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

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Let a=m(2,-1), b=n(3,1), so v=a+b=(2m+3n,-m+n).

Let be a 2×2 matrix:

( x₁ y₁ )

( x₂ y₂ )

Then Pv=a can be represented by the equations:

x₁(2m+3n)+y₁(-m+n)=2m

x₂(2m+3n)+y₂(-m+n)=-m

From these we can get 4 equations by matching the coefficients of m and n.

(1) 2x₁-y₁=2

(2) 3x₁+y₁=0

(3) 2x₂-y₂=-1

(4) 3x₂+y₂=0

Add (1) and (2): 5x₁=2, x₁=⅖, so y₁=⅘-2=-⁶⁄₅.

Add (3) and (4): 5x₂=-1, x₂=-⅕, so y₂=-⅖+1=⅗.

Now we have P=

⅕( 2 -6 )

    (-1  3 )

by Top Rated User (762k points)

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