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Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 6.6 Exercises

Creating elementary matrices.

In each case, create an elementary matrix of the stated size that represents the specified operation. Then also create its inverse.

1.

\(2 \times 2 \text{;}\) swap rows
Answer.
\begin{align*} E \amp = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \text{,} \\ \inv{E} \amp = \text{same.} \end{align*}

2.

\(2 \times 2 \text{;}\) multiply first row by \(-3\)
Answer.
\begin{align*} E \amp = \begin{abmatrix}{rr} -3 \amp 0 \\ 0 \amp 1 \end{abmatrix} \text{,} \\ \inv{E} \amp = \begin{abmatrix}{rr} -\frac{1}{3} \amp 0 \\ 0 \amp 1 \end{abmatrix} \text{.} \end{align*}

3.

\(2 \times 2 \text{;}\) multiply second row by \(\pi\)
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp \pi \end{bmatrix} \text{,} \\ \inv{E} \amp = \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1/\pi \end{bmatrix} \text{.} \end{align*}

4.

\(2 \times 2 \text{;}\) add \(4\) times the first row to the second
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \\ 4 \amp 1 \end{bmatrix} \text{,} \\ \inv{E} \amp = \begin{abmatrix}{rr} 1 \amp 0 \\ -4 \amp 1 \end{abmatrix} \text{.} \end{align*}

5.

\(2 \times 2 \text{;}\) subtract \(3/2\) times the second row from the first
Answer.
\begin{align*} E \amp = \begin{abmatrix}{rr} 1 \amp -\frac{3}{2} \\ 0 \amp 1 \end{abmatrix} \text{,} \\ \inv{E} \amp = \begin{abmatrix}{rr} 1 \amp \frac{3}{2} \\ 0 \amp 1 \end{abmatrix} \text{.} \end{align*}

6.

\(3 \times 3 \text{;}\) swap first and third rows
Answer.
\begin{align*} E \amp = \begin{bmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \end{bmatrix} \text{,} \\ \inv{E} \amp = \text{same.} \end{align*}

7.

\(3 \times 3 \text{;}\) swap second and third rows
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix} \text{,} \\ \inv{E} \amp = \text{same.} \end{align*}

8.

\(3 \times 3 \text{;}\) multiply second row by \(5/4\)
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp \frac{5}{4} \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \\ \inv{E} \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp \frac{4}{5} \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}

9.

\(3 \times 3 \text{;}\) add \(4\) times the first row to the second
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 4 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \\ \inv{E} \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ -4 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \text{.} \end{align*}

10.

\(3 \times 3 \text{;}\) subtract \(\sqrt{2}\) times the third row from the first
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp -\sqrt{2} \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \\ \inv{E} \amp = \begin{bmatrix} 1 \amp 0 \amp \sqrt{2} \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}

11.

\(4 \times 4 \text{;}\) swap third and fourth rows
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \end{bmatrix} \text{,}\\ \inv{E} \amp = \text{same.} \end{align*}

12.

\(4 \times 4 \text{;}\) multiply third row by \(1/7\)
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp \frac{1}{7} \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,}\\ \inv{E} \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 7 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}

13.

\(4 \times 4 \text{;}\) add \(2\) times the first row to the fourth
Answer.
\begin{align*} E \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 2 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,}\\ \inv{E} \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ \frac{1}{2} \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}

14. Non-elementary operations.

Suppose \(A\) is a \(4 \times \ell\) matrix, where \(\ell\) is an arbitrary but unknown positive integer.

(a)

Create a matrix \(E\) so that the result of \(E A\) is the same as applying the row operation \(R_2 \to 3 R_2 + 5 R_4\) to \(A\text{.}\)
Answer.
\(\displaystyle E = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \amp 5 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)

(c)

Express \(E\) as as a product of elementary matrices.
Answer 1.
\(\displaystyle E = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp \frac{5}{3} \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer 2.
\(\displaystyle E = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 5 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 3 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)

Recognizing elementary matrices.

Decide whether each matrix is elementary. If it is, state the corresponding row operation. If it is not, express it as a product of elementary matrices.

15.

\(\displaystyle \begin{bmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Elementary: swap first and second rows.

16.

\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 5 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Elementary: add \(5\) times the third row to the first.

17.

