+0  
 
+2
336
2
avatar+505 

Every vector v can be expressed uniquely in the form a + b where a is a scalar multiple of \(\begin{pmatrix} 2 \\ -1 \end{pmatrix}, \) and b is a scalar multiple of \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}.\)Find the matrix P such that Pv = a for all vectors v.

 

I tried to combine the two scalars, multiplied by a and b, and I got \(\begin{bmatrix} 2a + 3b \\ -a + b \end{bmatrix}\). I don't know how to get the matrix from this. Is it just \(\begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix}\)?\(\)

 Mar 14, 2022
 #1
avatar
0

Actually

\(\mathbf{P} = \begin{pmatrix} 2 & -1 \\ 3 & 1 \end{pmatrix}\)

 Mar 15, 2022
 #2
avatar+397 
+2

You have   \(\displaystyle \textbf{v}=\left[\begin{array}{c} 2a+3b\\ -a+b \end{array}\right]\) and require that  \(\displaystyle P\, \textbf{v}=\left[\begin{array}{c} 2a \\ -a \end{array} \right]\, .\)

So, let  \(\displaystyle P=\left[\begin{array}{cc} r & s \\ t & u \end{array}\right]\) say, multiply out the rhs and equate components across the equation to find r, s, t and u.

You should find that \(\displaystyle P = \left[ \begin{array}{cc} 2/5 & -6/5 \\ -1/5 & 3/5 \end{array} \right]\, .\)

 Mar 16, 2022

2 Online Users