# Linear Algebra determine the conditions

1. Aug 29, 2010

### _Bd_

1. The problem statement, all variables and given/known data

determine the conditions on a b c and d such that the matrix
a b
c d
will be row equivalent to the given matrix:
1 0
0 1

and

1 0
0 0

2. Relevant equations

3. The attempt at a solution

I have no idea what its asking, I mean on no. 1:
a = 1 b = 0 c = 0 and d =1
but what does that mean? I tried taking the resultant

for the first one would be 1(1) - 0(0) = 1

which would be ad - bc = 1 ??

2. Aug 29, 2010

### Staff: Mentor

This looks to me like two separate problems, or one problem with two parts.
Determine the conditions on a b c and d such that the matrix
a b
c d
will be row equivalent to the given matrix:
a)
1 0
0 1

b)
1 0
0 0

By "resultant" I think you mean determinant, and you're on the right track with that approach.

3. Aug 29, 2010

### _Bd_

yes, they are 2 problems, and yes I meant the determinant :P
still if im on the right approach. . .thats as far as I can go. . .is that the answer?
a)

b)

Last edited: Aug 29, 2010
4. Aug 29, 2010

### Staff: Mentor

You should start a new thread for the other problem.

For the original problem, suppose the matrix is
[2 1]
[3 0]

Will it be row-equivalent to the identity matrix or to the other one (the one that has all zeroes except for the upper left entry)?

5. Aug 29, 2010

### _Bd_

2(0) - 3(1) = -3 . . .Im guessing no? (its not zero nor 1)
or if you do some opperations
-r1 + r2 =
[1 -1]
[3 0]
and r_3 - 3r1

[1 -1]
[0 3]
divide by 3 on row 2
[1 -1]
[0 1]

[1 0]
[0 1]
what I dont understand is what
row equivalency is? unless it means that its like "multiples" (or row-operation-wise) of each other?

if so then what you mentioned is row equivalent to a) ??
still how would I write the answer?

6. Aug 30, 2010

### HallsofIvy

Okay, now do the same that to the matrix given: actually row-reduce the given matrix!

From
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$
as long as $a\ne 0$ we can divide the first row by a, then subtract c times the first row from the second to get
$$\begin{bmatrix}1 & \frac{b}{a} \\ 0 & d- \frac{bc}{a}\end{bmatrix}$$

If $d- bc/a\ne 0$ (which is the same as saying det= ad- bc= 0), we have the second form. If not, we can divide the second row by that and get the first form.

If a= 0, we can swap the rows, getting
$$\begin{bmatrix}c & d \\ a & b\end{bmatrix}$$
Now what?