\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 5 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Elementary: add \(5\) times the first row to the third.

18.

\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 5 \\ 0 \amp 1 \amp 0 \\ 5 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Not elementary:
\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 5 \\ 0 \amp 1 \amp 0 \\ 5 \amp 0 \amp 1 \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 5 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -24 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 5 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}

19.

\(\displaystyle \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp \pi \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Elementary: multiply the second row by \(\pi\text{.}\)

20.

\(\displaystyle \begin{bmatrix} 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)
Answer.
Not elementary:
\begin{equation*} \begin{bmatrix} 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} = \begin{bmatrix} 2 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 3 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}

21.

\(\displaystyle \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp -3 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix}\)
Answer.
Elementary: subtract \(3\) times the third row from the second.

22.

\(\displaystyle \begin{bmatrix} 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \end{bmatrix}\)
Answer 1.
Not elementary:
\begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \end{bmatrix} = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
Answer 2.
Not elementary:
\begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \end{bmatrix} = \begin{bmatrix} 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 0 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}

Computing inverses and decomposing into elementary matrices.

For each matrix:
  1. Use Procedure 6.3.7 to compute the inverse, or to determine that the matrix is not invertible.
  2. If the matrix is invertible, use multiplication to verify that your inverse is correct.
  3. If the matrix is invertible, express both it and its inverse as a product of elementary matrices.

23.

\(\displaystyle \begin{abmatrix}{rr} 2 \amp 5 \\ -1 \amp -2 \end{abmatrix}\)
Solution.
  1. We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:
    \begin{equation*} \begin{abmatrix}{rr|cc} 2 \amp 5 \amp 1 \amp 0 \\ -1 \amp -2 \amp 0 \amp 0 \end{abmatrix} \rowredarrow \begin{abmatrix}{cc|rr} 1 \amp 0 \amp -2 \amp -5 \\ 0 \amp 1 \amp 1 \amp 2 \end{abmatrix}\text{.} \end{equation*}
    To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle - R_1 \)
    3. \(\displaystyle R_2 - 2 R_1 \)
    4. \(\displaystyle R_1 - 2 R_2 \)
    From the reduced augmented matrix, we see that the inverse is
    \begin{equation*} \begin{abmatrix}{rr} -2 \amp -5 \\ 1 \amp 2 \end{abmatrix}\text{.} \end{equation*}
    Interestingly, in this case the inverse is equal to the negative of the original. That is, if we label the original matrix as \(A\text{,}\) then \(\inv{A} = - A\text{.}\) This also implies that \(A^2 = - I\text{,}\) so that in some sense we could say that \(A\) is a “square root” of \(-1\text{!}\)
  2. To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
  3. Create the elementary matrix for each of the steps in our row reduction process, in order:
    \begin{align*} E_1 \amp = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \text{,} \amp E_2 \amp = \begin{abmatrix}{rr} -1 \amp 0 \\ 0 \amp 1 \end{abmatrix} \text{,} \amp E_3 \amp = \begin{abmatrix}{rr} 1 \amp 0 \\ -2 \amp 1 \end{abmatrix} \text{,} \amp E_4 \amp = \begin{abmatrix}{rr} 1 \amp -2 \\ 0 \amp 1 \end{abmatrix} \text{.} \end{align*}
    These elementary matrices, in order from right to left, generate the inverse matrix:
    \begin{gather*} \inv{A} = E_4 E_3 E_2 E_1 \text{,} \\ \\ \begin{abmatrix}{rr} -2 \amp -5 \\ 1 \amp 2 \end{abmatrix} = \begin{bmatrix} 1 \amp -2 \\ 0 \amp 1 \end{bmatrix} \begin{abmatrix}{rr} 1 \amp 0 \\ -2 \amp 1 \end{abmatrix} \begin{abmatrix}{rr} -1 \amp 0 \\ 0 \amp 1 \end{abmatrix} \begin{abmatrix}{rr} 0 \amp 1 \\ 1 \amp 0 \end{abmatrix} \text{.} \end{gather*}
    The inverses of the elementary matrices, in the reverse order, generate the original matrix. To create those inverse elementary matrices, we use the reverse row operations. (Keep in mind that some operations reverse themselves.)
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle - R_1 \)
    3. \(\displaystyle R_2 + 2 R_1 \)
    4. \(R_1 + 2 R_2 \text{.}\)
    Therefore we have
    \begin{align*} \inv{E}_1 \amp = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \text{,} \amp \inv{E}_2 \amp = \begin{abmatrix}{rr} -1 \amp 0 \\ 0 \amp 1 \end{abmatrix} \text{,} \amp \inv{E}_3 \amp = \begin{bmatrix} 1 \amp 0 \\ 2 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_4 \amp = \begin{bmatrix} 1 \amp 2 \\ 0 \amp 1 \end{bmatrix} \text{,} \end{align*}
    and can decompose:
    \begin{gather*} A = \inv{E}_1 \inv{E}_3 \inv{E}_3 \inv{E}_4 \text{,} \\ \\ \begin{abmatrix}{rr} 2 \amp 5 \\ -1 \amp -2 \end{abmatrix} = \begin{bmatrix} 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} \begin{abmatrix}{rr} -1 \amp 0 \\ 0 \amp 1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \\ 2 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 2 \\ 0 \amp 1 \end{bmatrix} \text{.} \end{gather*}
    Note: since alternative sequences of row operations could be used to reduce the original matrix, there exist alternative collections of elementary matrices that could be used to generate \(A\) and \(\inv{A}\text{.}\)

