Linear Algebra determine the conditions

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Homework Help Overview

The discussion revolves around determining the conditions on the variables a, b, c, and d such that the matrix formed by these variables is row equivalent to two specified matrices: the identity matrix and a matrix with a row of zeros. The subject area is linear algebra, specifically focusing on matrix row equivalence and determinants.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the meaning of row equivalence and the implications of determinants for the matrices in question. There is an exploration of whether the problem consists of two separate parts or one combined problem. Some participants attempt to apply the determinant condition to derive relationships between the variables.

Discussion Status

The discussion is ongoing, with participants sharing their interpretations and approaches. Some guidance has been offered regarding the row reduction process and the implications of the determinant, but there is no explicit consensus on the conditions required for the matrices to be row equivalent.

Contextual Notes

Participants note potential confusion regarding the definitions and implications of row equivalence, as well as the need for clarity on whether the problem is addressing two distinct scenarios or a single scenario with multiple conditions.

_Bd_
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Homework Statement



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

Homework Equations




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 ??
 
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_Bd_ said:

Homework Statement



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

Homework Equations




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 ??

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.
 
yes, they are 2 problems, and yes I meant the determinant :P
still if I am on the right approach. . .thats as far as I can go. . .is that the answer?
a)
ad - bc = 1b)
ad - bc = 0
 
Last edited:
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)?
 
Mark44 said:
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)?

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]

add row_1 + row_2
[1 0]
[0 1]
what I don't 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?
 
Okay, now do the same that to the matrix given: actually row-reduce the given matrix!

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

If [itex]d- bc/a\ne 0[/itex] (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
[tex]\begin{bmatrix}c & d \\ a & b\end{bmatrix}[/tex]
Now what?
 

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