(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the following theorm:

Given any nxn matrix A, a mxm matrix B, the m+n x m+n matrix C:

[A|X]

[0|B]

where the top left corner of C is A, the top right (X) is any mxn matrix, the bottom right is a nxm 0 matrix and the bottom right is B.

Prove:

|C| = |A|*|B| (|C| is the determinant of C...)

I found a proof but the book did something different and i want to check if my way is correct.

2. Relevant equations

1) any square matrix that can't be reduced to the I matrix has a determinant of 0. (because it can be reduced to a matrix with a 0 row or coloum)

2) multiplying a row or coloum by a scalar changes the determinant by a factor of 1 over the scalar.

2) changing rows changes the sign of the determinant

3) adding the multiple of a row doesn't change the D.

3. The attempt at a solution

Here's my proof:

Lets say that both A and B can be reduced to the I matrix. if A can be reduced than it can be done with by operating only with the coloums using the 3 basic operations. the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'. so the determinant of A is just a (|A| = a) - this is because the determinant of I is 1.

The same argument can be made with B - this time operating on the rows giving the determinant of b.

Since we reduced A by operating on the coloums and B with the row, we can do all those operations on C without the operation on one of them effecting the rest of the matrix. Applying these operations to C changes the determinant by a*b and since the resaulting matrix is a trianguler matrix with a diagonal of only ones - the determinant of C is a*b. And so:

|C| = a*b = |A| * |B|.

If either A or B can't be reduced then one of them has a 0 determinant so |A|*|B| = 0

and by an argument similar to the above |C| can also be reduced to a matrix with either a 0 row or 0 coloum so |C| = 0.

Q.E.D

Is this a correct proof?

Thanks.

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# Homework Help: Linear Algebra Proof

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