24.

\(\displaystyle \begin{abmatrix}{rrr} 0 \amp 1 \amp 1 \\ 1 \amp - 2 \amp - 1 \\ 7 \amp -18 \amp -12 \end{abmatrix}\)
Solution.
  1. We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:
    \begin{equation*} \begin{abmatrix}{rrr|ccc} 0 \amp 1 \amp 1 \amp 1 \amp 0 \amp 0 \\ 1 \amp - 2 \amp - 1 \amp 0 \amp 1 \amp 0 \\ 7 \amp -18 \amp -12 \amp 0 \amp 0 \amp 1 \end{abmatrix} \rowredarrow \begin{abmatrix}{ccc|rrr} 1 \amp 0 \amp 0 \amp 6 \amp -6 \amp 1 \\ 0 \amp 1 \amp 0 \amp 5 \amp -7 \amp 1 \\ 0 \amp 0 \amp 1 \amp -4 \amp 7 \amp -1 \end{abmatrix}\text{.} \end{equation*}
    To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle R_3 - 7 R_1 \)
    3. \(\displaystyle R_1 + 2 R_2; R_3 + 4 R_2 \)
    4. \(\displaystyle - R_3 \)
    5. \(\displaystyle R_1 - R_3; R_2 - R_3 \)
    From the reduced augmented matrix, we see that the inverse is
    \begin{equation*} \begin{abmatrix}{rrr} 6 \amp -6 \amp 1 \\ 5 \amp -7 \amp 1 \\ -4 \amp 7 \amp -1 \end{abmatrix}\text{.} \end{equation*}
  2. To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
  3. Create the elementary matrix for each of the steps in our row reduction process, in order:
    \begin{align*} E_1 \amp = \begin{bmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp E_2 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ -7 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_3 \amp = \begin{bmatrix} 1 \amp 2 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp E_4 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 4 \amp 1 \end{bmatrix} \text{,} \end{align*}
    \begin{align*} E_5 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{abmatrix} \text{,} \amp E_6 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_7 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 1 \end{abmatrix} \text{.} \end{align*}
    These elementary matrices, in order from right to left, generate the inverse matrix:
    \begin{equation*} \inv{A} = E_7 E_6 E_5 E_4 E_3 E_2 E_1 \text{,} \end{equation*}
    \begin{align*} \begin{abmatrix}{rrr} 6 \amp -6 \amp 1 \\ 5 \amp -7 \amp 1 \\ -4 \amp 7 \amp -1 \end{abmatrix} = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 1 \end{abmatrix} \amp \begin{abmatrix}{rrr} 1 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 4 \amp 1 \end{bmatrix}\\ \amp \begin{bmatrix} 1 \amp 2 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ -7 \amp 0 \amp 1 \end{abmatrix} \begin{bmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{align*}
    The inverses of the elementary matrices, in the reverse order, generate the original matrix. To create those inverse elementary matrices, we use the reverse row operations. (Keep in mind that some operations reverse themselves.)
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle R_3 + 7 R_1 \)
    3. \(\displaystyle R_1 - 2 R_2; R_3 - 4 R_2 \)
    4. \(\displaystyle - R_3 \)
    5. \(\displaystyle R_1 + R_3; R_2 + R_3 \)
    Therefore we have
    \begin{align*} \inv{E}_1 \amp = \begin{bmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_2 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 7 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_3 \amp = \begin{abmatrix}{rrr} 1 \amp -2 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp \inv{E}_4 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp -4 \amp 1 \end{abmatrix} \text{,} \end{align*}
    \begin{align*} \inv{E}_5 \amp = \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{abmatrix} \text{,} \amp \inv{E}_6 \amp = \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_7 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}
    and can decompose:
    \begin{equation*} A = \inv{E}_1 \inv{E}_3 \inv{E}_3 \inv{E}_4 \inv{E}_5 \inv{E}_6 \inv{E}_7 \text{,} \end{equation*}
    \begin{align*} \begin{abmatrix}{rrr} 0 \amp 1 \amp 1 \\ 1 \amp - 2 \amp - 1 \\ 7 \amp -18 \amp -12 \end{abmatrix} = \begin{bmatrix} 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \amp \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 7 \amp 0 \amp 1 \end{bmatrix} \begin{abmatrix}{rrr} 1 \amp -2 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp -4 \amp 1 \end{abmatrix}\\ \amp \begin{abmatrix}{rrr} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{align*}
    Note: since alternative sequences of row operations could be used to reduce the original matrix, there exist alternative collections of elementary matrices that could be used to generate \(A\) and \(\inv{A}\text{.}\)

