<p>Let A and I be as follows.</p>
<p>A = [1 d]
[c b]
I=[1 0]
[0 1]</p>
<p>Prove that if b - cd != 0, then A is row equivalent to I</p>
<p>I’m CLUELESS as to WHERE TO START. Please help me</p>
<p>Let A and I be as follows.</p>
<p>A = [1 d]
[c b]
I=[1 0]
[0 1]</p>
<p>Prove that if b - cd != 0, then A is row equivalent to I</p>
<p>I’m CLUELESS as to WHERE TO START. Please help me</p>
<p>I tried simplifying to the matrix</p>
<p>[1 d]
[0 b - cd]</p>
<p>And have no clue what to do next.</p>
<p>A = {[1,d],[c,b]}</p>
<p>What you are doing is looking at the determinant. If the determinant of a matrix is non zero, then the row space is linearly independent, as well as the existance of an inverse matrix.</p>
<p>so the determinant of A = det(A) = 1<em>b - c</em>d = b-cd</p>
<p>if b-cd != 0, then there exists a unique inverse for A, A^-1. </p>
<p>More importantly, if b-cd != 0, then the rowspace of A is linearly independent, so you can apply the 3 matrix manipulation moves on it. There is a theorem that applying these moves keeps the system linearly independent.</p>
<p>Also, the most elegant way of doing this would be to realize that if b-cd != 0, both A and I form basis for a 2 dimensional vector. Thus they are row equivalent.</p>