Given a Matrix A, find a Product of Elementary Matrices that equals A

[WTFsandwich]

Homework Statement

Given

A = \left( \begin{array}{cc}<br /> 2 &amp; 1 \\<br /> 6 &amp; 4 \end{array} \right)

a) Express A as a product of elementary matrices.
b) Express the inverse of A as a product of elementary matrices.

Homework Equations

The Attempt at a Solution

Using the following EROs

Row2 --> Row2 - 3 * Row1
E_{1} = \left( \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> -3 &amp; 1 \end{array} \right)

Row1 --> 1/2 * Row1
E_{2} = \left( \begin{array}{cc}<br /> 1/2 &amp; 0 \\<br /> 0 &amp; 1 \end{array} \right)

Row1 --> Row1 - 1/2 * Row2
E_{3} = \left( \begin{array}{cc}<br /> 1 &amp; 0 \\<br /> -2 &amp; 1 \end{array} \right)

Multiplying all the Elementary matrices together, I got the Product

P = \left( \begin{array}{cc}<br /> 2 &amp; -1/2 \\<br /> -3 &amp; 1 \end{array} \right)

Which is A^-1.

 

[VeeEight]

Note that the inverse of A^-1 is A and also that given invertible A and B, (AB)^-1=B^-1A^-1

You have E₁E₂...E_n=A^-1 where E_i is an elementary matrix. So take the inverse of the whole thing

 

[WTFsandwich]

Note that the inverse of A^-1 is A and also that given invertible A and B, (AB)^-1=B^-1A^-1

You have E₁E₂...E_n=A^-1 where E_i is an elementary matrix. So take the inverse of the whole thing

If I'm understanding you correctly, I should take the inverses of all the elementary matrices and multiply those, and it should give me A?

Essentially, (E₁E₂...E_n)^-1 = A

 

[krtica]

Hey Sandwich,

Think of the matrix A as being equivalent to an identity matrix of the same size, but just manipulated by elementary row operations.

Vee is right, because if you multiply the inverse of A by A's corresponding elementary matrices, the product is the identity matrix. Try it out.

A=I(E1E2...En)^(-1)

Hope that helps!