25.

\(\displaystyle \begin{abmatrix}{rrr} 1 \amp 4 \amp -1 \\ -2 \amp -8 \amp -5 \\ -1 \amp -4 \amp 3 \end{abmatrix}\)
Solution.
We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:
\begin{equation*} \begin{abmatrix}{rrr|ccc} 1 \amp 4 \amp -1 \amp 1 \amp 0 \amp 0 \\ -2 \amp -8 \amp -5 \amp 0 \amp 1 \amp 0 \\ -1 \amp -4 \amp 3 \amp 0 \amp 0 \amp 1 \end{abmatrix} \rowredarrow \begin{abmatrix}{ccc|ccc} 1 \amp 4 \amp 0 \amp \frac{ 3}{2} \amp 0 \amp \frac{1}{2} \\ 0 \amp 0 \amp 1 \amp \frac{ 1}{2} \amp 0 \amp \frac{1}{2} \\ 0 \amp 0 \amp 0 \amp \frac{11}{2} \amp 1 \amp \frac{7}{2} \end{abmatrix}\text{.} \end{equation*}
This matrix is not fully reduced, but only on the right-hand side, which is irrelevant to our purposes.
To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
  1. \(\displaystyle R_2 + 2 R_1 \)
  2. \(\displaystyle R_3 + R_1 \)
  3. \(\displaystyle \frac{1}{2} R_3 \)
  4. \(\displaystyle R_2 \leftrightarrow R_3 \)
  5. \(\displaystyle R_1 + R_2; R_3 + 7 R_2 \)
Since the original matrix did not reduce to the identity, it is not invertible.

26.

