Section 2.4 Matrix Inverses

2.4A

 

Definition: An [latex]n \times n[/latex] matrix [latex]A[/latex] is said to be invertible if there is an [latex]n \times n[/latex] matrix [latex]C[/latex] such that [latex]CA=I_{n}[/latex] and [latex]AC=I_{n}[/latex] where [latex]I_{n}[/latex] is the [latex]n \times n[/latex] identity matrix.[latex]C[/latex] is the inverse of [latex]A[/latex]

 

Theorem: If [latex]B[/latex] and [latex]C[/latex] are both inverses of [latex]A[/latex], then [latex]B = C[/latex]

 

Fact: [latex]C[/latex] is uniquely determined by [latex]A[/latex]. We can write the unique inverse of [latex]A, A^{-1}[/latex]. Hence [latex]AA^{-1} = I_{n}[/latex] and [latex]A^{-1}A = I_{n}[/latex].

 

 

Example 1: Let [latex]A=\begin{bmatrix} 1 & -2\\ 3 & -2 \end{bmatrix}[/latex], show [latex]A^{-1}=\frac{1}{4}\begin{bmatrix} -2 & 2\\ -3 & 1 \end{bmatrix}[/latex].

 

 

Exercise 1: Let [latex]A=\begin{bmatrix} 2 & 3\\ 1 & -1 \end{bmatrix}[/latex], show [latex]A^{-1}=\frac{-1}{5}\begin{bmatrix} -1 & -3\\ -1 & 2 \end{bmatrix}[/latex].

 

Definition: Let [latex]A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}[/latex] then [latex]ad - bc[/latex] is called the determinant of [latex]A[/latex]. We write det[latex]A = ad - bc[/latex].

 

Theorem: Let [latex]A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}[/latex]. If det[latex]A = ad - bc \neq 0[/latex], then [latex]A[/latex] is invertible and

 

[latex]A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}[/latex].

 

If [latex]ad - bc = 0[/latex] then [latex]A[/latex] is not invertible.

Fact: [latex]A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}[/latex] is invertible if and only if [latex]ad - bc \neq 0[/latex]

 

 

Theorem: If [latex]A[/latex] is an invertible [latex]n \times n[/latex] matrix then for each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex], the equation [latex]A\vec{x} = \vec{b}[/latex] has the unique solution [latex]\vec{x} = A^{-1}\vec{b}[/latex].

Proof:

 

Example 2: Find the unique solution of [latex]A\vec{x} = \vec{b}[/latex] without using the row operations, where [latex]A=\begin{bmatrix} 4 & 3\\ -2 & -1 \end{bmatrix}[/latex] and [latex]\vec{b}=\begin{bmatrix}5\\-7\end{bmatrix}[/latex].

 

 

Exercise 2: Find the unique solution of [latex]A\vec{x} = \vec{b}[/latex] without using the row operations, where [latex]A=\begin{bmatrix} -2 & 1\\ 5 & 1 \end{bmatrix}[/latex] and [latex]\vec{b}=\begin{bmatrix}1\\2\end{bmatrix}[/latex].

 

Theorem:

(a)If [latex]A[/latex] is an invertible matrix, then [latex]A^{-1}[/latex] is invertible and [latex](A^{-1})^{-1} = A[/latex].

(b)If [latex]A[/latex] and [latex]B[/latex] are [latex]n \times n[/latex] invertible matrices, then so is [latex]AB[/latex] and the inverse of [latex]AB[/latex] is the product of the inverse of [latex]A[/latex] and [latex]B[/latex] in the reverse order, i.e. [latex](AB)^{-1} = B^{-1}A^{-1}[/latex].

(c)If [latex]A[/latex] is an invertible matrix, then so is [latex]A^T[/latex] and the inverse of [latex]A^T[/latex] is the transpose of [latex]A^{-1}[/latex], i.e. [latex](A^T)^{-1} = (A^{-1})^T[/latex].

Proof:

 

 

Example 3: Find the inverse of [latex]AB[/latex] by using the inverse of [latex]A[/latex] and the inverse of [latex]B[/latex] without finding [latex]AB[/latex] where [latex]A=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}[/latex] and [latex]B=\begin{bmatrix} 2 & 3\\ -1 & 2 \end{bmatrix}[/latex].

