Uncover the Power of Linear Algebra: Proving Invertibility of a 3x3 Matrix

[songofmisery]

let A be a 3x3 matrix. Suppose that for every row vector y=[y1 y2 y3] there exists a row vector x=[x1 x2 x3] such that xA=y. Show that A is invertable


i honestly have no idea where to even go with this. any help would be appreciated (:

 

[jbunniii]

Hint: A is invertible if and only if there is a 3x3 matrix B such that BA = ?

 

[songofmisery]

such that BA = I?

i read something about that in my textbook but i don't understand what I is. Is it just the inverse?

 

[songofmisery]

i know how to find an inverse, i just don't understand the part with row vectors and where it fits into the equation

 

[jbunniii]

No, I is the identity matrix:

I = \left[ \begin{array}{ccc}<br /> 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 \end{array} \right]

 

[jbunniii]

Try to write BA = I, one row at a time.

 

[songofmisery]

oh, that would make sense.

so could i say that B is x and I is y?

i feel like that's completely wrong.

 

[jbunniii]

No, B and I are 3x3 matrices, whereas x and y are 1x3.

If you're having trouble seeing what to do, I suggest naming the elements of the matrix B, for example as follows:

B = \left[ \begin{array}{ccc}<br /> a &amp; b &amp; c \\<br /> d &amp; e &amp; f \\<br /> g &amp; h &amp; i \end{array}\right]

Now, what's the first row of BA = I? It is of the form xA = y. What are x and y in this case?

 

[songofmisery]

x would be [a b c] and y would be [1 0 0]?

 

[jbunniii]

Right. So now, reverse the argument. You are given the fact that for every row vector y, there is a row vector x such that xA = y. So start by choosing y = [1 0 0], and writing the corresponding x = [a b c]. Now repeat for the remaining two rows. What do you end up with?