The term inverse you've met many times already with numbers: the number $\frac{1}{3}$ is the inverse of $3$ in the sense that $3 \times \frac{1}{3} \,=\, 1$, while the solution of the equation $3 x = 4$ is $x = \frac{4}{3}$ etc . We are now going to extend these ideas to matrices and use them to solve matrix equations $A {\bf x} = {\bf b}$, for instance. Recall that the Identity matrix $I_n$ is the $n \times n$ matrix $$I_n \ = \ \left[\begin{array}{cc} 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1\end{array}\right]\,.$$ It's called the identity because $A I_n = A$ for any $m \times n$ matrix $A$, while $I_n B = B$ for any $n \times p$ matrix $B$; in other words, it does for matrices what the number $1$ does for multiplication of numbers.
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Definition 9.1: an $n \times n$ matrix $A$ is said to
be Invertible when there exists an $n \times n$ matrix $B$
such that $AB = I_n$ and $BA = I_n$ where $I_n$ is the $n \times n$
identity matrix.
If $A$ is invertible it is often said to be Non-Singular, while $A$ is said to be Singular when it is not invertible. |
Notice that invertibility is defined only for square matrices because $A B = I_n = B A$ cannot hold unless $A, \,B$ are both $n \times n$. Other very useful properties also follow quickly from Definition 9.1.
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Inverse Property 9.1: (Uniqueness) if $A$ is invertible, then it can have only one inverse. This unique inverse is usually denoted by $A^{-1}$. Proof: if $A',\, A''$ are both inverses of $A$, then $$A'\,=\, A' I_n \,=\, A'(A A'')\qquad $$ $$\qquad (A'A)A'' \,=\, I_n A'' \,=\,A'',$$ so $A' = A''$. |
Inverse Property 9.2: (Double Inverse) if $A$ is invertible, then $(A^{-1})^{-1}\,=\,A.$ Proof: if $A$ is invertible, then $A^{-1} A = I_n = A A^{-1}$, so $$A^{-1}B \ = \ I_n \ = \ BA^{-1}$$ holds, i.e. $A^{-1}$ is invertible with $B = A$. Thus $(A^{-1})^{-1} \,=\, A$. |
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Inverse Property 9.3: if $A,\, B$ are invertible $n \times n$ matrices, then the product $AB$ is invertible and $$(AB)^{-1} \ = \ B^{-1}A^{-1}\,.$$ Proof: if $A,\, B$ are invertible, then $$(AB)(B^{-1}A^{-1}) \,=\, A(B B^{-1})A^{-1} \,=\, AI_n A^{-1} \,=\, I_n\,.$$ Similarly, $$ (B^{-1}A^{-1}) (AB) \ = \ I_n\,.$$ Thus $AB$ is invertible with $(AB)^{-1} \ = \ B^{-1}A^{-1}$. |
Inverse Property 9.4: (Matrix Equation) if $A$ is an invertible $n \times n$ matrix, then $A {\bf x} \,=\, {\bf b}$ has a unique solution ${\bf x} \, =\, A^{-1} {\bf b}$ for each ${\bf b}$ in ${\mathbb R}^n$. Proof: since $$A (A^{-1}{\bf b}) = (A A^{-1}){\bf b} = {\bf b}\,,$$ ${\bf x} = A^{-1}{\bf b}$ is one solution of $A {\bf x} = {\bf b}$. But if ${\bf y}$ also is a solution, then, $$ {\bf y}\,=\, I_n {\bf y} \,=\, (A^{-1}A){\bf y}\,=\, A^{-1}(A {\bf y}) \,=\, A^{-1} {\bf b} \,=\, {\bf x},$$ so ${\bf x} = A^{-1}{\bf b}$ is the only solution. |
Invertible $2 \times 2$ matrices are easy to describe. Recall that the DETERMINANT, $ \text{det}(A),$ of a $2 \times 2$ matrix $A$ is defined by $$ \det(A) \ = \ \det\left(\left[\begin{array}{cc} a & b\\ c & d\end{array}\right]\right) \ = \ ad - bc\,.$$
| Fundamental Theorem 9.1: let $A = \left[\begin{array}{cc} a & b\\ c & d\end{array}\right]$ be a $2 \times 2$ such that $\det(A) \ne 0$. Then $A$ is invertible and $$A^{-1} \ = \ \frac{1}{\det(A) }\left[\begin{array}{cc} d & -b\\ -c & a\end{array}\right]\,.$$ If $\,\det(A) = 0$, then $A$ is not invertible. |
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Problem 9.1: solve for $x,\,y$ in $$ \begin{array}{cc}2x\, +\, y & = \ 2\,,\\ 6x\, +\, 4y & = \ 4\,. \end{array}$$ Solution: As a matrix equation the system becomes $$ A\,{\bf x} \ = \ \left[\begin{array}{cc} 2 & 1\\ 6 & 4 \end{array}\right] \left[\begin{array}{cc} x\\ y \end{array}\right] \ =\ \left[\begin{array}{cc} 2\\ 4 \end{array}\right],$$ in which case, $$ {\bf x} \ = \ \left[\begin{array}{cc} x\\ y \end{array}\right] \ =\ A^{-1}\left[\begin{array}{cc} 2\\ 4 \end{array}\right].$$ |
Now
$$A^{-1} \ = \ \frac{1}{2}\left[\begin{array}{cc} 4 & -1\\
-6 & 2\end{array}\right] \ = \ \left[\begin{array}{cc} 2 & -\textstyle{\frac{1}{2}}\\
-3 & 1\end{array}\right].$$
But then,
$$ \left[\begin{array}{cc}
x\\ y \end{array}\right] \ = \
\left[\begin{array}{cc} 2 & -\textstyle{\frac{1}{2}}\\
-3 & 1\end{array}\right]\left[\begin{array}{cc}
2\\ 4 \end{array}\right] \ = \ \left[\begin{array}{cc}
2\\ -2 \end{array}\right].$$
Thus $x\,=\, 2,\ y\,=\, -2$.
