Question

can someone please help me with this question (2.27)
https://www.dropbox.com/s/mi87j0zz63p6b8r/Photo%20Mar%2008%2C%202%2011%2031%20PM.jpg?dl=0

Answer 1

what have you tried so far

Answer 2

i'm trying to apply the fact that A x A^-1= I

Answer 3

but then i dont see how to fit that in the rest of the story

Answer 4

thats really a good start!
keep going

Answer 5

ok so we know that A* x=b

Answer 6

that's all i have =/

Answer 7

You have\[A\begin{bmatrix} 2\\8\\4 \end{bmatrix} = \begin{bmatrix} 2\\0\\0 \end{bmatrix}\]

left multiply \(A^{-1}\) both sides, what do you get ?

Answer 8

You get \[\begin{bmatrix} 2\\8\\4 \end{bmatrix} = A^{-1}\begin{bmatrix} 2\\0\\0 \end{bmatrix}\]

yes ?

Answer 9

so A^-1 * b= x??

Answer 10

Its easy, we're almost done.
suppose the columns of \(A^{-1}\) are \(c_1, ~c_2, ~c_3\) :

\[\begin{align}\begin{bmatrix} 2\\8\\4 \end{bmatrix} &= A^{-1}\begin{bmatrix} 2\\0\\0 \end{bmatrix}\\~\\~\\
&=[c_1~~c_2~~c_3]\begin{bmatrix} 2\\0\\0 \end{bmatrix}\\~\\~\\
&=2c_1 + 0c_2+0c_3\\~\\~\\
&=2c_1
\end{align}\]

Answer 11

divide \(2\) both sides and you're done!

Answer 12

well, that makes evry more clear. Thank you a lot!

Answer 13

yw