\(\displaystyle \begin{abmatrix}{rrrr} 1 \amp 3 \amp 5 \amp -4 \\ 0 \amp 1 \amp 4 \amp -2 \\ 0 \amp 0 \amp -1 \amp 3 \\ -1 \amp -3 \amp -4 \amp 2 \end{abmatrix}\)
Solution.
  1. We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:
    \begin{align*} \amp \begin{abmatrix}{rrrr|cccc} 1 \amp 3 \amp 5 \amp -4 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 4 \amp -2 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp -1 \amp 3 \amp 0 \amp 0 \amp 1 \amp 0 \\ -1 \amp -3 \amp -4 \amp 2 \amp 1 \amp 0 \amp 0 \amp 1 \end{abmatrix}\\ \amp \rowredarrow \begin{abmatrix}{cccc|rrrr} 1 \amp 0 \amp 0 \amp 0 \amp 20 \amp -3 \amp 12 \amp 19 \\ 0 \amp 1 \amp 0 \amp 0 \amp -10 \amp 1 \amp -6 \amp -10 \\ 0 \amp 0 \amp 1 \amp 0 \amp 3 \amp 0 \amp 2 \amp 3 \\ 0 \amp 0 \amp 0 \amp 1 \amp 1 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{.} \end{align*}
    To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
    1. \(\displaystyle R_4 + R_1 \)
    2. \(\displaystyle R_1 - 3 R_2 \)
    3. \(\displaystyle R_3 \leftrightarrow R_4 \)
    4. \(\displaystyle R_1 + 7 R_3; R_2 - 4 R_3; R_4 + R_3 \)
    5. \(\displaystyle R_1 + 12 R_4; R_2 - 6 R_4; R_3 + 2 R_4 \)
    From the reduced augmented matrix, we see that the inverse is
    \begin{equation*} \begin{abmatrix}{rrrr} 20 \amp -3 \amp 12 \amp 19 \\ -10 \amp 1 \amp -6 \amp -10 \\ 3 \amp 0 \amp 2 \amp 3 \\ 1 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{.} \end{equation*}
  2. To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
  3. Create the elementary matrix for each of the steps in our row reduction process, in order:
    \begin{align*} E_1 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp E_2 \amp = \begin{abmatrix}{rrrr} 1 \amp -3 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_3 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \end{bmatrix} \text{,}\\ E_4 \amp = \begin{bmatrix} 1 \amp 0 \amp 7 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp E_5 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp -4 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_6 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \end{bmatrix} \text{,}\\ E_7 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 12 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_8 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp -6 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp E_9 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{.} \end{align*}
    These elementary matrices, in order from right to left, generate the inverse matrix:
    \begin{equation*} \inv{A} = E_9 E_8 E_7 E_6 E_5 E_4 E_3 E_2 E_1 \text{,} \end{equation*}
    \begin{align*} \begin{abmatrix}{rrrr} 20 \amp -3 \amp 12 \amp 19 \\ -10 \amp 1 \amp -6 \amp -10 \\ 3 \amp 0 \amp 2 \amp 3 \\ 1 \amp 0 \amp 1 \amp 1 \end{abmatrix} = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \amp \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp -6 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 12 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix}\\ \amp \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \end{abmatrix} \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp -4 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 7 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\\ \amp \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \end{bmatrix} \begin{abmatrix}{rrrr} 1 \amp -3 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{align*}
    The inverses of the elementary matrices, in the reverse order, generate the original matrix. To create those inverse elementary matrices, we use the reverse row operations. (Keep in mind that some operations reverse themselves.)
    1. \(\displaystyle R_4 - R_1 \)
    2. \(\displaystyle R_1 + 3 R_2 \)
    3. \(\displaystyle R_3 \leftrightarrow R_4 \)
    4. \(\displaystyle R_1 - 7 R_3; R_2 + 4 R_3; R_4 - R_3 \)
    5. \(\displaystyle R_1 - 12 R_4; R_2 + 6 R_4; R_3 - 2 R_4 \)
    Therefore we have
    \begin{align*} \inv{E}_1 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ -1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp \inv{E}_2 \amp = \begin{bmatrix} 1 \amp 3 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_3 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \end{bmatrix} \text{,}\\ \inv{E}_4 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp -7 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp \inv{E}_5 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 4 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_6 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \amp 1 \end{abmatrix} \text{,}\\ \inv{E}_7 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp -12 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{,} \amp \inv{E}_8 \amp = \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 6 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \text{,} \amp \inv{E}_9 \amp = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \text{.} \end{align*}
    and can decompose:
    \begin{equation*} A = \inv{E}_1 \inv{E}_3 \inv{E}_3 \inv{E}_4 \inv{E}_5 \inv{E}_6 \inv{E}_7 \inv{E}_8 \inv{E}_9 \text{,} \end{equation*}
    \begin{align*} \begin{abmatrix}{rrrr} 1 \amp 3 \amp 5 \amp -4 \\ 0 \amp 1 \amp 4 \amp -2 \\ 0 \amp 0 \amp -1 \amp 3 \\ -1 \amp -3 \amp -4 \amp 2 \end{abmatrix} = \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ -1 \amp 0 \amp 0 \amp 1 \end{abmatrix} \amp \begin{bmatrix} 1 \amp 3 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \end{bmatrix}\\ \amp \begin{abmatrix}{rrrr} 1 \amp 0 \amp -7 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 4 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 1 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \amp 1 \end{abmatrix}\\ \amp \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp -12 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix} \begin{bmatrix} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 6 \\ 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix} \begin{abmatrix}{rrrr} 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 1 \end{abmatrix}\text{.} \end{align*}
    Note: since alternative sequences of row operations could be used to reduce the original matrix, there exist alternative collections of elementary matrices that could be used to generate \(A\) and \(\inv{A}\text{.}\)

