Math 1410, Spring 2020

Matrix Inverses

February 25, 2020

Sean Fitzpatrick
University of Lethbridge

Recap

Warm-Up

\begin{align*} A \amp = \bbm 2\amp -1\\-1\amp 1\ebm, \amp B \amp = \bbm 3\amp 0 \amp -1\\-1\amp 2\amp 3\ebm \amp I_2 \amp = \bbm 1\amp 0\\0\amp 1\ebm\\ C \amp = \bbm 1 \amp 4\\-3\amp 2\\2\amp -2\ebm \amp D \amp = \bbm 1\amp 1\\1\amp 2\ebm \amp I_3 \amp = \bbm 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1\ebm \end{align*}

Compute the products:

\begin{equation*} AB, BA, BC, CB, AD, DA, I_2A, I_2B, BI_3, I_3C\text{.} \end{equation*}
Compare \(A(BC)\) with \((AB)C\)

Properties

Theorem:

Assuming each product below is defined,

  1. \(A(BC)=(AB)C\)

  2. \(A(B+C)=AB+AC\)

  3. \((A+B)C = AC+BC\)

  4. \(A(kB)=(kA)B = k(AB)\) for any scalar \(k\)

  5. \(I_mA = A\) and \(AI_n = A\)

The inverse of a matrix

Matrix inverses

For real numbers \(a\) and \(b\text{,}\) if \(a\neq 0\) and \(ax=b\text{,}\) we know \(x=\frac{b}{a}\text{.}\)

If \(A\) is a matrix, \(\vec{b}\) is a vector, and \(A\vec{x}=\vec{b}\text{,}\) can we similarly solve for \(\vec{x}\text{?}\)

Short answer: no. Longer answer: sometimes. Sort of.

Definition:

A square (\(n\times n\)) matrix \(A\) is invertible if \(AB=I_n=BA\) for some \(B\text{.}\)

We call \(B\) the inverse of \(A\text{,}\) and write \(B=A^{-1}\text{.}\)

Example

For \(A = \bbm 2\amp -1\\-1\amp 1\ebm\text{,}\) \(D = \bbm 1\amp 1\\1\amp 2\ebm\text{,}\) we saw that \(AD = I_2\) and \(DA=I_2\text{.}\) So \(D=A^{-1}\text{.}\)

Suppose \(A\bbm x\\y\ebm = \bbm 7\\-5\ebm\text{.}\) How can we use \(D\) to solve for \(\vec{x}=\bbm x\\y\ebm\text{?}\)

Computing the inverse

Block multiplication

Given \(A\text{,}\) an \(m\times n\) matrix, and \(B\text{,}\) an \(n\times p\) matrix, write \(B = \bbm B_1\amp B_2 \amp \cdots B_p\ebm\text{,}\) where the \(B_i\) are columns of \(B\text{.}\) Then

\begin{equation*} AB = A\bbm B_1\amp B_2\amp \cdots B_p\ebm = \bbm AB_1\amp AB_2\amp \cdots \amp AB_p\ebm\text{.} \end{equation*}

Suppose \(A,B\) are both \(n\times n\text{,}\) and \(AB = I_n\text{.}\) Then

\begin{equation*} AB_1 = \bbm 1\\0\\\vdots \\0\ebm, AB_2 = \bbm 0\\1\\\vdots \\0\ebm, \cdots, AB_n = \bbm 0\\0\\\vdots\\1\ebm\text{.} \end{equation*}
Finding \(B=A^{-1}\) amounts to solving (!) these systems of equations.

The inverse algorithm

Let \(A\) be an \(n\times n\) matrix.

  1. Form the augmented matrix \([\begin{array}{c|c}A\amp I_n\end{array}]\)

  2. Use row operations until you reach \([\begin{array}{c|c}R\amp B\end{array}]\text{,}\) where \(R\) is the RREF of \(A\text{.}\)

  3. If \(R\) has a row of zeros, then \(A\) is not invertible.

  4. If \(R=I_n\text{,}\) then \(B=A^{-1}\text{.}\)

Examples

Compute the inverses (if possible) of:

  • \begin{equation*} A = \bbm 3\amp -5\\2\amp 7\ebm \end{equation*}

  • \begin{equation*} B = \bbm 1\amp 3\amp -2\\-1\amp -2\amp 4\\5\amp 13\amp -14\ebm \end{equation*}

  • \begin{equation*} C = \bbm 1\amp 0\amp -4\\2\amp 1\amp -5\\0\amp 1\amp 2\ebm \end{equation*}

Invertible matrix theorem

Theorem:

Let \(A\) be an \(n\times n\) matrix. The following statements are all equivalent to the statement “\(A\) is invertible”:

  1. The rank of \(A\) is \(n\text{.}\)

  2. The RREF of \(A\) is equal to \(I_n\text{.}\)

  3. The system \(A\vec{x}=\vec{b}\) is consistent for every \(n\times 1\) vector \(\vec{b}\text{.}\)

  4. The only solution to the system \(A\vec{x}=\vec{0}\) is \(\vec{x}=\vec{0}\text{.}\)

  5. There is a matrix \(B\) such that \(AB=I_n\text{.}\)

  6. There is a matrix \(C\) such that \(CA=I_n\text{.}\)

Properties of the inverse

Let \(A\) and \(B\) be invertible matrices.

  1. The inverse of \(A\) is unique.

  2. \(A^{-1}\) is also invertible, and \((A^{-1})^{-1}=A\text{.}\)

  3. \(AB\) is invertible, and \((AB)^{-1}=B^{-1}A^{-1}\text{.}\)

  4. For any nonzero scalar \(k\text{,}\) \(kA\) is invertible, and \((kA)^{-1}=\frac1k A^{-1}\text{.}\)

  5. For any \(n\times 1\) vector \(\vec{b}\text{,}\) the unique solution to \(A\vec{x}=\vec{b}\) is \(\vec{x}=A^{-1}\vec{b}\text{.}\)

Examples

  1. Show that if \(A,B,C\) are invertible, then \((ABC)^{-1}=C^{-1}B^{-1}A^{-1}\text{.}\)

  2. Show that if \(A^3 = 4I_n\text{,}\) then \(A\) is invertible.

  3. Show that if \(A^2-3A+2I_n=0\text{,}\) then \(A\) is invertible.

  4. Simplify the expression \(B(AB)^{-1}(AB)^2B^{-1}\text{.}\)