← Back to Math 010

1.4 Inverses; Algebraic Properties of Matrices: Class Notes

Theorem 1.4.1 Properties of Matrix Arithmetic

There are two distributive laws, (d) and (e), because matrix multiplication is not commutative: $AB$ and $BA$ are usually different. Consider sizes:

If $A$ is $2\times 3$ and $B$ is $3\times 2$, then $AB$ is $2\times 2$, but $BA$ ($3\times 2$ times $2\times 3$) is $3\times 3$.

If $A$ is $3\times 3$ and $B$ is $3\times 2$, then $AB$ is defined, but $BA$ ($3\times 2$ times $3\times 3$) is not defined at all.

Proof of (h)

For $a(B+C)$ to be equal to $aB+aC$, they must have the same size and equal entries. If $B+C$ is defined (as assumed), then $B$ and $C$ have the same size, say $m\times n$. Scalar multiplication does not change the size of a matrix, so $a(B+C)$, $aB$, and $aC$ are all $m\times n$ matrices. Thus $aB+aC$ is an $m\times n$ matrix.

$(a(B+C){)}_{ij}$ is the entry in the $i$th row and $j$th column of the matrix on the left-hand side. Then

$$\begin{array}{ccc}\hfill {\displaystyle (a(B+C){)}_{ij}}& {\displaystyle =a(B+C{)}_{ij}}\hfill & {\displaystyle \text{scalar multiplication}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =a(b_{ij}+c_{ij})}\hfill & {\displaystyle \text{matrix addition}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =a{b}_{ij}+a{c}_{ij}}\hfill & {\displaystyle \text{distributive property of real numbers}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(aB{)}_{ij}+(aC{)}_{ij}}\hfill & {\displaystyle \text{definition of scalar multiplication}}\hfill \end{array}$$

Since corresponding entries are equal, $a(B+C)=aB+aC$.

Definition

Definition

Theorem 1.4.2 Properties of Zero Matrices

Some rules for real numbers do not carry over to matrices:

We can have $A\ne O$ and $B\ne O$ but $AB=O$.

Also, if $AC=BC$, it does not always follow that $A=B$. To disprove the statement “if $AC=BC$, then $A=B$”, a counterexample is enough: let $C=O$ and let $A\ne B$. Then $AC=O=BC$, but $A\ne B$.

Definition

$$I_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],I_3=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],I_n=\left[\begin{array}{cccc}1& 0& \cdots & 0\\ 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 1\end{array}\right]$$

Representing matrix multiplication. Let $A$ be $m\times k$ and $B$ be $k\times n$:

$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1k}\\ a_{21}& a_{22}& \cdots & a_{2k}\\ ⋮& ⋮& & ⋮\\ a_{i1}& a_{i2}& \cdots & a_{ik}\\ ⋮& ⋮& & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mk}\end{array}\right],B=\left[\begin{array}{cccccc}{b}_{11}& b_{12}& \cdots & b_{1j}& \cdots & b_{1n}\\ b_{21}& b_{22}& \cdots & b_{2j}& \cdots & b_{2n}\\ ⋮& ⋮& & ⋮& & ⋮\\ b_{k1}& b_{k2}& \cdots & b_{kj}& \cdots & b_{kn}\end{array}\right]$$

The entry in the $i$th row and $j$th column of $AB$ comes from row $i$ of $A$ and column $j$ of $B$. The matching inner subscripts run from 1 to $k$:

$$(AB{)}_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+\cdots +a_{ik}{b}_{kj}=\sum _{r=1}^k{a}_{ir}{b}_{rj}$$

For example, let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$. Then

$$A{I}_2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=I_{2}A$$

Theorem 1.4.3

Tip: No need to memorize theorem numbers; know what each theorem says.

Proof

An $n\times n$ matrix has at most $n$ pivots. If there are $n$ pivots in $R$, then $R=I_n$ by the definition of reduced row echelon form. If there are fewer than $n$ pivots, then at least one row does not have a leading 1. So at least one row is all zeros. (This uses the definitions of pivot and of reduced row echelon form.)

Definition

Why is the inverse unique? That is the next theorem.

Theorem 1.4.4 Uniqueness of a Matrix Inverse

Proof

If $B$ and $C$ are both inverses of $A$, then

$$\begin{array}{ccc}\hfill {\displaystyle B}& {\displaystyle =BI}\hfill & {\displaystyle \text{definition of }I}\hfill \\ \hfill {\displaystyle }& {\displaystyle =B(AC)}\hfill & {\displaystyle C\text{ is an inverse of }A}\hfill \\ \hfill {\displaystyle }& {\displaystyle =(BA)C}\hfill & {\displaystyle \text{associative property of matrix multiplication}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =IC}\hfill & {\displaystyle B\text{ is an inverse of }A}\hfill \\ \hfill {\displaystyle }& {\displaystyle =C}\hfill & {\displaystyle \text{definition of }I}\hfill \end{array}$$

Theorem 1.4.5

So to invert a $2\times 2$ matrix: swap the diagonal entries $a$ and $d$, change the signs of $b$ and $c$, and divide by $ad-bc$.

