TBIL-LA The Inverse of a Matrix (MX2)

Section 4.2 The Inverse of a Matrix (MX2)

Learning Outcomes

Determine if a matrix is invertible, and if so, compute its inverse.

Subsection 4.2.1 Warm Up

Activity 4.2.1.

Consider the matrices:

A = [\begin{array}{ccc} 1 & 5 & - 1 \\ 0 & 3 & 2 \end{array}], B = [\begin{array}{cccc} 7 & 2 & - 1 & 1 \\ 0 & 3 & 2 & - 2 \\ 1 & 1 & - 1 & - 3 \end{array}] .

Without using technology, what is the third column of the product

A B ?

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Subsection 4.2.2 Class Activities

Activity 4.2.2.

Let

A = [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}] .

Find a

3 \times 3

matrix

B

such that

B A = A,

that is,

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[\begin{array}{ccc} ? & ? & ? \\ ? & ? & ? \\ ? & ? & ? \end{array}] [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}] = [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}]

Check your guess using technology.

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Definition 4.2.3.

The identity matrix

I_{n}

(or just

I

when

n

is obvious from context) is the

n \times n

matrix

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I_{n} = [\begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & 1 \end{array}] .

It has a

1

on each diagonal element and a

0

in every other position.

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Fact 4.2.4.

For any square matrix

A,

I A = A I = A :

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[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}] = [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}] [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{ccc} 2 & 7 & - 1 \\ 0 & 3 & 2 \\ 1 & 1 & - 1 \end{array}]

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Activity 4.2.5.

Let

T : R^{n} \to R^{m}

be a linear map with standard matrix

A .

Sort the following items into three groups of statements: a group that means

T

is injective, a group that means

T

is surjective, and a group that means

T

is bijective.

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$T (\vec{x}) = \vec{b}$ has a solution for all $\vec{b} \in R^{m}$
$T (\vec{x}) = \vec{b}$ has a unique solution for all $\vec{b} \in R^{m}$
$T (\vec{x}) = \vec{0}$ has a unique solution.
The columns of $A$ span $R^{m}$
The columns of $A$ are linearly independent
The columns of $A$ are a basis of $R^{m}$
Every column of $RREF (A)$ has a pivot
Every row of $RREF (A)$ has a pivot
$m = n$ and $RREF (A) = I$

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Definition 4.2.6.

Let

T : R^{n} \to R^{n}

be a linear bijection with standard matrix

A .

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By item (B) from Activity 4.2.5 we may define an inverse map

T^{- 1} : R^{n} \to R^{n}

that defines

T^{- 1} (\vec{b})

as the unique solution

\vec{x}

satisfying

T (\vec{x}) = \vec{b},

that is,

T (T^{- 1} (\vec{b})) = \vec{b} .

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Furthermore, let

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A^{- 1} = [T^{- 1} ({\vec{e}}_{1}) \dots T^{- 1} ({\vec{e}}_{n})]

be the standard matrix for

T^{- 1} .

We call

A^{- 1}

the inverse matrix of

A,

and we also say that

A

is an invertible matrix.

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Activity 4.2.7.

Let

T : R^{3} \to R^{3}

be the linear bijection given by the standard matrix

A = [\begin{array}{ccc} 2 & - 1 & - 6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}] .

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(a)

To find

\vec{x} = T^{- 1} ({\vec{e}}_{1}),

we need to find the unique solution for

T (\vec{x}) = {\vec{e}}_{1} .

Which of these linear systems can be used to find this solution?

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$\begin{array}{cccc} 2 x_{1} & - 1 x_{2} & - 6 x_{3} & = x_{1} \\ 2 x_{1} & + 1 x_{2} & + 3 x_{3} & = 0 \\ 1 x_{1} & + 1 x_{2} & + 4 x_{3} & = 0 \end{array}$
$\begin{array}{cccc} 2 x_{1} & - 1 x_{2} & - 6 x_{3} & = x_{1} \\ 2 x_{1} & + 1 x_{2} & + 3 x_{3} & = x_{2} \\ 1 x_{1} & + 1 x_{2} & + 4 x_{3} & = x_{3} \end{array}$
$\begin{array}{cccc} 2 x_{1} & - 1 x_{2} & - 6 x_{3} & = 1 \\ 2 x_{1} & + 1 x_{2} & + 3 x_{3} & = 0 \\ 1 x_{1} & + 1 x_{2} & + 4 x_{3} & = 0 \end{array}$
$\begin{array}{cccc} 2 x_{1} & - 1 x_{2} & - 6 x_{3} & = 1 \\ 2 x_{1} & + 1 x_{2} & + 3 x_{3} & = 1 \\ 1 x_{1} & + 1 x_{2} & + 4 x_{3} & = 1 \end{array}$

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(b)

Use that system to find the solution

\vec{x} = T^{- 1} ({\vec{e}}_{1})

for

T (\vec{x}) = {\vec{e}}_{1} .

