# Linear Algebra: Proof

## Homework Statement

|A| + |B| = |C|
where
A=
[a_11 a_12 a_13]
[a_21 a_22 a_23]
[a_31 a_32 a_33]
B=
[b_11 b_12 b_13]
[b_21 b_22 b_23]
[b_31 b_32 b_33]
C=
[(a_11 + b_11) a_12 a_13]
[(a_21 + b_21) a_22 a_23]
[(a_31 + b_31) a_32 a_33]

?

## The Attempt at a Solution

First I tried taking the determinants of all 3 matrices but it got like WAY too messy and I didnt see anything close to them being equal (since i got a lot of terms that involved a_ij x b_ij and I didnt get them on the other ones.

Then I researched a little and came across that
A=
[a_11 a_12 a_13] = [a_11 a_12 a_13] + [ 0 a_12 a_13] + [0 a_12 a_13]
[a_21 a_22 a_23] [0 a_22 a_23] [a_21 a_22 a_23] [0 a_22 a_23]
[a_31 a_32 a_33] [0 a_32 a_33] [0 a_32 a_33] [a_31 a_32 a_33]

I think it goes somewhere around there but I keep getting lost and I dont know how to get around it.

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Mark44
Mentor

## Homework Statement

Prove that A + B = C
where
A=
[a_11 a_12 a_13]
[a_21 a_22 a_23]
[a_31 a_32 a_33]
B=
[b_11 b_12 b_13]
[b_21 b_22 b_23]
[b_31 b_32 b_33]
C=
[(a_11 + b_11) a_12 a_13]
[(a_21 + b_21) a_22 a_23]
[(a_31 + b_31) a_32 a_33]
If A and B are 3 x 3 matrices, then A + B is also a 3 x 3 matrix whose typical entry is cij = aij + bij.

For example, c11 = a11 + b11, which is what you have, but c12 = a12 + b12. Instead, you have a12 without b12.

The matrix you show for C can't be the sum of A and B.

?

## The Attempt at a Solution

First I tried taking the determinants of all 3 matrices but it got like WAY too messy and I didnt see anything close to them being equal (since i got a lot of terms that involved a_ij x b_ij and I didnt get them on the other ones.
Determinants don't enter into a problem about the sum of matrices. Since you seem to think that they are involved in this problem, it might be that you haven't given the exact problem statement.
Then I researched a little and came across that
A=
[a_11 a_12 a_13] = [a_11 a_12 a_13] + [ 0 a_12 a_13] + [0 a_12 a_13]
[a_21 a_22 a_23] [0 a_22 a_23] [a_21 a_22 a_23] [0 a_22 a_23]
[a_31 a_32 a_33] [0 a_32 a_33] [0 a_32 a_33] [a_31 a_32 a_33]

I think it goes somewhere around there but I keep getting lost and I dont know how to get around it.

wait let me say cause I didnt specify, sorry for the big mistake I meant to say
Proove that:

|A| + |B| = |C|

Mark44
Mentor
What you're trying to prove isn't true, so you're going to have a tough time proving it.

Here's a counter example.
$$A = \left[\begin{array}{c c c}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
$$B = \left[\begin{array}{c c c}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$$
$$C = \left[\begin{array}{c c c}4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
I picked these matrices to make calculating the determinants easier.
|A| = 1, |B| = 27, and |C| = 4
So |A| + |B| $\neq$ |C|

I don't think you copied the problem correctly, particularly matrix B.

HallsofIvy
Homework Helper
Actually, |A|+ |B|= |C| makes no sense if you give no more information about A, B, and C. Surely it is not true for any three matrices- for A and B fixed, |A|+ |B| is fixed and is not equal to |C| for all matrices |C|.

I suspect you mean "If A, B, and C are 3 by 3 matrices such that A+ B= C, then |A|+ |B|= |C|." That now makes sense (before, you had no "if" part) but, as Mark44 said, still is not true.

yes! OMG you guys are correct. . .I cant believe I overlooked it

the actual problem says the following:

|A| + |B| = |C|
where
A=
[a_11 a_12 a_13]
[a_21 a_22 a_23]
[a_31 a_32 a_33]
B=
[b_11 a_12 a_13]
[b_21 a_22 a_23]
[b_31 a_32 a_33]
C=
[(a_11 + b_11) a_12 a_13]
[(a_21 + b_21) a_22 a_23]
[(a_31 + b_31) a_32 a_33]

so the way I was going around was correct :) (I think. . .Im gonna start working it out but I believe i was on the right track )