Difference between revisions of "The Inverse of a Linear Transformation"

Revision as of 13:09, 29 September 2021

Inverse of an n-by-n matrix

An n-by-n matrix A is the inverse of n-by-n matrix B (and B the inverse of A) if BA = AB = I, where I is an identity matrix.

The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right. Row reduce this matrix and the right half will be the inverse. If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.

Example

Let $\mathrm {A} ={\begin{bmatrix}1&4&4\\2&5&8\\3&6&9\end{bmatrix}}$

We begin by expanding and partitioning A to include the identity matrix, and then proceed to row reduce A until we reach the identity matrix on the left-hand side.

{\begin{bmatrix}1&4&4&{\big |}&1&0&0\\2&5&8&{\big |}&0&1&0\\3&6&9&{\big |}&0&0&1\end{bmatrix}}\rightsquigarrow {\begin{bmatrix}1&4&4&{\big |}&1&0&0\\0&-3&0&{\big |}&-2&1&0\\0&-6&-3&{\big |}&-3&0&1\end{bmatrix}}\rightsquigarrow {\begin{bmatrix}1&4&4&{\big |}&1&0&0\\0&1&0&{\big |}&2/3&-1/3&0\\0&-6&-3&{\big |}&-3&0&1\end{bmatrix}}\rightsquigarrow

{\begin{bmatrix}1&0&4&{\big |}&-5/3&4/3&0\\0&1&0&{\big |}&2/3&-1/3&0\\0&0&-3&{\big |}&1&-2&1\end{bmatrix}}\rightsquigarrow {\begin{bmatrix}1&0&4&{\big |}&-5/3&4/3&0\\0&1&0&{\big |}&2/3&-1/3&0\\0&0&1&{\big |}&-1/3&2/3&-1/3\end{bmatrix}}\rightsquigarrow {\begin{bmatrix}1&0&0&{\big |}&-1/3&-4/3&4/3\\0&1&0&{\big |}&2/3&-1/3&0\\0&0&1&{\big |}&-1/3&2/3&-1/3\end{bmatrix}}

The matrix $\mathrm {B} ={\begin{bmatrix}-1/3&-4/3&4/3\\2/3&-1/3&0\\-1/3&2/3&-1/3\end{bmatrix}}$ is then the inverse of the original matrix A.

Inverse of a Linear Transformation

We now consider how to represent the inverse of a linear map. We start by recalling some facts about function inverses. Some functions have no inverse, or have an inverse on the left side or right side only.

Example: Where $\pi :\mathbb {R} ^{3}\to \mathbb {R} ^{2}$ is the projection map

{\begin{pmatrix}x\\y\\z\end{pmatrix}}\mapsto {\begin{pmatrix}x\\y\end{pmatrix}}

and $\eta :\mathbb {R} ^{2}\to \mathbb {R} ^{3}$ is the embedding

{\begin{pmatrix}x\\y\end{pmatrix}}\mapsto {\begin{pmatrix}x\\y\\0\end{pmatrix}}

the composition $\pi \circ \eta$ is the identity map on $\mathbb {R} ^{2}$ .

{\begin{pmatrix}x\\y\end{pmatrix}}{\stackrel {\eta }{\longmapsto }}{\begin{pmatrix}x\\y\\0\end{pmatrix}}{\stackrel {\pi }{\longmapsto }}{\begin{pmatrix}x\\y\end{pmatrix}}

We say $\pi$ is a left inverse map of $\eta$ or, what is the same thing, that $\eta$ is a right inverse map of $\pi$ . However, composition in the other order $\eta \circ \pi$ doesn't give the identity map— here is a vector that is not sent to itself under $\eta \circ \pi$ .

