Inverse Matrices

In the case of a square matrix, the matrix is invertible if the determinant of that Matrix is not equal to zero. The inverse of a matrix is unique. The inverse of the product of matrices is in reverse order,

i.e,

if you have three matrices A, B, C, the inverse of

$$(ABC{)}^{-1}=C^{-1}{B}^{-1}{A}^{-1}$$

Checking for correct row reduction

You can check if a matrix is invertible by row reducing and checking if it has a pivot in every column and every row, effectively becoming the identity matrix.

You can multiply the matrix by it’s inverse and you should find a form of rref?

If there exists a Vector x != 0 and Ax = 0, the null space of a is not just the zero Vector. As such A cannot be invertible

By contradiction, if A is invertible then there exists inverse of a such that A-1 x A = Identity matrix

$$x=A^{-1}Ax=A^{-1}xnonempty=\vec{0}$$

An n by m matrix is invertible if

• The determinant of A != 0
• Rank(A) = n
• RREF(A) = In
• There is a pivot in each row and each column of ref(A)
• All columns are LI
• All rows are LI
• Nul(A) = Nul(A^t) = {0} that t is supposed to be transposed?
• The matrix transformation T(x) = Ax is one-to-one and onto.

$$\begin{array}{c}A\vec{x}=\vec{y}=In\vec{y}\\ A^{-1}A\vec{x}=A^{-1}\vec{y}\\ In\vec{x}=A^{-1}\vec{y}\\ (A|In)\to (In|A^{-1})\end{array}$$

If a square matrix A is one-to-one and onto, then A is invertible with inverse A

such that A^-1 such that

$$A^{-1}A=A{A}^{-1}=I$$ $$[A|I]\to [I|A^{-1}]$$

By RREF.

In a square matrix of the form

$$A=\begin{array}{c}[ab]\\ [cd]\end{array},$$ $$A^{\prime }=\frac{1}{ad-bc}\begin{array}{c}\\ [d-b]\\ [-ca]\end{array}$$

Special Cases

The inverse of a diagonal matrix is simply the same matrix, with each value being 1/value.

The inverse of a permutation matrix is its transpose.