For what value of h will y be in span {v1 , v2, v3} if \(v_1 = \begin{bmatrix}1\\ -1\\ -2\end{bmatrix}\), \(v_2 = \begin{bmatrix}5\\ -4\\ -7\end{bmatrix}\), \(v_3 = \begin{bmatrix}-3\\ 1\\ 0\end{bmatrix}\) and \(y = \begin{bmatrix}-4\\ 3\\ h\end{bmatrix}\)

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If y is in span {v1, v2, v3} then it can be written as

x1v1 + x2v2 + x3v3 = y

where x1, x2, x3 are scalar

The augmented matrix is

\(\begin{bmatrix}1 & 5 & -3 & : & -4\\ -1 & -4 & 1 & : & 3\\ -2 & -7 & 0 & : & h\end{bmatrix}\)

Reducing to Echelong form

Apply R2 → R2 + R1,   R3 → R3 + 2R1

\(~\begin{bmatrix}1 & 5 & -3 & : & -4\\ 0 & 1 & -2 & : & -1\\ 0 & 3 & -6 & : & h-8\end{bmatrix}\)

Apply R3 → R3 – 3R2

\(~\begin{bmatrix}1 & 5 & -3 & : & -4\\ 0 & 1 & -2 & : & -1\\ 0 & 0 & 0 & : & h-8 + 3\end{bmatrix}\)

\(~\begin{bmatrix}1 & 5 & -3 & : & -4\\ 0 & 1 & -2 & : & -1\\ 0 & 0 & 0 & : & h-5\end{bmatrix}\)

It will be consistent when h-5 = 0

Therefore, h =5

Hence, y be in span {v1. v2. v3} when h = 5

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