I know that one easy way to find the inverse of a matrix is to reverse the two diagonal elements and change the sign of the two non-diagonal elements. Then if we divide by the determinant, we get the inverse matrix. But I'm confused how to find the determinant of this matrix (below) and also confused about why they have moved the $\epsilon$ from denominator to numerator.

Is there any easy way to find the inverse matrix?

Here is the matrix and its inverse matrix: enter image description here

  • $\begingroup$ Your recipe for inverting a matrix only works for $2 \times 2$ matrices... and you see it on the example you have given. $\endgroup$ – Jean Marie Oct 6 '16 at 3:42
  • $\begingroup$ By the way, $D$ is the "discrete derivation" matrix and $D^{-1}$ is "discrete integration" matrix $\endgroup$ – Jean Marie Oct 6 '16 at 3:46

The determinant of the large matrices filled with $\pm 1$'s is $1,$ because the determinant of a triangular matrix (matrix with all zeros either above or below the diagonal) is the product of diagonal entries. The $\epsilon$ is some unknown number that is applied to all the entries, i.e. your $\pm 1$'s are $\pm \frac{1}{\epsilon}$ in the first matrix, and the second matrix is all just $\epsilon$ nonzero entries. They have, then, "factored out" the $\frac{1}{\epsilon}$ and $\epsilon$ just as you would a constant term common to every element of a vector (or a common factor of every term of a polynomial, if you want an analogy).

To see why they wrote it with $\frac{1}{\epsilon}$ and $\epsilon,$ multiply the two matrices. The $\epsilon$ and $\frac{1}{\epsilon}$ cancel (you can move "outside" scalars around in a matrix product) leaving you with just the $\pm 1$ matrix (lets call it $A$ for short) and the matrix with the lower left triangle full of $1's$ (let's call it $A^{-1}$ for short).

Now, multiply $A*A^{-1}.$ Remember, we canceled those confusing $\epsilon$ and $\frac{1}{\epsilon}$ first.

What do we have? Well, the matrix $A$ tells you that, for row $1$ of the matrix its multiplying by on the left hand side, keep it the same (1 in the first column entry of row 1, zeros elsewhere).

The second row has $-1$ in the first column followed by $1$ in the second column This means to take the first row of the right hand matrix and subtract it off the second row, entry for entry, and re-record the second row with this result. That leaves you only with a $1$ in the second row, second column of the result.

The third row has a $0$ in the first column, $-1$ in the second column, and $1$ in the third column. This means, for the result's third row, you are to take the third row of $A^{-1},$ and add on the second row times $-1$ (i.e., subtract the second row from the third). That leaves us with a $1$ in the third row, third column of the result and zero elsewhere.

Continue in a like matter. What's your result?

  • $\begingroup$ Multiplying on the right by $A^{-1}$ tells you what to do with the columns of $A.$ to obtain your result. For instance, the first column of $A^{-1}$ is full of 1's. This means to take one copy of every column of $A$ and add them all together for the first column of the answer. Do it, and you see that all the entries have $1-1=0$ except for the first row entry, which is $1.$ For the second column, you have 1's except zero in the first row. That means take all columns except the first column of $A$ and add them for the second column of your answer (1 in second row, 0 elsewhere). $\endgroup$ – Thomas Rasberry Oct 6 '16 at 4:02

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