 

 

Exercise 3: Find the inverse of [latex]A^T[/latex] by using the inverse of [latex]A[/latex] without finding [latex]A^T[/latex] where [latex]A=\begin{bmatrix} 2 & 3\\ 1 & -1 \end{bmatrix}[/latex].

 

Group Work 1: Mark each statement True or False. Justify each answer.

a. In order for a matrix [latex]B[/latex] to be the inverse of [latex]A[/latex], the equation [latex]AB = I[/latex] and [latex]BA = I[/latex] must both be true.

 

b. If [latex]A[/latex] and [latex]B[/latex] are [latex]n \times n[/latex] and invertible then the inverse of [latex]AB[/latex] is
[latex]A^{-1}B^{-1}[/latex].

 

c. If [latex]A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}[/latex] and [latex]ab - cd \neq 0[/latex] then [latex]A[/latex] is invertible.

 

d. If [latex]A[/latex] is an invertible [latex]n \times n[/latex] matrix then the equation [latex]A\vec{x} = \vec{b}[/latex] is consistent for each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex].

 

e. If [latex]A[/latex] is invertible, then elementary row operations that reduce [latex]A[/latex] to the identity [latex]I_{n}[/latex] also reduce [latex]A^{-1}[/latex] to [latex]I_{n}[/latex].

 

f. If [latex]A[/latex] is invertible then the inverse of [latex]A^{-1}[/latex] is [latex]A[/latex].

 

Group Work 2: Suppose [latex]AB = AC[/latex] where [latex]B[/latex] and [latex]C[/latex] are [latex]n \times p[/latex] matrices and [latex]A[/latex] is invertible [latex]n \times n[/latex] matrix. Show that [latex]B = C[/latex]. Is this true in general? If this is not true in general, find an example.

 

Group Work 3: Let [latex]A = \begin{bmatrix} 1 & 2\\ 1 & 3\\ 1 & 5 \end{bmatrix}[/latex]. Construct a [latex]2 \times 3[/latex] matrix [latex]C[/latex] using only 1, −1 and 0 as entries, such that [latex]CA = I_{2}[/latex]. Then compute [latex]AC[/latex] and note that [latex]AC \neq I_{3}[/latex]

 

Group Work 4: In each case either prove the assertion or give an example showing that it is false.

a. If [latex]A \neq 0[/latex] is a square matrix, then [latex]A[/latex] is invertible.

 

b. If [latex]A[/latex] and [latex]B[/latex] are both invertible, then [latex]A + B[/latex] is invertible.

 

c. If [latex]A[/latex] and [latex]B[/latex] are both invertible, then [latex](A^{-1}B)^T[/latex] is invertible.

 

d. If [latex]A^2[/latex] is invertible, then [latex]A[/latex] is invertible.

 

2.4B

 

Question: How do we find the inverse of [latex]A[/latex] when [latex]A[/latex] is not a [latex]2 \times 2[/latex] matrix?

 

Theorem: An [latex]n \times n[/latex] matrix [latex]A[/latex] is invertible if and only if [latex]A[/latex] is row equivalent to [latex]I_{n}[/latex].

 

Example 1: Find [latex]A^{-1}[/latex] where [latex]A=\begin{bmatrix} 1 & 2 & 3\\ 0 & 1 & 2\\ 1 & -1 & 0 \end{bmatrix}[/latex].

 

 

Exercise 1: Find [latex]A^{-1}[/latex] where [latex]A=\begin{bmatrix} 2 & 0 & 1\\ 1 & 3 & 2\\ 0 & -1 & 2 \end{bmatrix}[/latex].

 

Theorem: Let [latex]A[/latex] be a [latex]n \times n[/latex] square matrix. Then the following statements are equivalent.

(a) [latex]A[/latex] is an invertible matrix.

(b) [latex]A[/latex] is row equivalent to the [latex]n \times n[/latex] identity matrix.

(c) [latex]A[/latex] has [latex]n[/latex] pivot positions or rank[latex]A = n[/latex].

