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Inverse of a Matrix

Definition. An $$n\times n$$ matrix $$A$$ is invertible if there an $$n\times n$$ matrix $$B$$ such that $AB=BA=I_n.$ This $$B$$ is called the inverse of $$A$$, denoted by $$A^{-1}$$, for which $$AA^{-1}=A^{-1}A=I_n.$$ An invertible matrix is also called a nonsingular matrix. A square matrix that is not invertible is called a singular matrix.

Example. For $$A=\left[\begin{array}{rrr}1&2\\4&6\end{array} \right]$$ and $$B=\left[\begin{array}{rrr}-3&1\\2&-0.5\end{array} \right]$$, $$AB=\left[\begin{array}{rrr}1&0\\0&1\end{array} \right] =BA.$$ So $$B=A^{-1}$$.

Theorem. Let $$A$$ and $$B$$ be two $$n\times n$$ invertible matrices. Then the following hold.

1. $$A^{-1}$$ is invertible and $$(A^{-1})^{-1}=A$$.

2. $$A^T$$ is invertible and $$(A^T)^{-1}= (A^{-1})^T$$.

3. For $$c\neq 0$$, $$cA$$ is invertible and $$(cA)^{-1}=\frac{1}{c}A^{-1}$$.

4. $$AB$$ is invertible and $$(AB)^{-1}=B^{-1}A^{-1}$$.

(a) and (c) are exercises. For (b) note that \begin{align*} A^T(A^{-1})^T & =(A^{-1}A)^T=I_n^T=I_n \text{ and }\\ (A^{-1})^TA^T &=(AA^{-1})^T=I_n^T=I_n. \end{align*} For (d) note that \begin{align*} (AB) (B^{-1}A^{-1}) & =A(BB^{-1}) A^{-1} =AI_n A^{-1} =A A^{-1}=I_n \text{ and }\\ (B^{-1}A^{-1}) (AB) & =B^{-1} (A^{-1}A) B =B^{-1} I_n B = B^{-1}B=I_n. \end{align*}

Example. For $$A=\left[\begin{array}{rrr}1&1\\3&4\end{array} \right]$$ and $$B=\left[\begin{array}{rrr}1&2\\2&5\end{array} \right]$$, $$A^{-1}=\left[\begin{array}{rrr}4&-1\\-3&1\end{array} \right]$$ and $$B^{-1}=\left[\begin{array}{rrr}5&-2\\-2&1\end{array} \right]$$. Verify $$(A^T)^{-1}=\left[\begin{array}{rrr}1&3\\1&4\end{array} \right]^{-1} = \left[\begin{array}{rrr}4&-3\\-1&1\end{array} \right] = (A^{-1})^T$$, $$(5A)^{-1}=\frac{1}{5} \left[\begin{array}{rrr}4&-1\\-3&1\end{array} \right] =\frac{1}{5} A^{-1}$$, and $(AB)^{-1}=\left[\begin{array}{rrr}3&7\\11&26\end{array} \right]^{-1} =\left[\begin{array}{rrr}26&-7\\-11&3\end{array} \right] =B^{-1}A^{-1}.$

How do we know a given square matrix $$A$$ is invertible? How do we find $$A^{-1}$$?

Theorem. Let $$A$$ be an $$n\times n$$ matrix. Then the following are equivalent.

1. $$A$$ is invertible.

2. $$A\overrightarrow{x}=\overrightarrow{b}$$ has a unique solution for each $$\overrightarrow{b}\in \mathbb R^n$$.