{\begin{pmatrix}0\\0\\1\end{pmatrix}}{\stackrel {\pi }{\longmapsto }}{\begin{pmatrix}0\\0\end{pmatrix}}{\stackrel {\eta }{\longmapsto }}{\begin{pmatrix}0\\0\\0\end{pmatrix}}

In fact, the projection $\pi$ has no left inverse at all. For, if $f$ were to be a left inverse of $\pi$ then we would have

{\begin{pmatrix}x\\y\\z\end{pmatrix}}{\stackrel {\pi }{\longmapsto }}{\begin{pmatrix}x\\y\end{pmatrix}}{\stackrel {f}{\longmapsto }}{\begin{pmatrix}x\\y\\z\end{pmatrix}}

for all of the infinitely many $z$ 's. But no function $f$ can send a single argument to more than one value.

(An example of a function with no inverse on either side is the zero transformation on $\mathbb {R} ^{2}$ .) Template:AnchorSome functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right. For instance, the map given by ${\vec {v}}\mapsto 2\cdot {\vec {v}}$ has the two-sided inverse ${\vec {v}}\mapsto (1/2)\cdot {\vec {v}}$ . In this subsection we will focus on two-sided inverses. The appendix shows that a function has a two-sided inverse if and only if it is both one-to-one and onto. Template:AnchorThe appendix also shows that if a function $f$ has a two-sided inverse then it is unique, and so it is called "the" inverse, and is denoted $f^{-1}$ . So our purpose in this subsection is, where a linear map $h$ has an inverse, to find the relationship between ${\rm {Rep}}_{B,D}(h)$ and ${\rm {Rep}}_{D,B}(h^{-1})$ (recall that we have shown, in Theorem II.2.21 of Section II of this chapter, that if a linear map has an inverse then the inverse is a linear map also).

Template:TextBox

Because of the correspondence between linear maps and matrices, statements about map inverses translate into statements about matrix inverses.

Template:TextBox

Template:AnchorHere is the arrow diagram giving the relationship between map inverses and matrix inverses. It is a special case of the diagram for function composition and matrix multiplication.

Beyond its place in our general program of seeing how to represent map operations, another reason for our interest in inverses comes from solving linear systems. A linear system is equivalent to a matrix equation, as here.

{\begin{array}{*{2}{rc}r}x_{1}&+&x_{2}&=&3\\2x_{1}&-&x_{2}&=&2\end{array}}\quad \Longleftrightarrow \quad {\begin{pmatrix}1&1\\2&-1\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}3\\2\end{pmatrix}}\qquad \qquad (*)

By fixing spaces and bases (e.g., $\mathbb {R} ^{2},\mathbb {R} ^{2}$ and ${\mathcal {E}}_{2},{\mathcal {E}}_{2}$ ), we take the matrix $H$ to represent some map $h$ . Then solving the system is the same as asking: what domain vector ${\vec {x}}$ is mapped by $h$ to the result ${\vec {d}}\,$ ? If we could invert $h$ then we could solve the system by multiplying ${\rm {Rep}}_{D,B}(h^{-1})\cdot {\rm {Rep}}_{D}({\vec {d}})$ to get ${\rm {Rep}}_{B}({\vec {x}})$ .

Template:TextBox

We finish by describing the computational procedure usually used to find the inverse matrix.

Template:TextBox

This procedure will find the inverse of a general $n\!\times \!n$ matrix. The $2\!\times \!2$ case is handy.

Template:TextBox Template:TextBox

We have seen here, as in the Mechanics of Matrix Multiplication subsection, that we can exploit the correspondence between linear maps and matrices. So we can fruitfully study both maps and matrices, translating back and forth to whichever helps us the most.

Over the entire four subsections of this section we have developed an algebra system for matrices. We can compare it with the familiar algebra system for the real numbers. Here we are working not with numbers but with matrices. We have matrix addition and subtraction operations, and they work in much the same way as the real number operations, except that they only combine same-sized matrices. We also have a matrix multiplication operation and an operation inverse to multiplication. These are somewhat like the familiar real number operations (associativity, and distributivity over addition, for example), but there are differences (failure of commutativity, for example). And, we have scalar multiplication, which is in some ways another extension of real number multiplication. This matrix system provides an example that algebra systems other than the elementary one can be interesting and useful.

