Independent and depentent matrix

Question

\begin{matrix} 3 & -6 & 6 \\ 2 & -2 & 6 \\ 2 & -2 & a \\ -3 & 7 & -5 \\ \end{matrix}

Determine all values of $a$ for which the following set of vectors is dependent or independent. You can select $always$, $'never$', '$a =$ ', or $'a ≠'$, then specify a value or comma-separated list of values.

After doing row operations i have gotten : \begin{matrix} 1 & -2 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & a/2 -5 \\ \end{matrix}

I know that set of vectors are independent when all the columns have pivots. I am not sure how to answer the question. I would say that it is independent when $a/2 -5=0$ so $ a=10$ and dependent when $a≠10.$ [![enter image description here][1]][1]

Answer 1

The problem is that there was an error in the row reduction. Here are the row reduction steps:

$$ \begin{bmatrix} 3&-6&6\\ 2&-2&6\\ 2&-2&a\\ -3&7&-5 \end{bmatrix} $$ $\frac{1}{3}R_1\rightarrow R_1$. $$ \begin{bmatrix} 1&-2&2\\ 2&-2&6\\ 2&-2&a\\ -3&7&-5 \end{bmatrix} $$ $-2R_1+R_2\rightarrow R_2$, $-2R_1+R_3\rightarrow R_3$, and $3R_1+R_4\rightarrow R_4$. $$ \begin{bmatrix} 1&-2&2\\ 0&2&2\\ 0&2&a-4\\ 0&1&1 \end{bmatrix} $$ $R_2\leftrightarrow R_4$. $$ \begin{bmatrix} 1&-2&2\\ 0&1&1\\ 0&2&a-4\\ 0&2&2 \end{bmatrix} $$ $-2R_2+R_3\rightarrow R_3$ and $-2R_2+R_4\rightarrow R_4$ $$ \begin{bmatrix} 1&-2&2\\ 0&1&1\\ 0&0&a-6\\ 0&0&0 \end{bmatrix} $$ This is in row echelon form regardless of the value of $a$. If $a$ is not $6$, then the last column has a pivot, and otherwise, the last column does not have a pivot.

Answer 2

A pivot is a leading nonzero entry after the matrix has been put in row echelon form. For this matrix, when $a=10$, the matrix is already in row echelon form. When $a\not=10$, you can use the third row to cancel out the $a/2-5$.

Now, circle your pivots. They are the leading nonzero entries. You should get three pivots. Look at the columns that these pivots are in, and you're done.