# Thread: Another linear dependent/independent question

1. ## Another linear dependent/independent question

For this augmented matrix, the answer is that it they are linearly dependent. We are asked to find whether the coefficients are linearly independent or dependent?
For some reason I though considering there is no solution that they would be independent.

Could someone please show me why. Thanks

$
\left( {\begin{array}{*{20}c}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 0 & 3 \\
\end{array}} \right)
$

2. Originally Posted by Craka
For this augmented matrix, the answer is that it they are linearly dependent.
WHAT are linearly independent? What does "they" refer to?

[qute] We are asked to find whether the coefficients are linearly independent or dependent?[/quote]
The coefficients of WHAT?

For some reason I though considering there is no solution that they would be independent.

Could someone please show me why. Thanks

$
\left( {\begin{array}{*{20}c}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 0 & 3 \\
\end{array}} \right)
$

3. It's a augmented matrix, so the co-efficients are the first 3 columns.
I don't know how else to explain, that is what the question asks.

4. ## [unsolved] Another linear dependent/independent question

Still needing help with this, please. Sorry re-read question this morning, question is to find whether the columns of the coefficient part of the matrix are independent or dependent

5. Originally Posted by Craka

Still needing help with this, please. Sorry re-read question this morning, question is to find whether the columns of the coefficient part of the matrix are independent or dependent
they are linearly dependent. one reason is that the reduced form of a square matrix with linearly independent columns is always the identity matrix. but in your matrix (ignore the last column) the

$3 \times 3$ matrix is not the identity matrix (the last column is 0). another reason is that if the columns of a quare matrix are linearly independent, then the augmented matrix will always show a unique

solution but in your case there's no solution.