DLA Exercises

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.

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15.

[\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}]

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Answer.

Elementary: swap first and second rows.

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16.

[\begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]

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Answer.

Elementary: add

5

times the third row to the first.

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17.

[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 5 & 0 & 1 \end{matrix}]

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Answer.

Elementary: add

5

times the first row to the third.

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18.

[\begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 0 \\ 5 & 0 & 1 \end{matrix}]

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Answer.

Not elementary:

[\begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 0 \\ 5 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 5 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 24 \end{matrix}] [\begin{matrix} 1 & 0 & 5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] .

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19.

[\begin{matrix} 1 & 0 & 0 \\ 0 & π & 0 \\ 0 & 0 & 1 \end{matrix}]

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Answer.

Elementary: multiply the second row by

π .

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20.

[\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]

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Answer.

Not elementary:

[\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] .

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21.

[\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]

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Answer.

Elementary: subtract

3

times the third row from the second.

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22.

[\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}]

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Answer 1.

Not elementary:

[\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}] .

Answer 2.

Not elementary:

[\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] .

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Computing inverses and decomposing into elementary matrices.

For each matrix:

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

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23.

[\begin{array}{rr} 2 & 5 \\ - 1 & - 2 \end{array}]

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Solution.

We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:

$[\begin{array}{rrcc} 2 & 5 & 1 & 0 \\ - 1 & - 2 & 0 & 0 \end{array}] \to_{reduce}^{row} [\begin{array}{ccrr} 1 & 0 & - 2 & - 5 \\ 0 & 1 & 1 & 2 \end{array}] .$

To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
1. $R_{1} \leftrightarrow R_{2}$
2. $- R_{1}$
3. $R_{2} - 2 R_{1}$
4. $R_{1} - 2 R_{2}$
From the reduced augmented matrix, we see that the inverse is

$[\begin{array}{rr} - 2 & - 5 \\ 1 & 2 \end{array}] .$

Interestingly, in this case the inverse is equal to the negative of the original. That is, if we label the original matrix as $A,$ then $A^{- 1} = - A .$ This also implies that $A^{2} = - I,$ so that in some sense we could say that $A$ is a “square root” of $- 1!$
To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
Create the elementary matrix for each of the steps in our row reduction process, in order:

$\begin{aligned} E_{1} & = [\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}], & E_{2} & = [\begin{array}{rr} - 1 & 0 \\ 0 & 1 \end{array}], & E_{3} & = [\begin{array}{rr} 1 & 0 \\ - 2 & 1 \end{array}], & E_{4} & = [\begin{array}{rr} 1 & - 2 \\ 0 & 1 \end{array}] . \end{aligned}$

These elementary matrices, in order from right to left, generate the inverse matrix:

$\begin{matrix} A^{- 1} = E_{4} E_{3} E_{2} E_{1}, \\ [\begin{array}{rr} - 2 & - 5 \\ 1 & 2 \end{array}] = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}] [\begin{array}{rr} 1 & 0 \\ - 2 & 1 \end{array}] [\begin{array}{rr} - 1 & 0 \\ 0 & 1 \end{array}] [\begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array}] . \end{matrix}$
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. $R_{1} \leftrightarrow R_{2}$
2. $- R_{1}$
3. $R_{2} + 2 R_{1}$
4. $R_{1} + 2 R_{2} .$
Therefore we have

$\begin{aligned} E_{1}^{- 1} & = [\begin{array}{c} 0 & 1 \\ 1 & 0 \end{array}], & E_{2}^{- 1} & = [\begin{array}{rr} - 1 & 0 \\ 0 & 1 \end{array}], & E_{3}^{- 1} & = [\begin{array}{c} 1 & 0 \\ 2 & 1 \end{array}], & E_{4}^{- 1} & = [\begin{array}{c} 1 & 2 \\ 0 & 1 \end{array}], \end{aligned}$

and can decompose:

$\begin{matrix} A = E_{1}^{- 1} E_{3}^{- 1} E_{3}^{- 1} E_{4}^{- 1}, \\ [\begin{array}{rr} 2 & 5 \\ - 1 & - 2 \end{array}] = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{array}{rr} - 1 & 0 \\ 0 & 1 \end{array}] [\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix}] [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] . \end{matrix}$
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 $A^{- 1} .$

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24.

[\begin{array}{rrr} 0 & 1 & 1 \\ 1 & - 2 & - 1 \\ 7 & - 18 & - 12 \end{array}]

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Solution.