Difference between revisions of "The Inverse of a Linear Transformation"

Revision as of 13:09, 29 September 2021

Inverse of an n-by-n matrix

Example

Inverse of a Linear Transformation

Navigation menu

Personal tools

Namespaces

Variants

Views

More

Search

Navigation

Tools

@@ Line 1: / Line 1: @@
+==Inverse of an n-by-n matrix==
 An n-by-n matrix A is the inverse of n-by-n matrix B (and B the inverse of A) if BA = AB = I,
 where I is an identity matrix.
@@ Line 4: / Line 5: @@
 The inverse of an n-by-n matrix can be calculated by creating an n-by-2n matrix which has the original matrix on the left and the identity matrix on the right.  Row reduce this matrix and the right half will be the inverse.  If the matrix does not row reduce completely (i.e., a row is formed with all zeroes as its entries), it does not have an inverse.
-== Example ==
+=== Example ===
 Let <math> \mathrm{A} = \begin{bmatrix}
 & 4 & 4\\
@@ Line 74: / Line 75: @@
 \end{bmatrix}
 </math> is then the inverse of the original matrix A.
+==Inverse of a Linear Transformation==
+We now consider how to represent the inverse of a linear map. We start by recalling some facts about function
+inverses. Some functions have no inverse, or have an inverse on the left side
+or right side only.
+Example: Where
+<math> \pi:\mathbb{R}^3\to \mathbb{R}^2 </math> is the projection map
+:<math>
+\begin{pmatrix} x \\ y \\ z \end{pmatrix}
+\mapsto
+\begin{pmatrix} x \\ y \end{pmatrix}
+</math>
+and <math> \eta:\mathbb{R}^2\to \mathbb{R}^3 </math> is the embedding
+:<math>
+\begin{pmatrix} x \\ y \end{pmatrix}
+\mapsto
+\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}
+</math>
+the composition <math>\pi\circ \eta</math> is the identity map on <math>\mathbb{R}^2</math>.
+:<math>
+\begin{pmatrix} x \\ y \end{pmatrix}
+\stackrel{\eta}{\longmapsto}
+\begin{pmatrix} x \\ y \\ 0 \end{pmatrix}
+\stackrel{\pi}{\longmapsto}
+\begin{pmatrix} x \\ y \end{pmatrix}
+</math>
+We say <math>\pi</math> is a '''left inverse map'''
+of <math>\eta</math> or, what is the same thing,
+that <math>\eta</math> is a '''right inverse map'''
+of <math>\pi</math>.
+However, composition in the other order <math>\eta\circ \pi</math>
+doesn't give the identity map&mdash; here is a vector that is not
+sent to itself under <math>\eta\circ \pi</math>.
+:<math>
+\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
+\stackrel{\pi}{\longmapsto}
+\begin{pmatrix} 0 \\ 0 \end{pmatrix}
+\stackrel{\eta}{\longmapsto}
+\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
+</math>
+In fact, the projection
+<math>\pi</math> has no left inverse at all.
+For, if <math>f</math> were to be a left inverse of <math>\pi</math>
+then we would have
+:<math>
+\begin{pmatrix} x \\ y \\ z \end{pmatrix}
+\stackrel{\pi}{\longmapsto}
+\begin{pmatrix} x \\ y \end{pmatrix}
+\stackrel{f}{\longmapsto}
+\begin{pmatrix} x \\ y \\ z \end{pmatrix}
+</math>
+for all of the infinitely many <math>z</math>'s. But no function <math>f</math> can send a single argument to more than one value.
+(An example of a function with no inverse on either side
+is the zero transformation on <math>\mathbb{R}^2</math>.)
+{{anchor|def:two sided inverse}}Some functions have a
+'''two-sided inverse map''', another function
+that is the inverse of the first, both from the left and from the right.
