Inverting a Matrix

Section 3.3 Inverting a Matrix

In this section we will introduce the definition of an invertible matrix and discuss some properties of the inverse of a matrix. We will then introduce a method to determine whether or not a matrix is invertible. Additionally, when a matrix is invertible we will be able to calculate its inverse.

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Subsection 3.3.1 Invertible Matrices

With matrix multiplication defined in terms of the composition of linear transformations, we turn to a specific composition in this subsection. Specifically, we will think about matrices for linear transformations \(S\) and \(T\) when \(S\circ T\) is the identity transformation.

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In Definition 3.1.14, we called such linear transformations invertible. When such a linear transformation can be accomplished by matrix multiplication, we will refer to the connected matrix using this same term.

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Definition 3.3.1.

Let \(A \in M_n(\ff)\text{.}\) The matrix \(B\) is an inverse matrix for \(A\) if

\begin{equation*} AB = BA = I_n\text{.} \end{equation*}

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If a matrix \(A\) has an inverse, we say that \(A\) is invertible or non-singular. If \(A\) is a matrix for which no inverse matrix exists, we say that \(A\) is singular or not invertible.

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It may strike the reader as strange that only square matrices have a chance at being invertible. (That is, we have only defined invertibility for square matrices.) There is a good reason for this, which we will explore in the exercises.

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If we know an inverse \(B\) of a matrix \(A\text{,}\) then we can solve matrix-vector equations with ease:

\begin{align*} A\bfx \amp = \bfy\\ B(A\bfx) \amp = B\bfy\\ (BA)\bfx \amp = B\bfy\\ I \bfx \amp = B\bfy\\ \bfx \amp = B\bfy\text{.} \end{align*}

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The next three propositions present some properties of matrix inverses.

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Proposition 3.3.2.

If a matrix \(A \in M_n(\ff)\) has an inverse, that inverse is unique.

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Proof.

Let \(A \in M_n(\ff)\) and suppose that both \(B,C \in M_n(\ff)\) are inverses for \(A\text{.}\) Then we have

\begin{equation*} B = BI_n = B(AC) = (BA)C = I_nC = C\text{.} \end{equation*}

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This proposition allows us to refer to the inverse of a matrix \(A\) and to use the notation \(A^{-1}\) for that matrix.

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Proposition 3.3.3.

Suppose that \(A,B \in M_n(\ff)\) are both invertible. Then \(AB\) is invertible as well and

\begin{equation*} (AB)^{-1} = B^{-1}A^{-1}\text{.} \end{equation*}

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Proof.

If \(A\) and \(B\) are both invertible, then both \(A^{-1}\) and \(B^{-1}\) exist. Since matrix multiplication is associative, the following calculations show that the matrix \(B^{-1}A^{-1}\) satisfies the properties of the inverse of \(AB\text{,}\) thereby making \(AB\) invertible:

\begin{align*} (AB)(B^{-1}A^{-1}) \amp = A(BB^{-1})A^{-1} = A(I_n)A^{-1} = AA^{-1} = I_n;\\ (B^{-1}A^{-1})(AB) \amp = B^{-1}(A^{-1}A)B = B^{-1}(I_n)B = B^{-1}B = I_n\text{.} \end{align*}

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This final proposition states what may seem like an obvious fact, but which should still be justified. That justification is left to the exercises.

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Proposition 3.3.4.

Let \(A \in M_n(\ff)\) be invertible. Then \(A^{-1}\) is also invertible and \((A^{-1})^{-1} = A\text{.}\)

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While we are not yet ready to calculate the inverse of a matrix (stay tuned!), we can provide examples of invertible matrices and their inverses.

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Example 3.3.5.

Consider the following matrix \(A \in M_2(\rr)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 3 \amp 5 \\ 1 \amp 2 \end{bmatrix}\text{.} \end{equation*}

We can verify that \(A^{-1}\) is

\begin{equation*} A^{-1} = \begin{bmatrix} 2 \amp -5 \\ -1 \amp 3 \end{bmatrix} \end{equation*}

by computing the product in both orders and verifying that the result is \(I_2\) in both cases.

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Similarly, here is a \(3\times 3\) matrix over \(\ff_3\) which is invertible:

\begin{equation*} B = \begin{bmatrix} 1 \amp 2 \amp 1 \\ 0 \amp 2 \amp 0 \\ 1 \amp 2 \amp 0 \end{bmatrix}\text{.} \end{equation*}

The reader is encouraged to verify that the following matrix satisfies the properties of the inverse of \(B\text{:}\)

\begin{equation*} B^{-1} = \begin{bmatrix} 0 \amp 2 \amp 1 \\ 0 \amp 2 \amp 0 \\ 1 \amp 0 \amp 2 \end{bmatrix}\text{.} \end{equation*}

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Subsection 3.3.2 Elementary Matrices

The method of determining the invertibility of a matrix starts with a simple definition.

