# Calculate inverse of matrix

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?

• There are lots of zeros in that matrix. Perhaps you learned a way to simplify the computation in a matrix like this. – GEdgar Mar 8 '15 at 17:26
• Hint: compute the determinant via blocks. $\det A = (-1)(-1)=1.$ – Ivo Terek Mar 8 '15 at 17:31
• hint: this is a block matrix. the inverse is in term of the inverse of the smaller matrices. – abel Mar 8 '15 at 17:53

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, an d in such a case the inverse is $\,\begin{bmatrix}A^{-1} &0\\0&B^{-1}\end{bmatrix}$.

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

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.

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.

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

• take advantage of the block structure of $A$ – abel Mar 8 '15 at 17:53