Let v = c 1 a 1 + c 2 a 2 + c 3 a 3 ⇒ [ p q r ] T = c 1 × [ 1 0 − 1 ] T + c 2 × [ 1 1 1 ] T + c 3 × [ 1 0 0 ] T ⇒ [ p q r ] T = [ c 1 c 2 c 3 ] T × [ 1 1 1 0 1 0 − 1 1 0 ] T ⇒ [ p q r ] = [ 1 1 1 0 1 0 − 1 1 0 ] × [ c 1 c 2 c 3 ] By Row-reducing we get, ⇒ [ q − r q p + r − 2 q ] = [ 1 0 0 0 1 0 0 0 1 ] × [ c 1 c 2 c 3 ] ⇒ c 1 = q − r , c 2 = q , c 3 = p + r − 2 q \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 Let v = c 1 a 1 + c 2 a 2 + c 3 a 3 ⇒ ⎣ ⎡ p q r ⎦ ⎤ T = c 1 × ⎣ ⎡ 1 0 − 1 ⎦ ⎤ T + c 2 × ⎣ ⎡ 1 1 1 ⎦ ⎤ T + c 3 × ⎣ ⎡ 1 0 0 ⎦ ⎤ T ⇒ ⎣ ⎡ p q r ⎦ ⎤ T = ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ T × ⎣ ⎡ 1 0 − 1 1 1 1 1 0 0 ⎦ ⎤ T ⇒ ⎣ ⎡ p q r ⎦ ⎤ = ⎣ ⎡ 1 0 − 1 1 1 1 1 0 0 ⎦ ⎤ × ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ By Row-reducing we get, ⇒ ⎣ ⎡ q − r q p + r − 2 q ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ × ⎣ ⎡ c 1 c 2 c 3 ⎦ ⎤ ⇒ c 1 = q − r , c 2 = q , c 3 = p + r − 2 q
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 :-}\\ So, v = (p,q,r) as a linear combination of the basis vectors from B :-
v = ( q − r ) a 1 + q a 2 + ( p + r − 2 q ) a 3 \boxed{{v=(q-r)a_1+qa_2+(p+r-2q)a_3}} v = ( q − r ) a 1 + q a 2 + ( p + r − 2 q ) a 3
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