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Definition 3.3.6.

An elementary matrix is one that is formed by performing a single elementary row operation on an identity matrix.

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Because elementary matrices are related to elementary row operations, there are three types of elementary matrices. The following example provides one elementary matrix (over \(\rr\)) of each type.

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Example 3.3.7.

Our first elementary matrix results from switching the second and third rows in \(I_3\text{:}\)

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}\text{.} \end{equation*}

Next we will look at a matrix which comes about by adding 4 times the first row of \(I_2\) to the second row:

\begin{equation} \begin{bmatrix} 1 \amp 0 \\ 4 \amp 1 \end{bmatrix}\text{.}\tag{3.6} \end{equation}

Finally, we have a matrix which is formed by multiplying the second row of \(I_4\) by 7:

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 7 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}

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Multiplying by an elementary matrix has the effect of carrying out an elementary row operation. In other words, if the \(n\times n\) matrix \(E\) results from applying an elementary row operation to \(I_n\text{,}\) and if \(A\) is another \(n\times n\) matrix, then \(EA\) is the matrix \(A\) after this same elementary row operation has been applied. We will demonstrate this in an example before stating the relevant theorem. (The proof of the theorem is saved for the exercises.)

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Example 3.3.8.

Let \(A\) be the following matrix over \(\rr\text{:}\)

\begin{equation*} A = \begin{bmatrix} -2 \amp 1 \\ 3 \amp 2 \end{bmatrix}\text{.} \end{equation*}

If we label as \(E\) the matrix in (3.6), then we can calculate \(EA\text{:}\)

\begin{equation*} EA = \begin{bmatrix} 1 \amp 0 \\ 4 \amp 1 \end{bmatrix} \begin{bmatrix} -2 \amp 1 \\ 3 \amp 2 \end{bmatrix} = \begin{bmatrix} -2 \amp 1 \\ -5 \amp 6 \end{bmatrix}\text{.} \end{equation*}

The reader can verify that \(EA\) is the result of adding four times the first row of \(A\) to the second row of \(A\text{.}\)

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Theorem 3.3.9.

If the elementary matrix \(E\) results from performing an elementary row operation on \(I_n\text{,}\) and if \(A\) is an \(n\times n\) matrix, then \(EA\) is the matrix that results from applying this same elementary row operation to \(A\text{.}\)

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Each elementary row operation is “reversible” in the sense that there is another elementary row operation that reverses the work done by the first. (This appears as Exercise 1.2.9.) We can use this fact to establish the following useful proposition.

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Proposition 3.3.10.

Every elementary matrix is invertible.

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Proof.

Let \(E\) be an elementary matrix and let \(E'\) denote the elementary matrix which represents the reverse elementary row operation from \(E\text{.}\) By the definition of these matrices and Theorem 3.3.9, we see that

\begin{equation*} EE' = I_n, \hspace{6pt} \text{and} \hspace{6pt} E'E = I_n \text{.} \end{equation*}

This shows that \(E\) and \(E'\) are inverses of each other and, in particular, this proves that \(E\) is invertible.

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We will now connect elementary matrices to the RREF of a matrix in the following proposition. This is largely a restatement of Algorithm 1.2.13 using elementary matrices.

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Proposition 3.3.11.

If \(A \in M_{m,n}(\ff)\text{,}\) then there exists \(B \in M_{m,n}(\ff)\) in RREF and elementary matrices \(E_1,\ldots,E_k \in M_m(\ff)\) such that

\begin{equation*} A = E_1\cdots E_kB\text{.} \end{equation*}

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Proof.

Since each matrix can be reduced to a matrix in RREF, and since elementary row operations are accomplished by multiplying by elementary matrices, there exist elementary matrices \(D_1, \ldots, D_k \in M_m(\ff)\) such that

\begin{equation*} B = D_k\cdots D_1A\text{.} \end{equation*}

Since elementary matrices are invertible, by repeated application of Proposition 3.3.3 we see that \(D_k\cdots D_1\) is invertible and

\begin{equation*} (D_k\cdots D_1)^{-1} = D_1^{-1}\cdots D_k^{-1}\text{.} \end{equation*}

Then we have

\begin{align*} B \amp = (D_k\cdots D_1)A\\ (D_k\cdots D_1)^{-1}B \amp = (D_k\cdots D_1)^{-1}(D_k\cdots D_1)A \\ D_1^{-1}\cdots D_k^{-1}B \amp = A \text{.} \end{align*}

We note that each \(D_i^{-1}\) is an elementary matrix, and if we define \(E_i = D_i^{-1}\) for each \(i=1,\ldots,k\text{,}\) we have our result.