We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:

$[\begin{array}{rrrccc} 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & - 2 & - 1 & 0 & 1 & 0 \\ 7 & - 18 & - 12 & 0 & 0 & 1 \end{array}] \to_{reduce}^{row} [\begin{array}{cccrrr} 1 & 0 & 0 & 6 & - 6 & 1 \\ 0 & 1 & 0 & 5 & - 7 & 1 \\ 0 & 0 & 1 & - 4 & 7 & - 1 \end{array}] .$

To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
1. $R_{1} \leftrightarrow R_{2}$
2. $R_{3} - 7 R_{1}$
3. $R_{1} + 2 R_{2}; R_{3} + 4 R_{2}$
4. $- R_{3}$
5. $R_{1} - R_{3}; R_{2} - R_{3}$
From the reduced augmented matrix, we see that the inverse is

$[\begin{array}{rrr} 6 & - 6 & 1 \\ 5 & - 7 & 1 \\ - 4 & 7 & - 1 \end{array}] .$
To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
Create the elementary matrix for each of the steps in our row reduction process, in order:

$\begin{aligned} E_{1} & = [\begin{array}{c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}], & E_{2} & = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 7 & 0 & 1 \end{array}], & E_{3} & = [\begin{array}{c} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], & E_{4} & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{array}], \end{aligned}$

$\begin{aligned} E_{5} & = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}], & E_{6} & = [\begin{array}{rrr} 1 & 0 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], & E_{7} & = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{array}] . \end{aligned}$

These elementary matrices, in order from right to left, generate the inverse matrix:

$A^{- 1} = E_{7} E_{6} E_{5} E_{4} E_{3} E_{2} E_{1},$

$\begin{aligned} [\begin{array}{rrr} 6 & - 6 & 1 \\ 5 & - 7 & 1 \\ - 4 & 7 & - 1 \end{array}] = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{array}] & [\begin{array}{rrr} 1 & 0 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{array}] \\ [\begin{array}{c} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 7 & 0 & 1 \end{array}] [\begin{array}{c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}] . \end{aligned}$
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. $R_{1} \leftrightarrow R_{2}$
2. $R_{3} + 7 R_{1}$
3. $R_{1} - 2 R_{2}; R_{3} - 4 R_{2}$
4. $- R_{3}$
5. $R_{1} + R_{3}; R_{2} + R_{3}$
Therefore we have

$\begin{aligned} E_{1}^{- 1} & = [\begin{array}{c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}], & E_{2}^{- 1} & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 7 & 0 & 1 \end{array}], & E_{3}^{- 1} & = [\begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], & E_{4}^{- 1} & = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \end{array}], \end{aligned}$

$\begin{aligned} E_{5}^{- 1} & = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}], & E_{6}^{- 1} & = [\begin{array}{c} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}], & E_{7}^{- 1} & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] . \end{aligned}$

and can decompose:

$A = E_{1}^{- 1} E_{3}^{- 1} E_{3}^{- 1} E_{4}^{- 1} E_{5}^{- 1} E_{6}^{- 1} E_{7}^{- 1},$

$\begin{aligned} [\begin{array}{rrr} 0 & 1 & 1 \\ 1 & - 2 & - 1 \\ 7 & - 18 & - 12 \end{array}] = [\begin{array}{c} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}] & [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 7 & 0 & 1 \end{array}] [\begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \end{array}] \\ [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}] [\begin{array}{c} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] . \end{aligned}$
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 $A^{- 1} .$

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25.

[\begin{array}{rrr} 1 & 4 & - 1 \\ - 2 & - 8 & - 5 \\ - 1 & - 4 & 3 \end{array}]

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Solution.

We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:

[\begin{array}{rrrccc} 1 & 4 & - 1 & 1 & 0 & 0 \\ - 2 & - 8 & - 5 & 0 & 1 & 0 \\ - 1 & - 4 & 3 & 0 & 0 & 1 \end{array}] \to_{reduce}^{row} [\begin{array}{cccccc} 1 & 4 & 0 & \frac{3}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & 0 & \frac{11}{2} & 1 & \frac{7}{2} \end{array}] .

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).

$R_{2} + 2 R_{1}$
$R_{3} + R_{1}$
$\frac{1}{2} R_{3}$
$R_{2} \leftrightarrow R_{3}$
$R_{1} + R_{2}; R_{3} + 7 R_{2}$

Since the original matrix did not reduce to the identity, it is not invertible.

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26.

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

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Solution.