(d) The equation [latex]A\vec{x}=\vec{0}[/latex] has only the trivial solution.

(e) The equation [latex]A\vec{x}=\vec{b}[/latex] has at least one solution for each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex].

(f) There is an [latex]n \times n[/latex] matrix [latex]C[/latex] such that [latex]CA=I_{n}[/latex].

(g) There is an [latex]n \times n[/latex] matrix [latex]D[/latex] such that [latex]AD=I_{n}[/latex].

(h) [latex]A^T[/latex] is an invertible matrix.

 

Remark: The theorem only applies to a square matrix. If [latex]A[/latex] and [latex]B[/latex] are square matrix. If [latex]AB = I[/latex] then both [latex]A[/latex] and [latex]B[/latex] are invertible with [latex]A^{-1}=B[/latex] and [latex]B^{-1}=A[/latex].

 

 

Proof: (1) We know (a) [latex]\Leftarrow \Rightarrow[/latex] (b), (b) [latex]\Leftarrow \Rightarrow[/latex] (c), (c) [latex]\Leftarrow \Rightarrow[/latex] (d), and (a) [latex]\Rightarrow[/latex] (f). Once we show (f) [latex]\Rightarrow[/latex] (d), then a, b, c, d, f are equivalent.

(2) We know (a) [latex]\Leftarrow \Rightarrow[/latex] (b), (a) [latex]\Rightarrow[/latex] (g), once we show (g) [latex]\Rightarrow[/latex] (e) and (e) [latex]\Rightarrow[/latex] (b), then a, b, e, g are equivalent.

(3) We know (a) [latex]\Rightarrow[/latex] (h), once we show (h) [latex]\Rightarrow[/latex] (a) then a, h are equivalent.

With (1) to (3), we have all are equivalent.

 

Example 2: Show (f) implies (d) in the above theorem, i.e when there is an [latex]n \times n[/latex] matrix [latex]C[/latex] such that [latex]CA = I_{n}[/latex] then the equation [latex]A\vec{x} = \vec{0}[/latex] only has trivial solution.

 

 

Exercise 2: Show (g) implies (e) in the above theorem, i.e when there is an [latex]n \times n[/latex] matrix [latex]D[/latex] such that [latex]AD = I_{n}[/latex] then for all [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex] there is a solution for the equation [latex]A\vec{x} = \vec{b}[/latex].

 

Example 3: Show (h) implies (a), i.e. when [latex]A^T[/latex] is invertible then [latex]A[/latex] is invertible.

 

 

Exercise 3: Show (e) implies (b), i.e. when there is a solution of [latex]A\vec{x}=\vec{b}[/latex] for all \vec{b} in [latex]\mathbb{R}^n[/latex] then [latex]A[/latex] is row equivalent to [latex]I_{n}[/latex].

 

Example 4: Decide if [latex]A[/latex] is invertible [latex]A=\begin{bmatrix} 1 & 0 & -1\\ 0 & 2 & 3\\ 2 & 1 & -2 \end{bmatrix}[/latex].

 

 

Exercise 4: Decide if [latex]A[/latex] is invertible [latex]A=\begin{bmatrix} 2 & -1 & 1\\ 1 & 0 & 3\\ 0 & 1 & -2 \end{bmatrix}[/latex].

 

Definition: A transformation [latex]T: \mathbb{R}^n \rightarrow \mathbb{R}^n[/latex] is said to be invertible if there is a function [latex]S: \mathbb{R}^n \rightarrow \mathbb{R}^n[/latex] such that

[latex]S(T(\vec{x})) = \vec{x}[/latex] and [latex]T(S(\vec{x})) = \vec{x}[/latex]

or all [latex]\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex]. [latex]S[/latex] is called the inverse of [latex]T[/latex] and it is uniquely determined by [latex]T[/latex], so we write [latex]T^{-1}[/latex].

