Question

Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Column 1 2nd Column negative 2 3rd Column 2 4st Column 3 2nd Row 1st Column negative 2 2nd Column 4 3rd Column negative 4 4st Column 3 3rd Row 1st Column 0 2nd Column 1 3rd Column negative 1 4st Column 4 EndMatrix ]1 −2 2 3 −2 4 −4 3 0 1 −1 4 Choose the correct answer below. A. The columns of the matrix do form a linearly independent set because there are more entries in each vector than there are vectors in the set

Answer 1

Linearly Dependent for not all scalars are null.

Explanation:

Hi there!

1)When we have vectors like
`v_(1),v_(2),v_(3), ...` we call them linearly dependent if we have scalars
`a_(1),a_(2),a_(3),...` as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

`a_(1)\vec{v_(1)}+a_(2)\vec{v_(2)}+a_(3)\vec{v_(3)}+...a_(m)\vec{v_(m)}=0`

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

`\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}`

So each column of this Matrix is a vector. So we can write them as:

`\vec{v_(1)}=\left \langle 1,-2,1 \right \rangle,\vec{v_(2)}=\left \langle -2,4,-1 \right \rangle,\vec{v_(3)}=\left \langle 2,-4,4 \right \rangle\vec{v_(4)}=\left \langle 3,3,4 \right \rangle` Or

Now let's rewrite it as a system of equations:

`a_(1)\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_(2)\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_(3)\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_(4)\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}`

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

`\left\{\begin{matrix}1a_(1) &-2a_(2) &+2a_(3) &+3a_(4) &=0 \\ &1a_(2) &-1a_(3) &+4a_(4) &=0 \\ & & &9a_(4) &=0 \end{matrix}\right.\Rightarrow a_(1)=0, a_(2)=a_(3),a_(4)=0`

`S=\begin{bmatrix}0\\ a_(3)\\ a_(3)\\ 0\end{bmatrix}`