Calculate inverse of matrix

Question

If $$A=\begin{bmatrix} -5 & 1 & 0 & 0\\ -19 & 4 & 0 & 0\\ 0 & 0 & 1 & 2\\ 0 & 0 & 3 & 5\\ \end{bmatrix}, $$ how do I calculate $A^{-1}$? Is there any simple method other than reducing it to reduced row echelon form? By using cofactors and determinant?

Answer 1

This is a block-diagonal matrix with $2\times 2$ blocks: $\,\begin{bmatrix}A&0\\0&B\end{bmatrix}$. It is invertible if and only if $A,B$ are invertible, and in such a case the inverse is $\,\begin{bmatrix}A^{-1} &0\\0&B^{-1}\end{bmatrix}$.

This being said, Gauß's method gives: $$A^{-1}=\begin{bmatrix}-4 & 1\\-19&5\end{bmatrix},\quad B^{-1}=\begin{bmatrix}-5 & 2\\3&-1\end{bmatrix}.$$

Answer 2

Reducing it to row-echelon form should be quite simple since you will only need to do two row-operations. You will find that $A^{-1}$ is of the same structure as $A$, that the non-zero entries are in the top left $2\times 2$ square and the bottom right $2\times 2$ square.

In general, if you have a matrix in block-diagonal form you only need to find the inverse for each of the blocks and assemble the inverses to form the inverse of the whole matrix.

Answer 3

1. General method: You add an identity matrix and then transform your $A$ into an identity matrix, while simultaneously $I$ gets transformed into $A^{-1}$. $$ \left(A | I\right) \to \cdots \to \left(I | A^{-1} \right) $$

$$ \left( \begin{array}{cccc|cccc} -5 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ -19 & 4 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 5 & 0 & 0 & 0 & 1 \end{array} \right) \to \\ \left( \begin{array}{cccc|cccc} 1 & -1/5 & 0 & 0 & -1/5 & 0 & 0 & 0 \\ 0 & 1/5 & 0 & 0 & -19/5 & 5 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 1 & 0 \\ 0 & 0 & 3 & 5 & 0 & 0 & 0 & 1 \end{array} \right) \to \\ \left( \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & -4 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -19 & 5 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & -3 & 1 \end{array} \right) \to \\ \left( \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & -4 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & -19 & 5 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 3 & -1 \end{array} \right) $$

2. Special case block diagonal: Using this special case of $$ \left( \begin{matrix} A & 0 \\ 0 & B \end{matrix} \right) $$ we can apply this algorithm to the matrix blocks $$ \left( \begin{array}{cc|cc} -5 & 1 & 1 & 0 \\ -19 & 4 & 0 & 1 \\ \end{array} \right) \to \\ \left( \begin{array}{cc|cc} 1 & -1/5 & -1/5 & 0 \\ 0 & 1/5 & -19/5 & 1 \\ \end{array} \right) \to \\ \left( \begin{array}{cc|cc} 1 & 0 & -4 & 1 \\ 0 & 1 & -19 & 5 \\ \end{array} \right) $$ and similar for the second block $B$.

3. Special case $2\times 2$ matrix: Using this even more special case of $2\times 2$ blocks, one could use $$ A = \left( \begin{matrix} \alpha & \beta \\ \gamma & \delta \end{matrix} \right) \Rightarrow A^{-1} = \frac{1}{\alpha\delta - \beta \gamma} \left( \begin{matrix} \delta & -\beta \\ -\gamma & \alpha \end{matrix} \right) $$ This uses cofactors and determinant, by the way.

For example $$ B = \left( \begin{matrix} 1 & 2 \\ 3 & 5 \end{matrix} \right) \Rightarrow B^{-1} = \frac{1}{-1} \left( \begin{matrix} 5 & -2 \\ -3 & 1 \end{matrix} \right) = \left( \begin{matrix} -5 & 2 \\ 3 & -1 \end{matrix} \right) $$

Answer 4

Hint If $B,C,D,E$ are two by two matrices then

$$\begin{bmatrix} B & 0 \\ 0 & C\end{bmatrix}\begin{bmatrix} D& 0 \\ 0 & E\end{bmatrix}=\begin{bmatrix} BD & 0 \\ 0 & CE\end{bmatrix}$$

Thus $$\begin{bmatrix} B & 0 \\ 0 & C\end{bmatrix}^{-1}=\begin{bmatrix} B^{-1} & 0 \\ 0 & C^{-1}\end{bmatrix}$$

Now use the formula for the inverse of 2x2.