HW 5: Let A be an n × n matrix, and b be a vector in R^n . If the system Ax = b has infinitely many solutions, then A is not invertible.
TRUE. A is invertible if and only if it can be row reduced to the identity matrix, in which case any linear system has a unique solution.
To prove the statement, we can use the contrapositive. The contrapositive of the statement is:
If A is invertible, then the system Ax=b has a unique solution.
We know that if A is invertible, then there exists a unique solution to the system Ax=b. This is because we can solve for x by multiplying both sides of the equation by A inverse:
A^-1 Ax = A^-1 b
Since A^-1 A is the identity matrix, we have:
x = A^-1 b
Therefore, there exists a unique solution for x when A is invertible.
Now we need to prove that if the system Ax=b has infinitely many solutions, then A is not invertible.
Assume that the system Ax=b has infinitely many solutions. This means that there are multiple values of x that satisfy the equation Ax=b, or in other words, the null space of the matrix A is not just the zero vector.
If A were invertible, then the null space would only contain the zero vector, as any non-zero vector in the null space would lead to a contradiction. This is because if Av=0 for some non-zero vector v, we can multiply both sides by A^-1 to get v = A^-1 0 = 0, which is a contradiction.
Therefore, if the system Ax=b has infinitely many solutions, then A cannot be invertible.
Thus, we have proven the statement:
If the system Ax = b has infinitely many solutions, then A is not invertible.
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