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Invertibility of $n \times n$ matrices for $n > 2$ is best left to technology (as Theorem 9.4 will soon show), but the special cases of diagonal and $3 \times 3$ triangular matrices are important conceptually and can be done by hand. In fact, direct calculation gives:
diagonal matrices: if the diagonal
entries $d_0,\, d_1,\, d_2$ are all non-zero, then
$$\left[\begin{array}{cc}
d_0 & 0 & 0 \\
0 & d_1 & 0 \\
0 & 0 & d_2 \end{array}\right]^{-1} \ = \ \left[\begin{array}{cc}
1/d_0 & 0 & 0 \\
0 & 1/d_1 & 0 \\
0 & 0 & 1/d_2 \end{array}\right]$$
triangular matrices:
$$\left[\begin{array}{cc}
1 & 0 & 0 \\
b & 1 & 0 \\
c & d & 1 \end{array}\right]^{-1} \ = \ \left[\begin{array}{cc}
1 & 0 & 0 \\
-b & 1 & 0 \\
db-c & -d & 1 \end{array}\right], \qquad \qquad \left[\begin{array}{cc}
1 & p & q \\
0 & 1 & r \\
0 & 0 & 1 \end{array}\right]^{-1} \ = \ \left[\begin{array}{cc}
1 & -p & pr-q \\
0 & 1 & -r \\
0 & 0 & 1 \end{array}\right]\,. $$
It's very helpful to remember these results!
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Problem 9.2: determine the inverse of the product $AB$ of the matrices $$A = \left[\begin{array}{cc}1 & 3 & -2\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{array}\right], \qquad B = \left[\begin{array}{cc}1 & 0 & 0\\ 2 & 1 & 0\\ -3 & 1 & 1 \end{array}\right].$$ Solution: by actually carrying out the multiplication of $A$ and $B$ we could determine $(AB)^{-1}$ by applying Theorem 9.4 to $\ [AB \ \ I_3]$ if technology is available. Since $A,\, B$ are $3 \times 3$ triangular matrices, however, the individual inverses $A^{-1}$ and $B^{-1}$ can be computed by hand and the inverse $(AB)^{-1}$ computed by hand also since $$(AB)^{-1} = B^{-1}A^{-1}\,.$$ |
Indeed, by earlier results for triangular matrices, $$ A^{-1} = \left[\begin{array}{cc}1 & -3 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right],\quad B^{-1} = \left[\begin{array}{cc}1 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & -1 & 1 \end{array}\right], $$ so that $$ B^{-1}A^{-1} \ = \ \left[\begin{array}{cc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 5 & -1 & 1\end{array}\right] \left[\begin{array}{cc} 1 & -3 & -4 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right] $$ $$ = \ \left[\begin{array}{cc} 1 & -3 & -4 \\ -2 & 7 & 10 \\ 5 & -16 & -21\end{array}\right] \ = \ (AB)^{-1}, $$ as can be confirmed with technology or by hand. |
It can be shown that an $n \times n$ matrix $A$ is invertible if and only if $\text{rref}(A) = I_n$. There are many such criteria equivalent to the invertibility of a matrix - every text usually scatters at least 25 of them throughout a text. We too start to collect them together for convenience of reference as well as for illustration of all the ideas we are developing.
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Invertible Matrix Theorem Part I. for an $n \times n$
matrix $A$ the following statements are equivalent$:$
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These criteria are worth remembering because they are used a lot in many situations (not just homework questions!!). Finally, let's introduce an algorithm for computing inverse matrices. It's proof requires modifications to Theorem 9.3. In practice, however, it is used mainly for matrices with $n$ reasonably small because there are important factorization methods based on Theorem 9.3 that require fewer operations (see the next Learning Module). Such methods are more efficient when $n$ is large.
| Fundamental Theorem 9.4: if $A$ is an $n \times n$ matrix and the augmented matrix $[A \ \ I_n]$ can be row reduced to $[ I_n\ \ B\,]$ for some $B$, then $A$ is invertible and $B = A^{-1}$. Otherwise, $A$ is singular. In matrix terms $$\text{rref}\,\big[A \ \ I_n\big] \ = \ \big[I_n \ \ A^{-1} \big]\,.$$ |
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Problem 9.3: determine $A^{-1}$ when $$A \ = \ \left[\begin{array}{cc}0 & 1 & 2\\ 1 & 0 & 3\\ 4 & -3 & 8 \end{array}\right].$$ Solution: since $$ \big[A \ \ I_3\big] \ = \ \left[\begin{array}{cc}0 & 1 & 2 & 1 & 0 & 0\\ 1 & 0 & 3 & 0 & 1 & 0\\ 4 & -3 & 8 & 0 & 0 & 1 \end{array}\right],$$ |
technology shows that $$\text{rref}\,[A \ \ I_3] = \left[\begin{array}{cc}1 & 0 & 0 & -9/2 & 7 & -3/2\\ 0 & 1 & 0 & -2 & 4 & -1\\ 0 & 0 & 1 & 3/2 & -2 & 1/2 \end{array}\right]. $$ Thus $A$ is invertible and $$ A^{-1} \ = \ \left[\begin{array}{cc} -9/2 & 7 & - 3/2\\ -2 & 4 & -1 \\ 3/2 & -2 & 1/2\end{array}\right].$$ |