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(c)

Similarly, solve

T (\vec{x}) = {\vec{e}}_{2}

to find

T^{- 1} ({\vec{e}}_{2}),

and solve

T (\vec{x}) = {\vec{e}}_{3}

to find

T^{- 1} ({\vec{e}}_{3}) .

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(d)

Use these to write

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A^{- 1} = [T^{- 1} ({\vec{e}}_{1}) T^{- 1} ({\vec{e}}_{2}) T^{- 1} ({\vec{e}}_{3})],

the standard matrix for

T^{- 1} .

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Activity 4.2.8.

Find the inverse

A^{- 1}

of the matrix

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A = [\begin{array}{cccc} 0 & 0 & 0 & - 1 \\ 1 & 0 & - 1 & - 4 \\ 1 & 1 & 0 & - 4 \\ 1 & - 1 & - 1 & 2 \end{array}]

by computing how it transforms each of the standard basis vectors for

R^{4} :

T^{- 1} ({\vec{e}}_{1}),

T^{- 1} ({\vec{e}}_{2}),

T^{- 1} ({\vec{e}}_{3}),

and

T^{- 1} ({\vec{e}}_{4}) .

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Activity 4.2.9.

Is the matrix

[\begin{array}{ccc} 2 & 3 & 1 \\ - 1 & - 4 & 2 \\ 0 & - 5 & 5 \end{array}]

invertible?

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Yes, because its transformation is a bijection.
Yes, because its transformation is not a bijection.
No, because its transformation is a bijection.
No, because its transformation is not a bijection.

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Observation 4.2.10.

n \times n

matrix

A

is invertible if and only if

RREF (A) = I_{n} .

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Activity 4.2.11.

Let

T : R^{2} \to R^{2}

be the bijective linear map defined by

T ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} 2 x - 3 y \\ - 3 x + 5 y \end{matrix}],

with the inverse map

T^{- 1} ([\begin{matrix} x \\ y \end{matrix}]) = [\begin{matrix} 5 x + 3 y \\ 3 x + 2 y \end{matrix}] .

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(a)

Compute

(T^{- 1} \circ T) ([\begin{matrix} - 2 \\ 1 \end{matrix}]) .

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(b)

A

is the standard matrix for

T

and

A^{- 1}

is the standard matrix for

T^{- 1},

find the

2 \times 2

matrix

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A^{- 1} A = [\begin{array}{ccc} ? & ? \\ ? & ? \end{array}] .

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Observation 4.2.12.

T^{- 1} \circ T = T \circ T^{- 1}

is the identity map for any bijective linear transformation

T .

Therefore

A^{- 1} A = A A^{- 1}

equals the identity matrix

I

for any invertible matrix

A .

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Subsection 4.2.3 Individual Practice

Activity 4.2.13.

Now that we have defined the inverse of a matrix, we have the ability to solve matrix equations. In the following equations,

A, B

all denote square matrices of the same size and

I

denotes the identity matrix. For each equation, solve for

X .

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(a)

A^{- 1} X A = B

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(b)

A X A^{- 1} = B

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(c)

A B X = I

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(d)

B A X = I

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Subsection 4.2.4 Videos

Figure 43. Video: Invertible matrices

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Figure 44. Video: Finding the inverse of a matrix

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Exercises 4.2.5 Exercises

Exercises available at https://tbil.org/preview/linear-algebra/exercises/#/bank/MX2/.

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Subsection 4.2.6 Mathematical Writing Explorations

Exploration 4.2.14.

Assume

A

is an

n \times n

matrix. Prove the following are equivalent. Some of these results you have proven previously.

$A$ row reduces to the identity matrix.
For any choice of $\vec{b} \in R^{n},$ the system of equations represented by the augmented matrix $[A | \vec{b}]$ has a unique solution.
The columns of $A$ are a linearly independent set.
The columns of $A$ form a basis for $R^{n} .$
The rank of $A$ is $n .$
The nullity of $A$ is 0.
$A$ is invertible.
The linear transformation $T$ with standard matrix $A$ is injective and surjective. Such a map is called an isomorphism.

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Exploration 4.2.15.

Assume $T$ is a square matrix, and $T^{4}$ is the zero matrix. Prove that $(I - T)^{- 1} = I + T + T^{2} + T^{3} .$ You will need to first prove a lemma that matrix multiplication distributes over matrix addition.
Generalize your result to the case where $T^{n}$ is the zero matrix.

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Subsection 4.2.7 Sample Problem and Solution

Sample problem Example B.1.19.

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