Solving systems using inverses.

In each case, solve the system using the inverse of the corresponding coefficient matrix.

27.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} \amp \amp y \amp + \amp z \amp = \amp 2 \\ x \amp - \amp 2 y \amp - \amp z \amp = \amp 6 \\ 7 x \amp - \amp 18 y \amp - \amp 12 z \amp = \amp -4 \end{sysofeqns}\)
Solution.
This system is represented in matrix form \(A \uvec{x} = \uvec{b} \) where
\begin{align*} A \amp = \begin{abmatrix}{rrr} 0 \amp 1 \amp 1 \\ 1 \amp - 2 \amp - 1 \\ 7 \amp -18 \amp -12 \end{abmatrix} \text{,} \amp \uvec{x} \amp = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} 2 \\ 6 \\ -4 \end{abmatrix} \text{.} \end{align*}
The coefficient matrix is identical to the matrix of Exercise 24 and has inverse
\begin{equation*} \begin{abmatrix}{rrr} 6 \amp -6 \amp 1 \\ 5 \amp -7 \amp 1 \\ -4 \amp 7 \amp -1 \end{abmatrix}\text{.} \end{equation*}
The system therefore has unique solution
\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrr} 6 \amp -6 \amp 1 \\ 5 \amp -7 \amp 1 \\ -4 \amp 7 \amp -1 \end{abmatrix} \begin{abmatrix}{r} 2 \\ 6 \\ -4 \end{abmatrix} = \begin{abmatrix}{r} -28 \\ -36 \\ 38 \end{abmatrix}\text{.} \end{equation*}
In this unique solution, \(x = -28\text{,}\) \(y = -36\text{,}\) and \(38\text{.}\)

28.

\(\displaystyle \begin{sysofeqns}{rcrcrcr} \amp \amp b \amp + \amp c \amp = \amp 9 \\ a \amp - \amp 2 b \amp - \amp c \amp = \amp 8 \\ 7 a \amp - \amp 18 b \amp - \amp 12 c \amp = \amp -3 \end{sysofeqns}\)
Solution.
This system has the same coefficient matrix as Exercise 27, but has variable column and column of constants
\begin{align*} \uvec{x} \amp = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} 9 \\ 8 \\ -2 \end{abmatrix} \text{.} \end{align*}
We can use the same inverse matrix computed in Exercise 24 to compute the unique solution
\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrr} 6 \amp -6 \amp 1 \\ 5 \amp -7 \amp 1 \\ -4 \amp 7 \amp -1 \end{abmatrix} \begin{abmatrix}{r} 9 \\ 8 \\ -2 \end{abmatrix} = \begin{abmatrix}{r} 3 \\ -14 \\ 23 \end{abmatrix}\text{.} \end{equation*}
In this unique solution, \(a = 3\text{,}\) \(b = -14\text{,}\) and \(c = 23\text{.}\)

29.

