+0

# done

0
121
2
+78

done

Mar 18, 2020
edited by rubikx2910  Apr 15, 2020

#1
+25243
+2

Let $$A$$ be a matrix, and let $$x$$ and $$y$$ be vectors such that neither is a scalar multiple of the other and such that
$$\mathbf{A} \mathbf{x} = \mathbf{y},\ \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}$$.
Then we have that $$\mathbf{A}^{-1} \mathbf{x} = a \mathbf{x} + b\mathbf{y}$$ for some scalars $$a$$ and $$b$$.
Find $$a$$ and $$b$$.

$$\begin{array}{|rcll|} \hline \mathbf{A} \mathbf{x} &=& \mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{x} &=& \mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{x} &=& \mathbf{A}^{-1}\mathbf{y} \\\\ \hline \mathbf{A} \mathbf{y} &=& \mathbf{x} + 2\mathbf{y} \quad &| \quad \times \mathbf{A}^{-1} \\ \mathbf{A}^{-1}\mathbf{A}\mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{A} = \text{Identity matrix} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{A}^{-1} \mathbf{y} \quad &| \quad \mathbf{A}^{-1}\mathbf{y} = \mathbf{x} \\ \mathbf{y} &=& \mathbf{A}^{-1}\mathbf{x} + 2\mathbf{x} \\ \mathbf{A}^{-1}\mathbf{x} &=& -2\mathbf{x}+\mathbf{y} \\ \hline \end{array}$$

$$\mathbf{a=-2,\ b=1}$$

Mar 18, 2020
edited by heureka  Mar 18, 2020
#2
+78
+1

done

rubikx2910  Mar 19, 2020
edited by rubikx2910  Apr 15, 2020