+For instance, the map given by
+<math>\vec{v}\mapsto 2\cdot \vec{v}</math> has the two-sided inverse
+<math>\vec{v}\mapsto (1/2)\cdot\vec{v}</math>.
+In this subsection we will focus on two-sided inverses.
+The appendix shows that a function
+has a two-sided inverse if and only if it is both one-to-one and onto.
+{{anchor|def:inverse function}}The appendix also shows that if a function <math>f</math> has a two-sided inverse then
+it is unique, and so it is called
+"the" inverse, and is denoted <math>f^{-1}</math>.
+So our purpose in this subsection is, where a linear map <math>h</math> has an inverse,
+to find the relationship between <math>{\rm Rep}_{B,D}(h)</math> and <math>{\rm Rep}_{D,B}(h^{-1})</math>
+(recall that we have shown, in [[Linear Algebra/Rangespace and Nullspace#th:OOHomoEquivalence|Theorem II.2.21]]<!--\ref{th:OOHomoEquivalence}-->
+of Section II of this chapter, that if a linear map has an inverse
+then the inverse is a linear map also).
+{{TextBox|1=
+;Definition 4.2{{anchor|def:inverse matrix}}:
+A matrix <math> G </math> is a '''left inverse matrix''' of the matrix <math> H </math> if <math> GH </math> is the identity matrix. It is a '''right inverse matrix''' if <math> HG </math> is the identity. A matrix <math>H</math> with a two-sided inverse is an '''invertible matrix'''. That two-sided inverse is called '''the inverse matrix''' and is denoted <math> H^{-1} </math>.
+}}
+Because of the correspondence between linear maps and matrices,
+statements about map inverses translate into statements about matrix inverses.
+{{TextBox|1=
+;Lemma 4.3{{anchor|le:LeftAndRightInvEqual}}:<!--\label{le:LeftAndRightInvEqual}-->
+If a matrix has both a left inverse and a right inverse then the two are equal.
+}}
+{{TextBox|1=
+;Theorem 4.4{{anchor|thm:invertable iff nonsingular}}:
+A matrix is invertible if and only if it is nonsingular.
+}}
+{{TextBox|1=
+;Proof:
+''(For both results.)'' Given a matrix <math>H</math>, fix spaces of appropriate dimension for the domain and codomain. Fix bases for these spaces. With respect to these bases, <math>H</math> represents a map <math>h</math>. The statements are true about the map and therefore they are true about the matrix.
+}}
+{{TextBox|1=
+;Lemma 4.5{{anchor|lem:ProdInvIsInv}}: <!--\label{lem:ProdInvIsInv}-->
+A product of invertible matrices is invertible&mdash; if <math> G </math> and <math> H </math> are invertible and if <math> GH </math> is defined then <math> GH </math> is invertible and <math> (GH)^{-1}=H^{-1}G^{-1} </math>.
+}}
+{{TextBox|1=
+;Proof:
+''(This is just like the prior proof except that it requires two maps.)''
+Fix appropriate spaces and bases and consider the represented maps <math> h </math> and
+<math> g </math>.
+Note that <math> h^{-1}g^{-1} </math> is a two-sided map inverse of <math> gh </math> since
+<math>
+(h^{-1}g^{-1})(gh)
+=
+h^{-1}(\mbox{id})h
+=h^{-1}h
+=\mbox{id}
+</math>
+and
+<math>
+(gh)(h^{-1}g^{-1})
+=
+g(\mbox{id})g^{-1}
+=gg^{-1}
+=\mbox{id}
+</math>.
+This equality is reflected in the matrices representing the maps, as required.
+}}
+{{anchor|arrow diagram}}Here is the arrow diagram giving the relationship
+between map inverses and matrix inverses.
+It is a special case
+of the diagram for function composition and matrix multiplication.
+<center>