We augment the matrix with the identity matrix and attempt to reduce to the identity on the left:

$\begin{aligned} [\begin{array}{rrrrcccc} 1 & 3 & 5 & - 4 & 1 & 0 & 0 & 0 \\ 0 & 1 & 4 & - 2 & 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 3 & 0 & 0 & 1 & 0 \\ - 1 & - 3 & - 4 & 2 & 1 & 0 & 0 & 1 \end{array}] \\ \to_{reduce}^{row} [\begin{array}{ccccrrrr} 1 & 0 & 0 & 0 & 20 & - 3 & 12 & 19 \\ 0 & 1 & 0 & 0 & - 10 & 1 & - 6 & - 10 \\ 0 & 0 & 1 & 0 & 3 & 0 & 2 & 3 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \end{array}] . \end{aligned}$

To achieve the reduction above, you can use the following sequence of operations (following Procedure 2.3.2).
1. $R_{4} + R_{1}$
2. $R_{1} - 3 R_{2}$
3. $R_{3} \leftrightarrow R_{4}$
4. $R_{1} + 7 R_{3}; R_{2} - 4 R_{3}; R_{4} + R_{3}$
5. $R_{1} + 12 R_{4}; R_{2} - 6 R_{4}; R_{3} + 2 R_{4}$
From the reduced augmented matrix, we see that the inverse is

$[\begin{array}{rrrr} 20 & - 3 & 12 & 19 \\ - 10 & 1 & - 6 & - 10 \\ 3 & 0 & 2 & 3 \\ 1 & 0 & 1 & 1 \end{array}] .$
To verify that we have the correct inverse, multiply it against the original matrix — the resulting product should be the identity matrix.
Create the elementary matrix for each of the steps in our row reduction process, in order:

$\begin{aligned} E_{1} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}], & E_{2} & = [\begin{array}{rrrr} 1 & - 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}], & E_{3} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}], \\ E_{4} & = [\begin{array}{c} 1 & 0 & 7 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{5} & = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - 4 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}], & E_{6} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}], \\ E_{7} & = [\begin{array}{rrrr} 1 & 0 & 0 & 12 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{8} & = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{9} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array}] . \end{aligned}$

These elementary matrices, in order from right to left, generate the inverse matrix:

$A^{- 1} = E_{9} E_{8} E_{7} E_{6} E_{5} E_{4} E_{3} E_{2} E_{1},$

$\begin{aligned} [\begin{array}{rrrr} 20 & - 3 & 12 & 19 \\ - 10 & 1 & - 6 & - 10 \\ 3 & 0 & 2 & 3 \\ 1 & 0 & 1 & 1 \end{array}] = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array}] & [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] [\begin{array}{rrrr} 1 & 0 & 0 & 12 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] \\ [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}] [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - 4 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 7 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] \\ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}] [\begin{array}{rrrr} 1 & - 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] . \end{aligned}$
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. $R_{4} - R_{1}$
2. $R_{1} + 3 R_{2}$
3. $R_{3} \leftrightarrow R_{4}$
4. $R_{1} - 7 R_{3}; R_{2} + 4 R_{3}; R_{4} - R_{3}$
5. $R_{1} - 12 R_{4}; R_{2} + 6 R_{4}; R_{3} - 2 R_{4}$
Therefore we have

$\begin{aligned} E_{1}^{- 1} & = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 1 & 0 & 0 & 1 \end{array}], & E_{2}^{- 1} & = [\begin{array}{c} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}], & E_{3}^{- 1} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}], \\ E_{4}^{- 1} & = [\begin{array}{rrrr} 1 & 0 & - 7 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{5}^{- 1} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}], & E_{6}^{- 1} & = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 1 & 1 \end{array}], \\ E_{7}^{- 1} & = [\begin{array}{rrrr} 1 & 0 & 0 & - 12 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{8}^{- 1} & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}], & E_{9}^{- 1} & = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 1 \end{array}] . \end{aligned}$

and can decompose:

$A = E_{1}^{- 1} E_{3}^{- 1} E_{3}^{- 1} E_{4}^{- 1} E_{5}^{- 1} E_{6}^{- 1} E_{7}^{- 1} E_{8}^{- 1} E_{9}^{- 1},$

$\begin{aligned} [\begin{array}{rrrr} 1 & 3 & 5 & - 4 \\ 0 & 1 & 4 & - 2 \\ 0 & 0 & - 1 & 3 \\ - 1 & - 3 & - 4 & 2 \end{array}] = [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 1 & 0 & 0 & 1 \end{array}] & [\begin{array}{c} 1 & 3 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array}] \\ [\begin{array}{rrrr} 1 & 0 & - 7 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array}] [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 1 & 1 \end{array}] \\ [\begin{array}{rrrr} 1 & 0 & 0 & - 12 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] [\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 2 \\ 0 & 0 & 0 & 1 \end{array}] . \end{aligned}$
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 $A^{- 1} .$

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Solving systems using inverses.

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

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27.

