# Diagonalizing matrices and powers

I'm wondering about the approach to another problem.

$B = \left( \begin{array}{ccc} a & 2a & 3a \\ 4b & 5b & 6b \\ 7c & 8c & 9c \end{array} \right)$

and I want to find $|B^4|$ i.e. the determinant of B to the 4th.

I was thinking of getting $B^k = S$$\Lambda^k$$S^{-1}$ where $S$ is the eigenvector matrix. Is that the only approach, or is there a better way to solve this problem?

$B^T$ provides an interesting matrix that I can manipulate more easily, but I'm not sure that helps me get to what I need.

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The determinant is multiplicative so you only need to compute $\det B$ and this is straightforward using the fact that the determinant is multilinear in the rows. Why is the transpose easier to manipulate? – Qiaochu Yuan Dec 13 '11 at 1:55

There's no need to think about eigenvectors/values for this problem.

Here's a hint. If $B'$ is obtained from $B$ by doing a row operation, then $|B'|$ is a certain multiple of $|B|$. Here are the three types of row operation with their effects on determinants (in each line, $B'$ should be regarded as after the row operation has been performed on $B$).

Row Swap: $|B'| = -|B|$.

Row Scaling: (multiply a row by a constant $k$). $|B'| = k|B|$.

Add multiple of one row to a different row: $|B'| = |B|$.

For your matrix, notice that each row has a particular nice form. Find a series of row operations that relate your matrix to a simpler one. I can see how to obtain a matrix with no unknown constants through row operations. Once you get to that stage, use the fact that $|A^p| = |A|^p$.

Hope this helps!

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Thanks for the tip! I can just multiply each row by 1/var and solve a much simpler problem. In general, are there other properties of determinants that are handy to know besides those three? – John Doe Dec 13 '11 at 2:02
Well there are tons of useful properties, but you'll want to consult a standard text on linear algebra to explore these. – Shaun Ault Dec 13 '11 at 12:56