Are the following set of vectors linearly independent?
`vec(a) = - 2hat(i) - 4hat(k), vec(b) = hat(i) -2hat(j) - hat(k) , vec(c) = hat(i) - 4hat(j) + 3hat(k)`.

Answer

To determine if the vectors \(\vec{a} = -2\hat{i} - 4\hat{k}\), \(\vec{b} = \hat{i} - 2\hat{j} - \hat{k}\), and \(\vec{c} = \hat{i} - 4\hat{j} + 3\hat{k}\) are linearly independent, we can use the determinant method. Here are the steps to solve the problem:

### Step 1: Write the vectors in component form
The vectors can be expressed in terms of their components:
- \(\vec{a} = (-2, 0, -4)\)
- \(\vec{b} = (1, -2, -1)\)
- \(\vec{c} = (1, -4, 3)\)

### Step 2: Form a matrix with the vectors as rows
We will form a matrix \(M\) using the components of the vectors:
\[ M = \begin{bmatrix} -2 & 0 & -4 \\ 1 & -2 & -1 \\ 1 & -4 & 3 \end{bmatrix} \]

### Step 3: Calculate the determinant of the matrix
To check for linear independence, we need to calculate the determinant of matrix \(M\). The determinant can be calculated using the formula:
\[ \text{det}(M) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is represented as:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix:
- \(a = -2\), \(b = 0\), \(c = -4\)
- \(d = 1\), \(e = -2\), \(f = -1\)
- \(g = 1\), \(h = -4\), \(i = 3\)

Calculating the determinant:
\[ \text{det}(M) = -2((-2)(3) - (-1)(-4)) - 0 + (-4)(1(-4) - (-2)(1)) \]
Calculating each term:
1. \((-2)(3) - (-1)(-4) = -6 - 4 = -10\)
2. The second term is \(0\).
3. For the third term: \(1(-4) - (-2)(1) = -4 + 2 = -2\)

Putting it all together:
\[ \text{det}(M) = -2(-10) + 0 - 4(-2) = 20 + 8 = 28 \]

### Step 4: Conclusion
Since the determinant \(\text{det}(M) = 28\) is not equal to zero, the vectors \(\vec{a}, \vec{b}, \vec{c}\) are linearly independent.