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Subsection 3.3.3 Finding the Inverse of a Matrix

We will now move on to develop an algorithm for finding the inverse of a matrix (when one exists). We need a lemma before stating our most important result of the section.

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Lemma 3.3.12.

If \(A \in M_n(\ff)\) is invertible, then for every \(\mathbf{b} \in \ff^n\text{,}\) the equation \(A\bfx = \mathbf{b}\) has a unique solution.

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Proof.

Let \(\bfb\) be any vector in \(\ff^n\text{.}\) Since \(A\) is invertible, \(A^{-1}\) exists, and we can show that \(\bfx = A^{-1}\bfb\) is a solution to the equation \(A\bfx = \bfb\text{:}\)

\begin{equation*} A(A^{-1}\bfb) = (AA^{-1})\bfb = I_n\bfb = \bfb\text{.} \end{equation*}

To show that this solution is unique, suppose \(\bfv\) is another solution to this equation, so \(A\bfv = \bfb\text{.}\) Then we have

\begin{align*} A\bfv \amp = \bfb\\ A^{-1}(A\bfv) \amp = A^{-1}\bfb\\ (A^{-1}A)\bfv \amp = A^{-1}\bfb\\ I_n\bfv \amp = A^{-1}\bfb\\ \bfv \amp = A^{-1}\bfb\text{.} \end{align*}

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Theorem 3.3.13.

A matrix \(A \in M_n(\ff)\) is invertible if and only if \(A\) is row equivalent to \(I_n\text{.}\) When \(A\) is invertible, any sequence of elementary row operations that reduces \(A\) to \(I_n\) also transforms \(I_n\) into \(A^{-1}\text{.}\)

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Proof.

If \(A\) is invertible, then by Lemma 3.3.12 the equation \(A\bfx = \bfb\) has a unique solution for every \(\bfb \in \ff^n\text{.}\) Now Theorem 1.3.6 means that the RREF of \(A\) has a pivot in each of its \(n\) columns. Since \(A\) is square, this means the RREF has a pivot in each row as well, meaning that the RREF of \(A\) must be \(I_n\text{.}\)

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Conversely, suppose that \(A\) is row equivalent to \(I_n\text{.}\) By Proposition 3.3.11, there exist elementary matrices \(E_1,\ldots,E_k\) such that

\begin{equation} A = E_1\cdots E_kI_n\text{.}\tag{3.7} \end{equation}

This means that \(A = E_1\cdots E_k\text{,}\) and since the product of invertible matrices is invertible, this proves that \(A\) is invertible.

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If we multiply both sides of (3.7) by \((E_1\cdots E_k)^{-1}\text{,}\) we get

\begin{equation*} E_k^{-1}\cdots E_1^{-1}A = I_n\text{,} \end{equation*}

which shows the sequence of elementary row operations (through multiplication by elementary matrices) used to transform \(A\) into \(I_n\text{.}\) On the other hand, if we take the equation \(A = E_1\cdots E_k\) from the previous paragraph and invert both sides, we get

\begin{equation*} A^{-1} = E_k^{-1}\cdots E_1^{-1}\text{,} \end{equation*}

which we can easily adjust to

\begin{equation*} A^{-1} = E_k^{-1}\cdots E_1^{-1}I_n\text{.} \end{equation*}

This establishes the final claim in the theorem.

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This theorem provides an algorithm for us to determine when a matrix is invertible and, in the case it is invertible, to calculate its inverse.

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Algorithm 3.3.14.

In order to determine whether or not a matrix \(A \in M_n(\ff)\) is invertible, follow these steps.

Reduce the matrix \([A \mid I_n]\) to its RREF.
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If the RREF has the form \([I_n \mid B]\text{,}\) then \(A\) is invertible and \(B = A^{-1}\text{.}\)
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If the RREF does not have \(I_n\) in its left \(n\) columns, then \(A\) is not invertible.
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We will end this section with several examples in which we work through this algorithm.

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Example 3.3.15.