3. The RREF of $$A$$ is $$I_n$$.

(b) $$\iff$$ (c) $$A\overrightarrow{x}=\overrightarrow{b}$$ has a unique solution for each $$\overrightarrow{b}\in \mathbb R^n$$ if and only if each column of the RREF of $$A$$ has a leading 1 if and only if the RREF of $$A$$ is $$I_n$$.
(a) $$\implies$$ (b) Suppose $$A$$ is invertible. Let $$\overrightarrow{b}\in \mathbb R^n$$. Then $$A\overrightarrow{x}=\overrightarrow{b} \implies \overrightarrow{x}=A^{-1}\overrightarrow{b}$$.
(b) $$\implies$$ (a) Suppose $$A\overrightarrow{x}=\overrightarrow{b}$$ has a unique solution for each $$\overrightarrow{b}\in \mathbb R^n$$. Let $$A\overrightarrow{v_i}=\overrightarrow{e_i}$$ for $$i=1,2,\ldots,n$$. Then $A[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}] =[A\overrightarrow{v_1}\: A\overrightarrow{v_2}\:\cdots A\overrightarrow{v_n}] =[\overrightarrow{e_1}\:\overrightarrow{e_2}\:\cdots\overrightarrow{e_n}]=I_n.$ To show $$A^{-1}=[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]$$, it suffices to show $$[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]A=I_n$$. Since $$A[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]=I_n$$, $A[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]A =I_nA=A.$ Let $$\overrightarrow{b_i}$$ be the $$i$$th column of $$[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]A$$ for $$i=1,2,\ldots,n$$. Then $$A\overrightarrow{b_i}=\overrightarrow{a_i}$$. But $$A\overrightarrow{e_i}=\overrightarrow{a_i}$$. By the uniqueness of solution of $$A\overrightarrow{x}=\overrightarrow{a_i}$$, $$\overrightarrow{b_i}=\overrightarrow{e_i}$$ for $$i=1,2,\ldots,n$$. Thus $[\overrightarrow{v_1}\:\overrightarrow{v_2}\:\cdots\overrightarrow{v_n}]A =[\overrightarrow{e_1}\:\overrightarrow{e_2}\:\cdots\overrightarrow{e_n}]=I_n.$

To find $$A^{-1}$$ for an invertible matrix $$A$$, we investigate how row operations on $$A$$ are obtained from premultiplying $$A$$ by elementary matrices.

Definition. An $$n\times n$$ elementary matrix is obtained by applying an elementary row operation on $$I_n$$.

Example.

1. $$E_{ij}$$ is obtained by $$R_i\leftrightarrow R_j$$ on $$I_n$$. Note that $$E_{ij}A$$ is obtained by $$R_i\leftrightarrow R_j$$ on $$A$$. \begin{align*} A=\left[\begin{array}{rrr}0&2&4\\1&-3&0\\-1&3&1\end{array} \right]\xrightarrow{R_1\leftrightarrow R_2} \left[\begin{array}{rrr}1&-3&0\\0&2&4\\-1&3&1\end{array} \right] &=\left[\begin{array}{rrr}0&1&0\\1&0&0\\0&0&1\end{array} \right] \left[\begin{array}{rrr}0&2&4\\1&-3&0\\-1&3&1\end{array} \right]\\ &=E_{12}A. \end{align*}

2. For $$c\neq 0$$, $$E_{i}(c)$$ is obtained by $$cR_i$$ on $$I_n$$. Note that $$E_{i}(c)A$$ is obtained by $$cR_i$$ on $$A$$. \begin{align*} E_{12}A=\left[\begin{array}{rrr}1&-3&0\\0&2&4\\-1&3&1\end{array} \right]\xrightarrow{\frac{1}{2} R_2} \left[\begin{array}{rrr}1&-3&0\\0&1&2\\-1&3&1\end{array} \right] &=\left[\begin{array}{rrr}1&0&0\\0&\frac{1}{2}&0\\0&0&1\end{array} \right] \left[\begin{array}{rrr}1&-3&0\\0&2&4\\-1&3&1\end{array} \right]\\ &=E_2\left(\frac{1}{2}\right) E_{12}A. \end{align*}

3. $$E_{ij}(c)$$ is obtained by $$cR_i+R_j$$ on $$I_n$$. Note that $$E_{ij}(c)A$$ is obtained by $$cR_i+R_j$$ on $$A$$. \begin{align*} E_2\left(\frac{1}{2}\right) E_{12}A=\left[\begin{array}{rrr}1&-3&0\\0&1&2\\-1&3&1\end{array} \right]\xrightarrow{R_1+ R_3} \left[\begin{array}{rrr}1&-3&0\\0&1&2\\0&0&1\end{array} \right] &=\left[\begin{array}{rrr}1&0&0\\0&1&0\\1&0&1\end{array} \right] \left[\begin{array}{rrr}1&-3&0\\0&1&2\\-1&3&1\end{array} \right]\\ &=E_{13}(1)E_2\left(\frac{1}{2}\right) E_{12}A. \end{align*}

Remark. Elementary matrices are invertible. Moreover, $$E_{ij}^{-1}=E_{ij}$$, $$E_{i}(c)^{-1}=E_{i}\left(\frac{1}{c}\right)$$ for $$c\neq 0$$, and $$E_{ij}(c)^{-1}=E_{ij}(-c)$$.

