Question #155938

Let B =(a1, a2, a3) be an ordered basis of 

R3 with a1 = (1, 0, -1), a2 = (1, 1, 1), 

a3 = (1, 0, 0). Write the vector v = (p,q,r)) as 

a linear combination of the basis vectors 

from B


1
Expert's answer
2021-01-18T15:50:22-0500

Let v=c1a1+c2a2+c3a3[pqr]T=c1×[101]T+c2×[111]T+c3×[100]T  [pqr]T=[c1c2c3]T×[111010110]T  [pqr]=[111010110]×[c1c2c3]  By Row-reducing we get,[qrqp+r2q]=[100010001]×[c1c2c3]  c1=qr,  c2=q,  c3=p+r2q\text{Let } v=c_1a_1+c_2a_2+c_3a_3\\ \Rightarrow \begin{bmatrix}p\\q\\r\end{bmatrix}^T=c_1\times\begin{bmatrix}1\\0\\-1\end{bmatrix}^T+c_2\times\begin{bmatrix}1\\1\\1\end{bmatrix}^T+c_3\times\begin{bmatrix}1\\0\\0\end{bmatrix}^T\\\;\\ \Rightarrow \begin{bmatrix}p\\q\\r\end{bmatrix}^T=\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}^T\times\begin{bmatrix}1&1&1\\0&1&0\\-1&1&0\end{bmatrix}^T\\\;\\ \Rightarrow \begin{bmatrix}p\\q\\r\end{bmatrix}=\begin{bmatrix}1&1&1\\0&1&0\\-1&1&0\end{bmatrix}\times \begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}\\\;\\ \text{By Row-reducing we get,}\\ \Rightarrow \begin{bmatrix}q-r\\q\\p+r-2q\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\times \begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}\\\;\\ \Rightarrow c_1=q-r, \;c_2=q,\;c_3=p+r-2q





So, v = (p,q,r) as a linear combination of the basis vectors from B :-\text{So, v = (p,q,r) as a linear combination of the basis vectors from B :-}\\



v=(qr)a1+qa2+(p+r2q)a3\boxed{{v=(q-r)a_1+qa_2+(p+r-2q)a_3}}


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