 

Theorem: Let [latex]T: \mathbb{R}^n \rightarrow \mathbb{R}^n[/latex] be a transformation and let [latex]A[/latex] be the standard matrix of [latex]T[/latex] such that [latex]T(\vec{x}) = A(\vec{x})[/latex]. Then [latex]T[/latex] is invertible if and only if [latex]A[/latex] is invertible. The inverse transform of [latex]T[/latex] is defined by [latex]S(\vec{x})=A^{-1}\vec{x}[/latex].

 

Proof:

 

Example 5: Let [latex]T[/latex] be a transformation from [latex]\mathbb{R}^2[/latex] to [latex]\mathbb{R}^2[/latex] defined by [latex]T(x_{1}, x_{2}) = (2x_{1} - x_{2}, -x_{1} + 3x_{2})[/latex]. Show [latex]T[/latex] is invertible and find [latex]T^{-1}[/latex].

 

 

Exercise 5: Let [latex]T[/latex] be a transformation from [latex]\mathbb{R}^2[/latex] to [latex]\mathbb{R}^2[/latex] defined by [latex]T(x_{1}, x_{2}) = (-x_{1} + 2x_{2}, x_{1} - 4x_{2})[/latex]. Show [latex]T[/latex] is invertible and find [latex]T^{-1}[/latex].

 

Theorem: All the following matrices are square matrices of the same size.

(a) [latex]I_{n}[/latex] is invertible and [latex]I^{-1}=I[/latex].

(b) If [latex]A[/latex] is invertible, so is [latex]A^k[/latex] for any [latex]k > 0[/latex], and [latex](A^k)^{-1} = (A^{-1})^k[/latex].

(c) If [latex]A[/latex] is invertible and [latex]a \neq 0[/latex] is a number, then [latex]aA[/latex] is invertible and [latex](aA)^{-1} = \frac{1}{a}A^{-1}[/latex].

 

Example 6: Show the statement (c) above.

 

 

Exercise 6: Show [latex](A^k)^{-1} = (A^{-1})^k[/latex] if [latex]A[/latex] is invertible.

 

Group Work 1: Mark each statement True or False. Justify each answer. All
matrices are [latex]n \times n[/latex] matrices.

a. If [latex]A\vec{x} = \vec{0}[/latex] has only trivial solution, then [latex]A[/latex] is row equivalent to [latex]I_{n}[/latex].

 

b. If [latex]A^T[/latex] is not invertible then [latex]A[/latex] is not invertible.

 

c. If there is an [latex]n \times n[/latex] matrix [latex]D[/latex] such that [latex]AD = I_{n}[/latex] then [latex]DA = I_{n}[/latex].

 

d. If the transformation [latex]\vec{x} \rightarrow A\vec{x}[/latex] maps [latex]\mathbb{R}^n[/latex] into [latex]\mathbb{R}^n[/latex] then [latex]A[/latex] is row equivalent to [latex]I_{n}[/latex].

 

e. If [latex]A\vec{x} = \vec{0}[/latex] has nontrivial solution, then [latex]A[/latex] has fewer than [latex]n[/latex] pivot positions.

 

f. If there is a [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex] such that the equation [latex]A\vec{x} = \vec{b}[/latex] is consistent then the solution is unique.

 

Group Work 2: Show that if there is a vector [latex]\vec{b}[/latex] such that [latex]A\vec{x} = \vec{b}[/latex] has more than one solution then the [latex]n \times n[/latex] matrix [latex]A[/latex] is not invertible.

 

Group Work 3: For square matrices [latex]A[/latex] and [latex]B[/latex]. Show that [latex]A = B[/latex] if and only if [latex]A^{-1}B = I[/latex].

 

Group Work 4: In each case either prove the assertion or give an example
showing that it is false

a. [latex]A[/latex] has no inverse when [latex]A[/latex] has a row of zero.

 

b. If [latex]AB = B[/latex] for some [latex]B \neq 0[/latex] then [latex]A[/latex] is invertible.

 

c. If [latex]A[/latex] is invertible then [latex]A^2 \neq 0[/latex].

 

d. If [latex]AB = 0[/latex] then one of [latex]A[/latex] or [latex]B[/latex] must be zero.

 

e. If [latex]AB = 0[/latex] and [latex]A[/latex] is invertible then [latex]B = 0[/latex].