+[[Image:Linalg_matinv_arrow.png|x200px]]
+</center>
+Beyond its place in our general program of
+seeing how to represent map operations,
+another reason for our interest in inverses comes from solving
+linear systems.
+A linear system is equivalent to a matrix equation, as here.
+:<math>
+\begin{array}{*{2}{rc}r}
+x_1  &+  &x_2  &=  &3  \\
+x_1  &-  &x_2  &=  &2
+\end{array}
+\quad\Longleftrightarrow\quad
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}
+\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}
+=
+\begin{pmatrix} 3 \\ 2 \end{pmatrix}
+\qquad\qquad (*)</math>
+By fixing spaces and bases (e.g., <math>\mathbb{R}^2,\mathbb{R}^2</math> and <math>\mathcal{E}_2,\mathcal{E}_2</math>),
+we take the matrix <math>H</math> to represent some map <math>h</math>.
+Then solving the system is the same as
+asking: what domain vector <math>\vec{x}</math> is mapped by <math>h</math> to the result
+<math>\vec{d}\,</math>?
+If we could invert <math>h</math> then we could solve the system
+by multiplying <math>{\rm Rep}_{D,B}(h^{-1})\cdot{\rm Rep}_{D}(\vec{d})</math>
+to get <math>{\rm Rep}_{B}(\vec{x})</math>.
+{{TextBox|1=
+;Example 4.6{{anchor|ex:InverseByLinSys}}:  <!--\label{ex:InverseByLinSys}-->
+We can find a left inverse for the matrix just given
+:<math>
+\begin{pmatrix}
+m  &n  \\
+p  &q
+\end{pmatrix}
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}
+=
+\begin{pmatrix}
+  &0  \\
+  &1
+\end{pmatrix}
+</math>
+by using Gauss' method to solve the resulting linear system.
+:<math>
+\begin{array}{*{4}{rc}r}
+m  &+  &2n  &    &   &   &    &=  &1     \\
+m  &-  &n   &    &   &   &    &=  &0     \\
+&   &    &    &p  &+  &2q  &=  &0     \\
+&   &    &    &p  &-  &q   &=  &1
+\end{array}
+</math>
+Answer: <math> m=1/3 </math>, <math> n=1/3 </math>, <math> p=2/3 </math>, and <math> q=-1/3 </math>.
+This matrix is actually the two-sided inverse of <math>H</math>,
+as can easily be checked.
+With it we can solve the system (<math>*</math>) above by
+applying the inverse.
+:<math>
+\begin{pmatrix} x \\ y \end{pmatrix}
+=\begin{pmatrix}
+/3  &1/3  \\
+/3  &-1/3
+\end{pmatrix}
+\begin{pmatrix} 3 \\ 2 \end{pmatrix}
+=\begin{pmatrix} 5/3 \\ 4/3 \end{pmatrix}
+</math>
+}}
+{{TextBox|1=
+;Remark 4.7:
+Why solve systems this way, when
+Gauss' method takes less arithmetic
+(this assertion can be made precise by counting the
+number of arithmetic operations,
+as computer algorithm designers do)?
+Beyond its conceptual appeal of fitting into our program of
+discovering how to represent the various map operations,
+solving linear systems by using the matrix inverse has
+at least two advantages.
+First, once the work of finding an inverse has been done,
+solving a system with the
+same coefficients but different constants is easy and fast: if
+we change the entries on the right of the system (<math>*</math>)
+then we get a related problem
+:<math>
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}
+\begin{pmatrix} x \\ y \end{pmatrix}
+=
+\begin{pmatrix} 5 \\ 1 \end{pmatrix}
+</math>
+with a related solution method.
+:<math>
+\begin{pmatrix} x \\ y \end{pmatrix}
+=
+\begin{pmatrix}
+/3  &1/3  \\
+/3  &-1/3
+\end{pmatrix}
+\begin{pmatrix} 5 \\ 1 \end{pmatrix}
+=
+\begin{pmatrix} 2 \\ 3 \end{pmatrix}
+</math>
+In applications, solving many systems having the same matrix of