Consider the following matrix \(A \in M_3(\rr)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp 0 \amp -3 \\ 1 \amp -1 \amp -2 \\ 0 \amp 3 \amp 2 \end{bmatrix}\text{.} \end{equation*}

To determine whether or not this matrix is invertible, we row reduce the matrix \([A \mid I_3]\text{:}\)

\begin{equation*} \left[\begin{array}{@{}ccc|ccc@{}} 2 \amp 0 \amp -3 \amp 1 \amp 0 \amp 0 \\ 1 \amp -1 \amp -2 \amp 0 \amp 1 \amp 0 \\ 0 \amp 3 \amp 2 \amp 0 \amp 0 \amp 1 \end{array}\right] \sim \left[\begin{array}{@{}ccc|ccc@{}} 1 \amp 0 \amp 0 \amp -4 \amp 9 \amp 3 \\ 0 \amp 1 \amp 0 \amp 2 \amp -4 \amp -1 \\ 0 \amp 0 \amp 1 \amp -3 \amp 6 \amp 2 \end{array}\right]\text{.} \end{equation*}

We see that \(A\) is invertible and that

\begin{equation*} A^{-1} = \begin{bmatrix} -4 \amp 9 \amp 3 \\ 2 \amp -4 \amp -1 \\ -3 \amp 6 \amp 2 \end{bmatrix}\text{.} \end{equation*}

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Example 3.3.16.

Consider the following matrix \(A \in M_4(\qq)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp 3 \amp 0 \amp 1 \\ 1 \amp 1 \amp -3 \amp 7 \\ 0 \amp -2 \amp 1 \amp 0 \\ -1 \amp -4 \amp 2 \amp -2 \end{bmatrix}\text{.} \end{equation*}

We now row reduce \([A \mid I_4]\) to determine whether or not \(A\) is invertible. We find that \([A\mid I_4]\) is row equivalent to

\begin{equation*} \left[\begin{array}{@{}cccc|cccc@{}} 1 \amp 0 \amp 0 \amp 2 \amp 0 \amp 0 \amp 2 \amp -1 \\ 0 \amp 1 \amp 0 \amp -1 \amp 0 \amp -1/5 \amp -1/5 \amp -1/5 \\ 0 \amp 0 \amp 1 \amp -2 \amp 0 \amp -2/5 \amp 3/5 \amp -2/5 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp 3/5 \amp -17/5 \amp 13/5 \end{array}\right]\text{.} \end{equation*}

This calculation shows that \(A\) is not invertible.

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Example 3.3.17.

Consider the following matrix \(A \in M_2(\ff_3)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 1 \amp 2 \\ 2 \amp 0 \end{bmatrix}\text{.} \end{equation*}

We now row reduce \([A \mid I_2]\text{:}\)

\begin{equation*} \left[\begin{array}{@{}cc|cc@{}} 1 \amp 2 \amp 1 \amp 0 \\ 2 \amp 0 \amp 0 \amp 1 \end{array}\right] \sim \left[\begin{array}{@{}cc|cc@{}} 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 1 \amp 2 \amp 2 \end{array}\right]\text{.} \end{equation*}

This proves that \(A\) is invertible and that

\begin{equation*} A^{-1} = \begin{bmatrix} 0 \amp 2 \\ 2 \amp 2 \end{bmatrix}\text{.} \end{equation*}

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Reading Questions 3.3.4 Reading Questions

1.

Consider the following matrix defined over \(\rr\text{:}\)

\begin{equation*} A = \begin{bmatrix} 2 \amp 1 \\ -1 \amp -1 \end{bmatrix}\text{.} \end{equation*}

Write down elementary matrices \(E_1, \ldots, E_n\) which reduce \(A\) to \(I_2\text{.}\) In other words, find elementary matrices \(E_1, \ldots, E_n\) such that \(I_2 = E_n\cdots E_1A\text{.}\)

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2.

Consider the following matrix defined over \(\ff_5\text{:}\)

\begin{equation*} A = \begin{bmatrix} 4 \amp 0 \amp 4 \\ 3 \amp 3 \amp 1 \\ 1 \amp 0 \amp 3 \end{bmatrix}\text{.} \end{equation*}

Determine whether or not \(A\) is invertible. Explain your answer. If \(A\) is invertible, find the inverse.

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Exercises 3.3.5 Exercises

1.

