# write B as a linear combination of the other matrices, if possible. B=begin{bmatrix}2 & 5 0 & 3 end{bmatrix} , A_1=begin{bmatrix}1 & 2 -1 & 1 end{bmatrix} , A_2=begin{bmatrix}0 &1 2 & 1 end{bmatrix}

write B as a linear combination of the other matrices, if possible.
$B=\left[\begin{array}{cc}2& 5\\ 0& 3\end{array}\right],{A}_{1}=\left[\begin{array}{cc}1& 2\\ -1& 1\end{array}\right],{A}_{2}=\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$
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hajavaF
Step 1
Given $B=\left[\begin{array}{cc}2& 5\\ 0& 3\end{array}\right],{A}_{1}=\left[\begin{array}{cc}1& 2\\ -1& 1\end{array}\right],{A}_{2}=\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$
Here we write B is a linear combination of the other matrices .
Step 2
Let us consider the relation
are arbitary constant
$⇒\left[\begin{array}{cc}2& 5\\ 0& 3\end{array}\right]={C}_{1}\left[\begin{array}{cc}1& 2\\ -1& 1\end{array}\right]+{C}_{2}\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$
$⇒\left[\begin{array}{cc}2& 5\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}{C}_{1}+0\cdot {C}_{2}& 2{C}_{1}+{C}_{2}\\ -{C}_{1}+2{C}_{2}& {C}_{1}+{C}_{2}\end{array}\right]$
$⇒{C}_{1}=2,-\left(1\right)2{C}_{1}+{C}_{2}=5,-\left(2\right)-{C}_{1}+2{C}_{2}=0-\left(3\right)$

we get
$2x2+{C}_{2}=5$
$⇒{C}_{2}=5-4=1$
Also satisfy the equation (3) and (4)
Step 3
Hence From (A)
$B=2{A}_{1}+{A}_{2}$
$⇒B=2\left[\begin{array}{cc}1& 2\\ -1& 1\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$
Hence B is a linear combination of .
Jeffrey Jordon