+coefficients is common.
+Another advantage of inverses is that we can
+explore a system's sensitivity to changes in the constants.
+For example, tweaking the <math>3</math> on the right of the system (<math>*</math>) to
+:<math>
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}
+\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}
+=
+\begin{pmatrix} 3.01 \\ 2 \end{pmatrix}
+</math>
+can be solved with the inverse.
+:<math>
+\begin{pmatrix}
+/3  &1/3  \\
+/3  &-1/3
+\end{pmatrix}
+\begin{pmatrix} 3.01 \\ 2 \end{pmatrix}
+=
+\begin{pmatrix} (1/3)(3.01)+(1/3)(2) \\ (2/3)(3.01)-(1/3)(2) \end{pmatrix}
+</math>
+to show that <math> x_1 </math> changes by <math> 1/3 </math> of the tweak while <math> x_2 </math> moves by <math> 2/3 </math> of that tweak. This sort of analysis is used, for example, to decide how accurately data must be specified in a linear model to ensure that the solution has a desired accuracy.
+}}
+We finish by describing the computational procedure
+usually used to find the inverse matrix.
+{{TextBox|1=
+;Lemma 4.8{{anchor|lem:ComputeInvMat}}: <!--\label{lem:ComputeInvMat}-->
+A matrix is invertible if and only if it can be written as the product of elementary reduction matrices. The inverse can be computed by applying to the identity matrix the same row steps, in the same order, as are used to Gauss-Jordan reduce the invertible matrix.
+}}
+{{TextBox|1=
+;Proof:
+A matrix <math>H</math> is invertible if and only if it is nonsingular and thus
+Gauss-Jordan reduces to the identity.
+By [[Linear Algebra/Mechanics of Matrix Multiplication#cor:ReducViaMatrices|Corollary 3.22]]<!--\ref{cor:ReducViaMatrices}--> this reduction can
+be done with elementary matrices
+<math> R_r\cdot R_{r-1}\dots R_1\cdot H=I </math>.
+This equation gives the two halves of the result.
+First, elementary matrices are invertible and their inverses are also
+elementary.
+Applying <math>R_r^{-1}</math> to the left of both sides of that equation, then
+<math>R_{r-1}^{-1}</math>, etc., gives <math>H</math> as the product of
+elementary matrices <math>H=R_1^{-1}\cdots R_r^{-1}\cdot I</math>
+(the <math>I</math> is here to cover the trivial <math>r=0</math> case).
+Second, matrix inverses are unique and so comparison of the above equation with <math>H^{-1}H=I</math> shows that <math>H^{-1}=R_r\cdot R_{r-1}\dots R_1\cdot I</math>. Therefore, applying <math>R_1</math> to the identity, followed by <math>R_2</math>, etc., yields the inverse of <math>H</math>.
+}}
+{{TextBox|1=
+;Example 4.9:
+To find the inverse of
+:<math>
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}
+</math>
+we do Gauss-Jordan reduction, meanwhile performing the same operations on
+the identity.
+For clerical convenience we write the matrix and the identity side-by-side,
+and do the reduction steps together.
+:<math>\begin{array}{rcl}
+\left(\begin{array}{cc|cc}
+  &1   &1  &0  \\
+  &-1  &0  &1
+\end{array}\right)
+&\xrightarrow[]{-2\rho_1+\rho_2}
+&\left(\begin{array}{cc|cc}
+  &1   &1  &0  \\
+  &-3  &-2 &1
+\end{array}\right)                 \\
+&\xrightarrow[]{-1/3\rho_2}
+&\left(\begin{array}{cc|cc}
+  &1   &1   &0     \\
+  &1   &2/3 &-1/3
+\end{array}\right)                   \\
+&\xrightarrow[]{-\rho_2+\rho_1}
+&\left(\begin{array}{cc|cc}
+  &0   &1/3 &1/3  \\