\(\displaystyle \begin{sysofeqns}{rcrcrcrcr} x_1 \amp + \amp 3 x_2 \amp + \amp 5 x_3 \amp - \amp 4 x_4 \amp = \amp 0 \\ \amp \amp x_2 \amp + \amp 4 x_3 \amp - \amp 2 x_4 \amp = \amp -4 \\ \amp \amp \amp - \amp x_3 \amp + \amp 3 x_4 \amp = \amp 1 \\ -x_1 \amp - \amp 3 x_2 \amp - \amp 4 x_3 \amp + \amp 2 x_4 \amp = \amp -3 \end{sysofeqns}\)
Solution.
This system is represented in matrix form \(A \uvec{x} = \uvec{b} \) where
\begin{align*} A \amp = \begin{abmatrix}{rrrr} 1 \amp 3 \amp 5 \amp -4 \\ 0 \amp 1 \amp 4 \amp -2 \\ 0 \amp 0 \amp -1 \amp 3 \\ -1 \amp -3 \amp -4 \amp 2 \end{abmatrix} \text{,} \amp \uvec{x} \amp = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} 0 \\ -4 \\ 1 \\ -3 \end{abmatrix} \text{.} \end{align*}
The coefficient matrix is identical to the matrix of Exercise 26 and has inverse
\begin{equation*} \begin{abmatrix}{rrrr} 20 \amp -3 \amp 12 \amp 19 \\ -10 \amp 1 \amp -6 \amp -10 \\ 3 \amp 0 \amp 2 \amp 3 \\ 1 \amp 0 \amp 1 \amp 1 \end{abmatrix}\text{.} \end{equation*}
The system therefore has unique solution
\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrrr} 20 \amp -3 \amp 12 \amp 19 \\ -10 \amp 1 \amp -6 \amp -10 \\ 3 \amp 0 \amp 2 \amp 3 \\ 1 \amp 0 \amp 1 \amp 1 \end{abmatrix} \begin{abmatrix}{r} 0 \\ -4 \\ 1 \\ -3 \end{abmatrix} = \begin{abmatrix}{r} -33 \\ 20 \\ -7 \\ -2 \end{abmatrix}\text{.} \end{equation*}
In this unique solution, \(x_1 = -33\text{,}\) \(x_2 = 20\text{,}\) \(x_3 = -7\text{,}\) and \(x_4 = -2\text{.}\)

30.

\(\displaystyle \begin{sysofeqns}{rcrcrcrcr} w \amp + \amp 3 x \amp + \amp 5 y \amp - \amp 4 z \amp = \amp -2 \\ \amp \amp x \amp + \amp 4 y \amp - \amp 2 z \amp = \amp 1 \\ \amp \amp \amp - \amp y \amp + \amp 3 z \amp = \amp 3 \\ -w \amp - \amp 3 x \amp - \amp 4 y \amp + \amp 2 z \amp = \amp 0 \end{sysofeqns}\)
Solution.
This system has the same coefficient matrix as Exercise 29, but has variable and constant columns
\begin{align*} \uvec{x} \amp = \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} \text{,} \amp \uvec{b} \amp = \begin{abmatrix}{r} -2 \\ 1 \\ 3 \\ 0 \end{abmatrix} \text{.} \end{align*}
We can use the same inverse matrix computed in Exercise 26 to compute the unique solution
\begin{equation*} \uvec{x} = \inv{A} \uvec{b} = \begin{abmatrix}{rrrr} 20 \amp -3 \amp 12 \amp 19 \\ -10 \amp 1 \amp -6 \amp -10 \\ 3 \amp 0 \amp 2 \amp 3 \\ 1 \amp 0 \amp 1 \amp 1 \end{abmatrix} \begin{abmatrix}{r} -2 \\ 1 \\ 3 \\ 0 \end{abmatrix} = \begin{abmatrix}{r} -7 \\ 3 \\ 0 \\ 1 \end{abmatrix}\text{.} \end{equation*}
In this unique solution, \(w = -7\text{,}\) \(x = 3\text{,}\) \(y = 0\text{,}\) and \(z = 1\text{.}\)

31. Properties of row equivalence.

Say that matrix \(B\) is row equivalent to matrix \(A\) if there exists some sequence of elementary row operations that can be applied to \(A\) to end up with result \(B\text{.}\)
Use the ideas of Subsection 6.3.3 to help verify the following properties of the row equivalence relation.

(a) Reflexivity.

Verify that the property of being row equivalent is reflexive: every matrix is row equivalent to itself.

(b) Symmetry.

Verify that the property of being row equivalent is symmetric: it is always true that if \(B\) is row equivalent to \(A\) then it must also be that \(A\) is row equivalent to \(B\text{.}\)

(c) Transitivity.

Verify that the property of being row equivalent is transitive: it is always true that if \(C\) is row equivalent to \(B\) and \(B\) is row equivalent to \(A\) then it must also be that \(C\) is row equivalent to \(A\text{.}\)
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