How to find the inverse of a large symmetric matrix?

Question

I am looking for ways to quickly determine the inverse of symmetric Matrices, up to 6x6. My lecturer was solving the following question on the board but didn't explain how he was finding the inverse in his head so quickly so I am hoping someone can help me

Let $$A = \left[\begin{matrix} 4 & 0 & 0 & 0 & 2 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 2 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 \end{matrix}\right]$$

Find $A^{-1}$

It took him maybe about a minute to solve and write the solution on the board and said that we should be able to do it in about the same time.

Many thanks in advance

Answer 1

The elementary algorithm usually taught for finding an inverse is to row-reduce your matrix, applying the same row operations to the identity matrix. When your matrix is reduced to the identity, then what started as the identity will be your inverse.

In this case I want to subtract half of row $1$ from row $5$, which will get rid of the $2$ below the diagonal, and turn the $4$ at position $(5,5)$ into a $3$. Then I want to subtract $\frac23$ of row $5$ from row $1$, which will get rid of the $2$ over the diagonal. Finally, I multiply the rows by the constants $\frac14,1,\frac14,\frac15,\frac13,\frac16$. Applying these three steps to $I_6$:

$$\begin{bmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\-\frac12&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix} \rightarrow \begin{bmatrix}\frac43&0&0&0&-\frac23&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\-\frac12&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix} \rightarrow \begin{bmatrix}\frac13&0&0&0&-\frac16&0\\0&1&0&0&0&0\\0&0&\frac14&0&0&0\\0&0&0&\frac15&0&0\\-\frac16&0&0&0&\frac13&0\\0&0&0&0&0&\frac16\end{bmatrix}$$

I don't know if that's what your professor did, but in a matrix so sparse with non-zero entries, it works.

Alternatively, the only rows/columns that make this different from a diagonal matrix are numbers $1$ and $5$. Thus, the diagonal entries for row/columns $2,3,4,6$ simply need to be replaced with their reciprocals, while the four entries where row/columns $1$ and $5$ interact are given by the $2\times 2$:

$$\begin{bmatrix}4&2\\2&4\end{bmatrix}$$

which has determinant $12$. By the usual trick for inverting a $2\times 2$, its inverse is $\frac1{12}\begin{bmatrix}4&-2\\-2&4\end{bmatrix}$.

Answer 2

$$\left[\begin{matrix} 4 & 0 & 0 & 0 & 2 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 2 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 \end{matrix}\right]\xrightarrow[]{R_5\to R_5-\frac12R_1}\left[\begin{matrix} 4 & 0 & 0 & 0 & 2 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 \end{matrix}\right]\xrightarrow[]{C_5\to C_5-\frac12C_1}\left[\begin{matrix} 4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 \end{matrix}\right]$$ Now as the reduced matrix is a diagonal one, we can easily find its inverse to be $$\left[\begin{matrix} \frac14 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac14 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac15 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac13 & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac16 \end{matrix}\right]$$ Now we apply the inverse column and row operations, namely $C_5\to C_5+\frac12C_1$ and $R_5\to R_5+\frac12R_1$, to get $A^{-1}$ as $$\left[\begin{matrix} \frac14 & 0 & 0 & 0 & \frac18 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac14 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac15 & 0 & 0 \\ \frac18 & 0 & 0 & 0 & \frac{11}{24} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac16 \end{matrix}\right]$$