+  &1   &2/3 &-1/3
+\end{array}\right)
+\end{array}
+</math>
+This calculation has found the inverse.
+:<math>
+\begin{pmatrix}
+  &1  \\
+  &-1
+\end{pmatrix}^{-1}
+=
+\begin{pmatrix}
+/3 &1/3  \\
+/3 &-1/3
+\end{pmatrix}
+</math>
+}}
+{{TextBox|1=
+;Example 4.10{{anchor|exam:ThreeByThreeMatInv}}: <!--\label{exam:ThreeByThreeMatInv}-->
+This one happens to start with a row swap.
+:<math>\begin{array}{rcl}
+\left(\begin{array}{ccc|ccc}
+  &3  &-1  &1  &0  &0  \\
+  &0  &1   &0  &1  &0  \\
+  &-1 &0   &0  &0  &1
+\end{array}\right)
+&\xrightarrow[]{\rho_1\leftrightarrow\rho_2}
+&   \left(\begin{array}{ccc|ccc}
+  &0  &1   &0  &1  &0  \\
+  &3  &-1  &1  &0  &0  \\
+  &-1 &0   &0  &0  &1
+\end{array}\right)                             \\
+&\xrightarrow[]{-\rho_1+\rho_3}
+&   \left(\begin{array}{ccc|ccc}
+  &0  &1   &0  &1  &0  \\
+  &3  &-1  &1  &0  &0  \\
+  &-1 &-1  &0  &-1 &1
+\end{array}\right)                             \\
+&\vdots                                  \\
+&\xrightarrow[]{}
+&   \left(\begin{array}{ccc|ccc}
+  &0  &0   &1/4  &1/4  &3/4  \\
+  &1  &0   &1/4  &1/4  &-1/4 \\
+  &0  &1   &-1/4 &3/4  &-3/4
+\end{array}\right)
+\end{array}
+</math>
+}}
+{{TextBox|1=
+;Example 4.11:
+A non-invertible matrix is detected by the fact that the left half won't
+reduce to the identity.
+:<math>
+\left(\begin{array}{cc|cc}
+  &1   &1  &0  \\
+  &2   &0  &1
+\end{array}\right)
+\xrightarrow[]{-2\rho_1+\rho_2}
+\left(\begin{array}{cc|cc}
+  &1   &1  &0  \\
+  &0   &-2 &1
+\end{array}\right)
+</math>
+}}
+This procedure will find the inverse of a general <math>n \! \times \! n</math> matrix.
+The <math>2 \! \times \! 2</math> case is handy.
+{{TextBox|1=
+;Corollary 4.12{{anchor|cor:TwoByTwoInv}}:<!--\label{cor:TwoByTwoInv}-->
+The inverse for a <math> 2 \! \times \! 2 </math> matrix exists and equals
+:<math>
+\begin{pmatrix}
+a  &b  \\
+c  &d
+\end{pmatrix}^{-1}
+=
+\frac{1}{ad-bc}
+\begin{pmatrix}
+d  &-b \\
+-c  &a
+\end{pmatrix}
+</math>
+if and only if <math> ad-bc\neq 0 </math>.
+}}
+{{TextBox|1=
+;Proof:
+This computation is [[#exer:TwoByTwoInv|Problem 10]]<!--\ref{exer:TwoByTwoInv}-->.
+}}
+We have seen here, as in the Mechanics of Matrix Multiplication subsection,
+that we can exploit the correspondence between
+linear maps and matrices.
+So we can fruitfully study both maps and matrices, translating back and forth
+to whichever helps us the most.
+Over the entire four subsections of
+this section we have developed an algebra system for matrices.
+We can compare it with the familiar algebra system for the real numbers.
+Here we are working not with numbers but with matrices.
+We have matrix addition and subtraction operations,
+and they work in much the same
+way as the real number operations, except that they only combine same-sized
+matrices.
+We also have a matrix multiplication operation
+and an operation inverse to multiplication.
+These are somewhat like the familiar real number operations
+(associativity, and distributivity over addition, for example), but
+there are differences (failure of commutativity, for example).
+And, we have scalar multiplication, which is in some ways another extension
+of real number multiplication.
+This matrix system provides an example that algebra
+systems other than the
+elementary one can be interesting and useful.