Theorem. Let $$A$$ be an $$n\times n$$ invertible matrix. A sequence of elementary row operations that reduces $$A$$ to $$I_n$$ also reduces $$I_n$$ to $$A^{-1}$$.

Since $$A$$ is invertible, the RREF of $$A$$ is $$I_n$$. Suppose $$I_n$$ is obtained from $$A$$ by successively premultiplying by elementary matrices $$E_1,E_2,\ldots,E_k$$, i.e., $E_kE_{k-1}\cdots E_1A=I_n.$ Postmultiplying by $$A^{-1}$$, we get $E_kE_{k-1}\cdots E_1AA^{-1}=I_nA^{-1} \implies E_kE_{k-1}\cdots E_1 I_n=A^{-1}.$

Gauss-Jordan elimination:
Find the RREF of $$[A\;|\;I_n]$$. If the the RREF of $$A$$ is $$I_n$$, then $$A$$ is invertible and the RREF of $$[A\;|\;I_n]$$ is $$[I_n\;|\;A^{-1}]$$. Otherwise $$A$$ is not invertible.

Example. $[A\;|\;I_3]=\left[\begin{array}{rrr|rrr}0&2&4&1&0&0\\1&-3&0&0&1&0\\-1&3&1&0&0&1\end{array} \right] \xrightarrow{R_1\leftrightarrow R_2} \left[\begin{array}{rrr|rrr}1&-3&0&0&1&0\\0&2&4&1&0&0\\-1&3&1&0&0&1\end{array} \right]$ $\xrightarrow{R_1+ R_3} \left[\begin{array}{rrr|rrr}1&-3&0&0&1&0\\0&2&4&1&0&0\\0&0&1&0&1&1\end{array} \right] \xrightarrow{-4R_3+ R_2} \left[\begin{array}{rrr|rrr}1&-3&0&0&1&0\\0&2&0&1&-4&-4\\0&0&1&0&1&1\end{array} \right]$ $\xrightarrow{\frac{1}{2} R_2} \left[\begin{array}{rrr|rrr}1&-3&0&0&1&0\\0&1&0&\frac{1}{2}&-2&-2\\0&0&1&0&1&1\end{array} \right] \xrightarrow{3R_2+ R_1} \left[\begin{array}{rrr|rrr}1&0&0&\frac{3}{2}&-5&-6\\0&1&0&\frac{1}{2}&-2&-2\\0&0&1&0&1&1\end{array} \right] =[I_3\;|\;A^{-1}]$ Thus $$A^{-1}=\left[\begin{array}{rrr}\frac{3}{2}&-5&-6\\ \frac{1}{2}&-2&-2\\0&1&1\end{array} \right]$$. Notice how elementary matrices $$E_{12}$$, $$E_{13}(1)$$, $$E_{32}(-4)$$, $$E_2\left(\frac{1}{2}\right)$$, $$E_{21}(3)$$ are successively applied on $$A$$ to get $$I_3$$:
$E_{21}(3) E_2\left(\frac{1}{2}\right) E_{32}(-4) E_{13}(1) E_{12}A= I_3.$ Verify that the product of those elementary matrices is $$A^{-1}$$:
$A^{-1}=E_{21}(3) E_2\left(\frac{1}{2}\right) E_{32}(-4) E_{13}(1) E_{12}.$
Remark. For an $$m\times n$$ matrix $$A$$ there is a generalized inverse called the Moore-Penrose inverse, denoted by $$A^+$$, which can be found using the singular-value decomposition of $$A$$.

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