#8

Use Theorem 1.4.5 to compute the inverse.

$$\left[\begin{array}{cc}6& 4\\ -2& -1\end{array}\right]$$

Solution

$$ad-bc=6(-1)-(-2)(4)=2$$

$${\displaystyle A^{-1}=\frac{1}{2}\left[\begin{array}{cc}-1& -4\\ 2& 6\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}& -2\\ 1& 3\end{array}\right]}$$

Check:

$${\displaystyle A^{-1}A=\left[\begin{array}{cc}-\frac{1}{2}& -2\\ 1& 3\end{array}\right]\left[\begin{array}{cc}6& 4\\ -2& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]✓}$$

Theorem 1.4.6

$A$ and $B$ are invertible, so $A^{-1}$ and $B^{-1}$ exist.

Proof

Consider $(AB)(B^{-1}{A}^{-1})$:

$$\begin{array}{ccc}\hfill {\displaystyle (AB)(B^{-1}{A}^{-1})}& {\displaystyle =A(B{B}^{-1})A^{-1}}\hfill & {\displaystyle \text{associative property}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =AI{A}^{-1}}\hfill & {\displaystyle \text{definition of inverse}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =A{A}^{-1}}\hfill & {\displaystyle \text{definition of }I}\hfill \\ \hfill {\displaystyle }& {\displaystyle =I}\hfill & {\displaystyle \text{definition of inverse}}\hfill \end{array}$$

Since multiplying $(AB)$ by $(B^{-1}{A}^{-1})$ yields $I$, $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

A similar argument is used to show $(B^{-1}{A}^{-1})(AB)=I$, which you are encouraged to complete.

Note: if $A$, $B$, and $C$ are invertible, then $(ABC{)}^{-1}=C^{-1}{B}^{-1}{A}^{-1}$, and so on for longer products.

Theorem 1.4.7 Properties of Negative Exponents

In (c), $k^{-1}$ is the reciprocal of the number $k$, while $A^{-1}$ is the inverse matrix.

#15

Use the given information to find $A$.

$$(7A{)}^{-1}=\left[\begin{array}{cc}-3& 7\\ 1& -2\end{array}\right]$$

Solution

By Theorem 1.4.7(c), ${\displaystyle (7A{)}^{-1}=\frac{1}{7}{A}^{-1}}$, so

$${\displaystyle \frac{1}{7}{A}^{-1}=\left[\begin{array}{cc}-3& 7\\ 1& -2\end{array}\right]⟹A^{-1}=\left[\begin{array}{cc}-21& 49\\ 7& -14\end{array}\right]}$$

By Theorem 1.4.7(a), $A={\left(A^{-1}\right)}^{-1}$. Use Theorem 1.4.5:

$${\displaystyle A={\left[\begin{array}{cc}-21& 49\\ 7& -14\end{array}\right]}^{-1}=\frac{1}{(-21)(-14)-(49)(7)}\left[\begin{array}{cc}-14& -49\\ -7& -21\end{array}\right]}$$

$${\displaystyle =\frac{1}{-49}\left[\begin{array}{cc}-14& -49\\ -7& -21\end{array}\right]=\left[\begin{array}{cc}\frac{2}{7}& 1\\ \frac{1}{7}& \frac{3}{7}\end{array}\right]}$$

Another way: invert both sides first. $7A={[(7A{)}^{-1}]}^{-1}$, so

$${\displaystyle 7A={\left[\begin{array}{cc}-3& 7\\ 1& -2\end{array}\right]}^{-1}=\frac{1}{6-7}\left[\begin{array}{cc}-2& -7\\ -1& -3\end{array}\right]=\left[\begin{array}{cc}2& 7\\ 1& 3\end{array}\right]⟹A=\frac{1}{7}\left[\begin{array}{cc}2& 7\\ 1& 3\end{array}\right]}$$

#23

Find all values of $a$, $b$, $c$, and $d$ (if any) for which the matrices $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ and $B=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ commute.

Solution

The matrices commute when $AB=BA$.

$$AB=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& a\\ 0& c\end{array}\right]$$

$$BA=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}c& d\\ 0& 0\end{array}\right]$$

Setting them equal, corresponding entries must match:

$$\left[\begin{array}{cc}0& a\\ 0& c\end{array}\right]=\left[\begin{array}{cc}c& d\\ 0& 0\end{array}\right]$$

The first entry of row 1 and the last entry of row 2 both say $c=0$, and the second entry of row 1 says $a=d$. Nothing restricts $b$.

