Define linearly independent set of vectors with an example. Show that the vectors (1, 4, 3), (0, 3, 1) and (3, -5, 4) are linearly independent. Do they form a basis? Justify.

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The given three vector are

V1 = (1, -4, 3), V2 = (0, 3, 1) and V3 = (3, -5, 4)

to check its independency

We have an augmented form

\(~\begin{bmatrix}1 & -4 & 3 & : & 0\\ 0 & 3 & 1 & : & 0\\ 3 & -5 & 4 & : & 0\end{bmatrix}\)

R3 → R3 – 3R1

\(~\begin{bmatrix}1 & -4 & 3 & : & 0\\ 0 & 3 & 1 & : & 0\\0 & 7 & -5 & : & 0\end{bmatrix}\)

\(R_2 \to \frac{1}{3} R_2\)

\(~\begin{bmatrix}1 & -4 & 3 & : & 0\\ 0 & 1 & \frac{1}{3} & : & 0\\0 & 7 & -5 & : & 0\end{bmatrix}\)

R1 → R1 + 4R2,   R3 → R3 – 7R2

\(~\begin{bmatrix}1 & 0 & \frac{13}{3} & : & 0\\ 0 & 1 & \frac{1}{3} & : & 0\\0 & 0 & \frac{-22}{3}& : & 0\end{bmatrix}\)

\(R_3 \to \frac{-3}{22} R_3\)

\(~\begin{bmatrix}1 & 0 & \frac{13}{3} & : & 0\\ 0 & 1 & \frac{1}{3} & : & 0\\0 & 0 & 1& : & 0\end{bmatrix}\)

\(R_1 \to R_1 – \frac{13}{2}R_3\) ,  \(R_1 \to R_2 – \frac{1}{3}R_3\)

\(~\begin{bmatrix}1 & 0 & 0 & : & 0\\ 0 & 1 & 0 & : & 0\\0 & 0 & 1& : & 0\end{bmatrix}\)

Hence, It is linearly independent.

Yes, they form basis because there is pivot element in each Row.

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