# Invertible Matrix

• September 17th 2009, 12:15 PM
noles2188
Invertible Matrix
For which values of the constants b and c is the following matrix invertible? (The following is all one matrix)

[ 0 1 b]
[-1 0 c]
[-b -c 0]
• September 17th 2009, 12:26 PM
Krizalid
it's a very straightforward problem, you only need to compute its determinant and set it different from zero.
• September 17th 2009, 12:28 PM
Jameson
Quote:

Originally Posted by noles2188
For which values of the constants b and c is the following matrix invertible? (The following is all one matrix)

[ 0 1 b]
[-1 0 c]
[-b -c 0]

I'll suggest some things before one of the wonderful members gives better advice.

You know that invertible matrices must have a non-zero determinant, so maybe calculating that can lead to some conclusions. The columns and rows must also be linearly independent, so I would think about that. Hope that helps some...
• September 17th 2009, 12:48 PM
noles2188
well the thing is, we haven't learned how to do 3x3 determinants yet. I know how to do it, but we've just been trying to get reduced row echelon form to check invertibility. Any other suggestions?
• September 17th 2009, 04:25 PM
Jameson
Quote:

Originally Posted by noles2188
well the thing is, we haven't learned how to do 3x3 determinants yet. I know how to do it, but we've just been trying to get reduced row echelon form to check invertibility. Any other suggestions?

Well if you get the reduced row echelon form of the matrix, you can easily check if the rows and columns are linearly independent. If this matrix is invertible the rank must be 3.
• September 17th 2009, 06:05 PM
seld
well if you're going to reduce row echelon form it, you can always set it up like:

$
\left [\begin{array}{ccc}
0 & 1 & b \\
-1 & 0 & c \\
-b & -c & 0 \end{array} \Bigg|\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array} \right]
$

And reduce that and you can see what values you need to set b and c to be, to make it invertible.

Though you automatically know that b and c cannot both be zero and $b \ne c$ otherwise it'd be not invertible.