Prove : If an n*n matrix A can be expressed as a product of elementary matrices, Ax = b is consistent for every n*1 matrix b.
My thoughts on the question :
Since A can be expressed as a product of elementary matrices and elementary matrices are invertible, A can be expressed as a product of invertible matrices. Hence, A is invertible.
Now let us consider Ax = b
Since A is invertible and if A^-1 is it's inverse matrix,
(A^-1)Ax = (A^-1)b
x = (A^-1)b
Now how do we show that, Ax = b is consistent for every n*1 matrix b, in this case(when A is an n*n matrix which can be expressed as a product of elementary matrices)
Thank you!
don't consistent matrices have non-zero determinants?
If A is invertible, its determinant is non-zero.
To prove that Ax = b is consistent for every n*1 matrix b when A can be expressed as a product of elementary matrices, we need to show that for any given b, there exists at least one solution x that satisfies the equation.
Let's consider the given equation Ax = b. We know that A is invertible since it can be expressed as a product of elementary matrices. This means that A has a unique inverse, denoted by A^-1.
Now, let's multiply both sides of the equation by A^-1:
A^-1(Ax) = A^-1b
Since matrix multiplication is associative, we can simplify:
I_n x = A^-1b
Where I_n represents the identity matrix of size n. The product of A^-1 and A results in the identity matrix, so:
x = A^-1b
This shows that for any given b, there exists a solution x that satisfies the equation Ax = b. Therefore, the system Ax = b is consistent for every n*1 matrix b when A can be expressed as a product of elementary matrices.
To prove that the equation Ax = b is consistent for every n*1 matrix b, given that matrix A can be expressed as a product of elementary matrices, we can follow these steps:
Step 1: Recall that the equation Ax = b is consistent if and only if the matrix equation Ax = 0 has only the trivial solution x = 0.
Step 2: Assume that there exists a non-zero vector x such that Ax = 0.
Step 3: Since A can be expressed as a product of elementary matrices, we can express A as A = E1 * E2 * ... * Ek, where each Ei is an elementary matrix.
Step 4: Now consider the matrix equation (E1 * E2 * ... * Ek) * x = 0.
Step 5: We can rewrite the above equation as E1 * (E2 * (E3 * (... (Ek * x) ...))) = 0.
Step 6: Since elementary matrices are invertible, each of these elementary matrices Ei * (E(i+1) * (E(i+2) * (... (Ek * x) ...))) = 0 implies that (E(i+1) * (E(i+2) * (... (Ek * x) ...))) = 0.
Step 7: Continuing this process, we find that (Ek * x) = 0.
Step 8: However, since Ek is invertible, Ek * x = 0 implies x = 0.
Step 9: Therefore, assuming there exists a non-zero vector x such that Ax = 0, we have reached a contradiction, as it implies x = 0.
Step 10: Hence, the matrix equation Ax = b is consistent for every n*1 matrix b.
In summary, by expressing matrix A as a product of invertible elementary matrices, we showed that the equation Ax = b is consistent for every n*1 matrix b.