inverses and IMT

general inverse info

if a is an invertible nxn matrix, then for every b in R^n, the equation Ax = b has exactly ONE solution

this means the matrix is both one to one and onto

determinant of A

ad - bc
if ad - bc = 0, then A is not invertible
the determinant of an invertible matrix IS NEVER 0

how to find the inverse of 2×2 matrix A

A^-1 = (1/det(A)) * (upper left (d) upper right -b lower left -c lower right a)

Suppose A and B are invertible nxn matrixes

1. A^-1 is invertible and (A^-1)^-1 = A

2. AB is invertible, and (AB)^-1 = B^-1A^-1

3. A^T is invertible and (A^T)^-1 = (A^-1) ^T

   1. A transpose is invertible and the inverse of A transpose equals the transpose of A inverse

invertible matrix theorem

for an nxn matrix A where T(x) = A(x)

• T is one to one

• A has a pivot in every column

• A has n pivots

• A has a pivot in each row

• T is onto

• A is invertible

For matrix transformation T: R^n → R^n

Any condition for one to one is equivalent to any condition for onto which is equivalent to invertible

How to compute A^-1 beyond 2×2 matrices

1. Put A = Identity matrix into RREF

2. If the result is Identity matrix = something, the matrix A is invertible and A^-1 = the something

3. otherwise A is NOT invertible

Suppose A and B are invertible nxn matrices

True or False: (A + B)² = A² + B² + 2AB

False. The order in which matrices are multiplied together must be respected.

(A +B)² = A² + AB + BA + B²

Suppose A and B are invertible nxn matrices

True or False: A + B is invertible

False. Two invertible matrices can sum to the 0 matrix, which is not invertible

Suppose A and B are invertible nxn matrices

True or False: (AB)^-1 = A^-1B^-1

False. (AB)^-1 = B^-1A^-1

Suppose A and B are invertible nxn matrices

True or False: A^7 is invertible

True. A power of an invertible matrix is invertible. The inverse of A^7 is (A^-1)^7

Suppose A and B are invertible nxn matrices

True or False: (I - A)(I + A) = In - A²

True

Suppose A is an nxn matrix

Yes, no, maybe: If the equation Ax = 0 has a nontrivial solution, then A has fewer than n pivots

Yes. When Ax = 0 has a nontrivial solution, this means there is at least one free variable. This implies that some column does not have a pivot, so there are fewer than n pivots.

Suppose A is an nxn matrix

Yes, no, maybe: If the linear transformation T(x) = Ax is one to one, then the columns of A form a linearly dependent set

No. If the linear transformation T(x) = Ax is one to one, this means the only vector that can be mapped to the zero vector is the zero vector, meaning the columns of A satisfies the definition of linear independence.

NO FREE VARIABLES

Suppose A is an nxn matrix

Yes, no, maybe: If the equation Ax = 0 has the trivial solution, then the columns of A son R^n

Maybe. The matrix equation Ax = 0 always has the trivial solution. However, the question does not say whether or not the trivial solution is the only solution to the equation. There could be nontrivial solutions in which case the columns of A are not linearly independent, and thus cannot span R^n.

Suppose A is an nxn matrix

Yes, no, maybe: If -A is not invertible, then A is also not invertible

Yes.

Suppose A is an nxn matrix

Yes, no, maybe: If the linear transformation T(x) = Ax is onto, then it is also one to one

Yes. If T is onto, then every row in A has a pivot. Since A is a square, every column also has a pivot, so T is one to one.

Suppose A is an nxn matrix

Yes, no, maybe: The product of any two invertible matrices is invertible.

Yes.

Suppose A is an nxn matrix

Yes, no, maybe: A square matrix with two identical columns can be invertible

No. this means the columns are linearly dependent and thus cannot be invertible

Suppose A is an nxn matrix

Yes, no, maybe: If A² is row equivalent to the nxn identity matrix, then the columns of A span R^n

Yes. If A² is row equivalent to the identity matrix, then A² is invertible by the IMT. Since A² is invertible, A is also invertible.

Suppose A is an nxn matrix

Yes, no, maybe: If A is invertible, then the equation Ax = b has exactly one solution for all b ion R^n

Yes. The Solution is x = A^-1b