matrix} 2 & 1 & 0 \\ -2 & 0 & 1 \\ 1 & 1 & 1 \\ \end{matrix}
$X* N=M{^2}+2X $ => $X*N-2X=M{^2} $ => $X*(N-2I)=M{^2} $ => $X=M{^2}*(N-2I){^{-1}} $ Did I accurately order the matrices here?
I calculated $M{^2}$ being equal to:
\begin{matrix} 20 & 16 & -28 \\ -5 & -1 & 13 \\ 6 & 6 & 6 \\ \end{matrix}
and $(N-I){^{-1}}$ is: \begin{matrix} -5 & 1 & 3 \\ 1 & 0 & 0 \\ 4 & -1 & -2 \\ \end{matrix}
And from here it follows that, $M{^2}(N-I){^{-1}}=X$ is equal to:
\begin{matrix} 196 & 48 & 116 \\ 76 & -18 & -41 \\ 0 & 0 & 6 \\ \end{matrix}