#25

Solve the system using an inverse matrix.

$$\begin{array}{cc}\hfill {\displaystyle 3x_1-2x_2}& {\displaystyle =-1}\hfill \\ \hfill {\displaystyle 4x_1+5x_2}& {\displaystyle =3}\hfill \end{array}$$

Solution

Write the system as a matrix equation, in the form $A\overrightarrow{x}=\overrightarrow{b}$:

$$\left[\begin{array}{cc}3& -2\\ 4& 5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}-1\\ 3\end{array}\right]$$

Multiply both sides on the left by $A^{-1}$:

$$A^{-1}A\overrightarrow{x}=A^{-1}\overrightarrow{b}⟹\overrightarrow{x}=A^{-1}\overrightarrow{b}$$

Tip: This is like solving $3x=15$ by multiplying both sides by ${\displaystyle \frac{1}{3}}$ to get $x=5$.

$${\displaystyle A=\left[\begin{array}{cc}3& -2\\ 4& 5\end{array}\right]⟹A^{-1}=\frac{1}{15+8}\left[\begin{array}{cc}5& 2\\ -4& 3\end{array}\right]}$$

$${\displaystyle \overrightarrow{x}=\frac{1}{23}\left[\begin{array}{cc}5& 2\\ -4& 3\end{array}\right]\left[\begin{array}{c}-1\\ 3\end{array}\right]=\frac{1}{23}\left[\begin{array}{c}1\\ 13\end{array}\right]}$$

#39

Simplify the expression assuming that $A$, $B$, $C$ and $D$ are invertible.

$$(AB{)}^{-1}\left(A{C}^{-1}\right){\left(D^{-1}{C}^{-1}\right)}^{-1}{D}^{-1}$$

Solution

Use Theorem 1.4.6 on $(AB{)}^{-1}$ and on ${\left(D^{-1}{C}^{-1}\right)}^{-1}$ (reverse the order and invert each factor), with ${\left(C^{-1}\right)}^{-1}=C$ and ${\left(D^{-1}\right)}^{-1}=D$:

$$=B^{-1}{A}^{-1}A{C}^{-1}CD{D}^{-1}$$

Now $A^{-1}A=I$, $C^{-1}C=I$, and $D{D}^{-1}=I$:

$$=B^{-1}\left(A^{-1}A\right)\left(C^{-1}C\right)\left(D{D}^{-1}\right)=B^{-1}III=B^{-1}$$

Theorem 1.4.8 Properties of the Transpose

In (e), if $A$ is $m\times k$ and $B$ is $k\times n$, then $B^T$ is $n\times k$ and $A^T$ is $k\times m$, so the order has to reverse for the sizes to match.

Proof of (b)

Let $A=\left[a_{ij}\right]$ and $B=\left[b_{ij}\right]$ be matrices of the same size. Then

$$\begin{array}{cc}\hfill {\displaystyle {((A+B{)}^T)}_{ij}}& {\displaystyle =(A+B{)}_{ji}=(A{)}_{ji}+(B{)}_{ji}}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\left(A^T\right)}_{ij}+{\left(B^T\right)}_{ij}}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={(A^T+B^T)}_{ij}}\hfill \end{array}$$

So $(A+B{)}^T=A^T+B^T$.

Tip: You try: prove (e) and Theorem 1.4.9 yourself before reading the proofs below.

Proof of (e)

Let $A=\left[a_{ij}\right]$ be an $n\times r$ matrix and $B=\left[b_{ij}\right]$ be an $r\times m$ matrix. ${((AB{)}^T)}_{ij}$ is the entry in the $i$th row and $j$th column of $(AB{)}^T$.

$$\begin{array}{cc}\hfill {\displaystyle {((AB{)}^T)}_{ij}}& {\displaystyle =(AB{)}_{ji}=\sum _{k=1}^r{a}_{jk}{b}_{ki}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\sum _{k=1}^r{b}_{ki}{a}_{jk}}\hfill & {\displaystyle \text{multiplication of real numbers is commutative}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\sum _{k=1}^r{\left(B^T\right)}_{ik}{\left(A^T\right)}_{kj}}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\left(B^T{A}^T\right)}_{ij}}\hfill \end{array}$$

Theorem 1.4.9

Proof

Since $A$ is invertible, $A^{-1}$ exists. By Theorem 1.4.8(e),

$$A^T{\left(A^{-1}\right)}^T={\left(A^{-1}A\right)}^T=I^T=I$$

because $I^T=I$. Likewise,

$${\left(A^{-1}\right)}^T{A}^T={\left(A{A}^{-1}\right)}^T=I^T=I$$

So $A^T$ is invertible, with inverse ${\left(A^{-1}\right)}^T$.