For each of the following matrices \(A\text{,}\) find the RREF of \(A\) (call it \(B\)), and elementary matrices \(E_1, \ldots, E_k\) such that \(B = E_k \cdots E_1 A\text{.}\)

\(A \in M_{2,3}(\rr)\text{:}\)

\begin{equation*} A = \begin{bmatrix} -4 \amp 6 \amp 1 \\ -5 \amp 2 \amp -1 \end{bmatrix} \end{equation*}

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\(A \in M_2(\ff_5)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 1 \amp 4 \\ 3 \amp 0 \end{bmatrix} \end{equation*}

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\(A \in M_2(\cc)\text{:}\)

\begin{equation*} A = \begin{bmatrix} -1-i \amp -3 \\ -3 \amp 2-3i \end{bmatrix} \end{equation*}

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2.

For each of the following matrices \(A\text{,}\) find the RREF of \(A\) (call it \(B\)), and elementary matrices \(E_1, \ldots, E_k\) such that \(B = E_k \cdots E_1 A\text{.}\)

\(A \in M_3(\rr)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 3 \amp 2 \amp -2 \\ 2 \amp -3 \amp 1 \\ -2 \amp -2 \amp 0 \end{bmatrix} \end{equation*}

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\(A \in M_3(\ff_3)\text{:}\)

\begin{equation*} A = \begin{bmatrix} 1 \amp 1 \amp 2 \\ 1 \amp 1 \amp 1 \\ 0 \amp 1 \amp 1 \end{bmatrix} \end{equation*}

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Answer.

The RREF of \(A\) is \(I_3\) and it takes nine elementary row operations to reduce \(A\text{.}\)
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The RREF of \(A\) is \(I_3\) and it takes six elementary row operations to reduce \(A\text{.}\)
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3.

For each of the following matrices \(A\) in \(M_3(\rr)\text{,}\) determine whether or not \(A\) is invertible using the algorithm from this section. In the cases where \(A\) is invertible, find the inverse.

\(\displaystyle A = \begin{bmatrix} -5 \amp -5 \amp 20 \\ 4 \amp -4 \amp 32 \\ 0 \amp -1 \amp 6 \end{bmatrix}\)
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\(\displaystyle A = \begin{bmatrix} -1 \amp -1 \amp -6 \\ -3 \amp 0 \amp -5 \\ 3 \amp 6 \amp 5 \end{bmatrix}\)
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4.

For each of the following matrices \(A\) in \(M_3(\ff_5)\text{,}\) determine whether or not \(A\) is invertible using the algorithm from this section. In the cases where \(A\) is invertible, find the inverse.

\(\displaystyle A = \begin{bmatrix} 4 \amp 3 \amp 0 \\ 2 \amp 3 \amp 1 \\ 2 \amp 4 \amp 3 \end{bmatrix}\)
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\(\displaystyle A = \begin{bmatrix} 0 \amp 3 \amp 4 \\ 4 \amp 0 \amp 4 \\ 4 \amp 3 \amp 3 \end{bmatrix} \)
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Answer.

The matrix \(A\) is invertible and the inverse of \(A\) is

\begin{equation*} \begin{bmatrix} 0 \amp 2 \amp 1 \\ 2 \amp 4 \amp 2 \\ 4 \amp 0 \amp 2 \end{bmatrix}\text{.} \end{equation*}

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The matrix \(A\) is not invertible since its RREF is

\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}

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5.

Use an inverse matrix to solve the following linear system over \(\ff_3\text{:}\)

\begin{align*} x+2y \amp = 1\\ x+y \amp = 2\text{.} \end{align*}

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6.

Use an inverse matrix to solve the following linear system over \(\ff_5\text{:}\)

\begin{align*} 2x + 4y \amp = 1\\ x + 3y \amp = 4\text{.} \end{align*}

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Answer.

If \(A = \begin{bmatrix} 2 \amp 4 \\ 1 \amp 3 \end{bmatrix}\text{,}\) then the inverse over \(\ff_5\) is \(A^{-1} = \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix}\text{.}\) Then we have

\begin{equation*} \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.} \end{equation*}

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7.

Use an inverse matrix to solve the following linear system over \(\rr\text{:}\)

\begin{align*} -4x + 2y + 4z \amp = 1\\ -2x + y -6z \amp = -2\\ -3x - y + 2z \amp = 3\text{.} \end{align*}

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Writing Exercises

8.

Let \(A \in M_n(\ff)\) be invertible. Prove that \(A^T\) is invertible and that \((A^T)^{-1} = (A^{-1})^T\text{.}\)

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9.

Let \(A \in M_n(\ff)\text{.}\) Prove that \(A\) is invertible if and only if \(A^T\) is invertible.

