Question

Express A and A^-1 as products of elementary matrices: A=[2 1 1; 1 2 1; 1 1 2].

Answer 1

we want to find a series of matrices E_1, E_2, ... E_n so that
\[E_nE_{n-1} \cdots_2E_1A = I\]
So
\[A^{-1} = E_nE_{n-1} \cdots_2E_1\]
and
\[A = E_n^{-1}E_{n-1}^{-1} \cdots E_2^{-1}E_1^{-1}\]

Answer 2

To find the elementary matrices, row reduce A to echelon form, but encode those row operations as a left matrix multiplication. For instance, the first row operation would be to multiply row 2 by -2 and add to row 1. This would be multipling A on the left by
\[\left(\begin{array}{ccc}
1&0&0\\
1&-2&0\\
0&0&1\\
\end{array}\right)\]

Answer 3

which would be E_1

Answer 4

but, I'll be honest. Seems like we could take advantage of symmetry in this problem.... just don't remember how.

Answer 5

I don't quite get what you're saying, so have that matrix and column of zeros on the right?

Answer 6

And try to row reduce that matrix?

Answer 7

Yes, you want to row reduce A to echelon form, but you want to "capture" each single row operation as a matrix.

Answer 8

so it'd be like [2 11; 1 2 1; 1 1 2] but everytime I do an operation I'd also have to do the same operation to [1 0 0; 0 1 0; 0 0 1]?

Answer 9

Yes, by record each change in the identity matrix as it's own matrix, E_1, E_2, ... and make sure you only do one row operation at a time.

Answer 10

Oh okay thank you!

Answer 11

\[E_1 = \left(\begin{array}{ccc}
1&0&0\\
1&-2&0\\
0&0&1\\
\end{array}\right)\]
\[E_2 = \left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
1&0&-2\\
\end{array}\right)\]

Answer 12

\[E_3 = \left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&1&-3\\
\end{array}\right)\]
\[E_4 = \left(\begin{array}{ccc}
1&0&0\\
0&1&0\\
0&0&\frac{1}{8}\\
\end{array}\right)\]

Answer 13

\[E_5 = \left(\begin{array}{ccc}
1&0&0\\
0&1&1\\
0&0&1\\
\end{array}\right)\]
\[E_6 = \left(\begin{array}{ccc}
1&0&0\\
0&-\frac{1}{3}&0\\
0&0&1\\
\end{array}\right)\]

Answer 14

\[E_7 = \left(\begin{array}{ccc}
1&0&-1\\
0&1&0\\
0&0&1\\
\end{array}\right)\]
\[E_8 = \left(\begin{array}{ccc}
1&-1&0\\
0&1&0\\
0&0&1\\
\end{array}\right)\]
\[E_9 = \left(\begin{array}{ccc}
\frac{1}{2}&0&0\\
0&1&0\\
0&0&1\\
\end{array}\right)\]

Answer 15

so
\[A^{-1} = E_9E_8E_7E_6E_5E_4E_3E_2E_1\]

Answer 16

nice job :)