so I was doing this question about matrix, everything was fine until here, the question goes like this.
$Let\:M=\begin{bmatrix}A & B \\0 & C \end{bmatrix} \text{ where}\:A,\:B,\:C\:\text{is}\:a\:2\:\text{by}\:2\:\text{matrix}.$
and i need to find $M^{-1}$ which is pretty straight forward because if M is a usual matrix where A, B, C is just some real numbers, $M^{-1}$ would be $\begin{bmatrix}1/A & -B/AC \\0 & 1/C \end{bmatrix}$ but in the case of A, B, C being a 2×2 matrix I'd have to express it in form of matrix multiplication.
At first I thought it would be $\begin{bmatrix}A^{-1} & -BA^{-1}C^{-1} \\0 & C^{-1} \end{bmatrix}$
but the solution suggests it to be $\begin{bmatrix}A^{-1} & -A^{-1}BC^{-1} \\0 & C^{-1} \end{bmatrix}$
Here I don't understand this, what decides the placement of $-A^{-1}$, $B$ and $C^{-1}$?
Many thanks.
Best Answer
Matrix multiplication is not commutative. When you're checking what produces the top right element of the product between $M$ and $M^-1$, see that with your answer the first term would be $A (-BA^{-1}C^{-1})$ while with their answer it would be $A (-A^{-1}BC^{-1})$. Notice that only the latter produces $-BC^{-1}$, which is the opposite of the second term ($BC^{-1}$) which produces the required zero matrix on the top right block of $MM^{-1} = I$.