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10.

Let \(E \in M_n(\ff)\) be an elementary matrix. Prove that \(E^T\) is also an elementary matrix.

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11.

Let \(A \in M_{m,n}(\ff)\text{.}\)

Suppose that \(A\) is left-invertible, meaning that there is an \(n\times m\) matrix \(B\) such that \(BA=I_n\text{.}\) Prove that \(m \ge n\text{.}\)
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Suppose that \(A\) is right-invertible, meaning that there is an \(n \times m\) matrix \(B\) such that \(AB = I_m\text{.}\) Prove that \(m \le n\text{.}\)
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Prove that any \(A\) which is invertible must be a square matrix.
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12.

Prove Proposition 3.3.4.

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Solution.

Let \(A \in M_n(\ff)\) be invertible. This means that \(A^{-1}\) exists and that \(AA^{-1} = A^{-1}A = I_n\text{.}\) These equalities involving \(A\) and \(A^{-1}\) show that \(A^{-1}\) is invertible and that \(A\) is a matrix which has the properties of the inverse of \(A^{-1}\text{.}\) Since the inverse of a matrix is unique, it must be that \((A^{-1})^{-1} = A\text{.}\)

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13.

Suppose \(AB=AC\text{,}\) where \(A\) is an \(n \times n\) matrix and \(B\) and \(C\) are \(n\times p\) matrices.

Show that if \(A\) is invertible, then \(B=C\text{.}\)
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Provide an example where \(AB=AC\) but \(A\) is not invertible and \(B \neq C\text{.}\)
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14.

Suppose that \((B-C)D=0\text{,}\) where \(B\) and \(C\) are \(m\times n\) matrices, \(D\) is an \(n\times n\) matrix, and \(0\) is the \(m\times n\) zero matrix.

Show that if \(D\) is invertible, then \(B=C\text{.}\)
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Provide an example where \((B-C)D=0\) but \(D\) is not invertible and \(B \neq C\text{.}\)
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Answer.

We suppose that \((B-C)D = 0\) where \(D\) is invertible. Then we have

\begin{align*} (B-C)D \amp = 0 \\ (B-C)D D^{-1} \amp = 0 D^{-1}\\ B-C \amp = 0\\ B \amp = C\text{.} \end{align*}

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Suppose that \(B = \begin{bmatrix} 1 \amp 3 \\ 1 \amp 2 \end{bmatrix}\text{,}\) \(C = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 0 \end{bmatrix}\text{,}\) and \(D = \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix}\text{.}\) Then, it is true that \(D\) is not invertible. We obviously have \(B \neq C\text{.}\) But we do have \(B - C = \begin{bmatrix} 0 \amp 2 \\ 0 \amp 2 \end{bmatrix}\) and then

\begin{equation*} (B-C)D = \begin{bmatrix} 0 \amp 2 \\ 0 \amp 2 \end{bmatrix} \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} = \begin{bmatrix} 0 \amp 0 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}

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15.

Suppose that \(A\) and \(B\) are \(n\times n\) matrices and that \(AB\) is invertible. Prove that \(A\) is invertible.

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16.

Suppose that \(A \in M_n(\ff)\) is upper triangular and invertible. (A matrix is upper triangular when all entries below the main diagonal are \(0\text{.}\)) Prove that \(A^{-1}\) is also upper triangular.

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Answer.

Suppose that \(A\) is invertible and upper triangular. Then we can use elementary row operations to reduce \(\left[\begin{array}{@{}c|c@{}} A \amp I \end{array}\right]\) to \(\left[\begin{array}{@{}c|c@{}} I \amp A^{-1} \end{array}\right]\text{.}\) But the elementary row operations that are needed for this reduction will be only the scaling operation and the replacement operation, where we add a multiple of one row to another higher up in the matrix. (Specifically, we will only need to add a multiple of a row \(i\) to a row \(j\) where \(i > j\text{.}\)) The elementary matrices that correspond to these specific elementary row operations are all upper triangular. And since the product of upper triangular matrices is upper triangular (this should probably be proved), this shows that \(A^{-1}\) will be upper triangular.

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17.

How many matrices in \(M_2(\ff_2)\) are invertible? What proportion of the matrices in \(M_2(\ff_2)\) are invertible?

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18.

How many matrices in \(M_2(\ff_3)\) are invertible? What proportion of the matrices in \(M_2(\ff_3)\) are invertible?

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19.

Prove Theorem 3.3.9. (You will need three cases.)

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