Question

# Prove that for a square n ×n matrix A, Ax = b (1) has one and...

Prove that for a square n ×n matrix A, Ax = b (1) has one and only one solution if and only if A is invertible; i.e., that there exists a matrix n ×n matrix B such that AB = I = B A.

NOTE 01: The statement or theorem is of the form P iff Q, where P is the statement “Equation (1) has a unique solution” and Q is the statement “The matrix A is invertible”. This means that you must prove two things: 1) P =⇒ Q and 2) Q =⇒ P. NOTE 02: For this problem the most mathematically sound (and ‘pleasing’) proof does not involve jumping to any of the statements in the list of non-singular equivalences, in, for example “LINEAR ALGEBRA IN A NUTSHELL” on the last page of the textbook for 22A “Introduction to Linear Algebra”, by Gilbert Strang unless you have derived that statement yourself directly from the fact that equation (1) has a unique solution. See if you can do the problem this way. If you just jump from one the statements in the list of non-singular equivalences, to another, without deriving the equivalence, you will receive 25% credit for this problem. HINT: What is the first step to one of the other non-singular equivalences, which you can take directly from equation (1)? In other words, find and prove a second non-singular equivalence such that equation (1) is essentially the definition of this second statement. Ask if you need help trying to prove the statement in Problem 01 this way

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