{\begin{array}{rcrcrcr} y & + & z & = & 2 \\ x & - & 2 y & - & z & = & 6 \\ 7 x & - & 18 y & - & 12 z & = & - 4 \end{array}

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Solution.

This system is represented in matrix form

A x = b

where

\begin{aligned} A & = [\begin{array}{rrr} 0 & 1 & 1 \\ 1 & - 2 & - 1 \\ 7 & - 18 & - 12 \end{array}], & x & = [\begin{array}{c} x \\ y \\ z \end{array}], & b & = [\begin{array}{r} 2 \\ 6 \\ - 4 \end{array}] . \end{aligned}

The coefficient matrix is identical to the matrix of Exercise 24 and has inverse

[\begin{array}{rrr} 6 & - 6 & 1 \\ 5 & - 7 & 1 \\ - 4 & 7 & - 1 \end{array}] .

The system therefore has unique solution

x = A^{- 1} b = [\begin{array}{rrr} 6 & - 6 & 1 \\ 5 & - 7 & 1 \\ - 4 & 7 & - 1 \end{array}] [\begin{array}{r} 2 \\ 6 \\ - 4 \end{array}] = [\begin{array}{r} - 28 \\ - 36 \\ 38 \end{array}] .

In this unique solution,

x = - 28,

y = - 36,

and

38 .

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28.

{\begin{array}{rcrcrcr} b & + & c & = & 9 \\ a & - & 2 b & - & c & = & 8 \\ 7 a & - & 18 b & - & 12 c & = & - 3 \end{array}

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Solution.

This system has the same coefficient matrix as Exercise 27, but has variable column and column of constants

\begin{aligned} x & = [\begin{array}{c} a \\ b \\ c \end{array}], & b & = [\begin{array}{r} 9 \\ 8 \\ - 2 \end{array}] . \end{aligned}

We can use the same inverse matrix computed in Exercise 24 to compute the unique solution

x = A^{- 1} b = [\begin{array}{rrr} 6 & - 6 & 1 \\ 5 & - 7 & 1 \\ - 4 & 7 & - 1 \end{array}] [\begin{array}{r} 9 \\ 8 \\ - 2 \end{array}] = [\begin{array}{r} 3 \\ - 14 \\ 23 \end{array}] .

In this unique solution,

a = 3,

b = - 14,

and

c = 23 .

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29.

{\begin{array}{rcrcrcrcr} x_{1} & + & 3 x_{2} & + & 5 x_{3} & - & 4 x_{4} & = & 0 \\ x_{2} & + & 4 x_{3} & - & 2 x_{4} & = & - 4 \\ - & x_{3} & + & 3 x_{4} & = & 1 \\ - x_{1} & - & 3 x_{2} & - & 4 x_{3} & + & 2 x_{4} & = & - 3 \end{array}

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Solution.

This system is represented in matrix form

A x = b

where

\begin{aligned} A & = [\begin{array}{rrrr} 1 & 3 & 5 & - 4 \\ 0 & 1 & 4 & - 2 \\ 0 & 0 & - 1 & 3 \\ - 1 & - 3 & - 4 & 2 \end{array}], & x & = [\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}], & b & = [\begin{array}{r} 0 \\ - 4 \\ 1 \\ - 3 \end{array}] . \end{aligned}

The coefficient matrix is identical to the matrix of Exercise 26 and has inverse

[\begin{array}{rrrr} 20 & - 3 & 12 & 19 \\ - 10 & 1 & - 6 & - 10 \\ 3 & 0 & 2 & 3 \\ 1 & 0 & 1 & 1 \end{array}] .

The system therefore has unique solution

x = A^{- 1} b = [\begin{array}{rrrr} 20 & - 3 & 12 & 19 \\ - 10 & 1 & - 6 & - 10 \\ 3 & 0 & 2 & 3 \\ 1 & 0 & 1 & 1 \end{array}] [\begin{array}{r} 0 \\ - 4 \\ 1 \\ - 3 \end{array}] = [\begin{array}{r} - 33 \\ 20 \\ - 7 \\ - 2 \end{array}] .

In this unique solution,

x_{1} = - 33,

x_{2} = 20,

x_{3} = - 7,

and

x_{4} = - 2 .

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30.

{\begin{array}{rcrcrcrcr} w & + & 3 x & + & 5 y & - & 4 z & = & - 2 \\ x & + & 4 y & - & 2 z & = & 1 \\ - & y & + & 3 z & = & 3 \\ - w & - & 3 x & - & 4 y & + & 2 z & = & 0 \end{array}

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Solution.

This system has the same coefficient matrix as Exercise 29, but has variable and constant columns

\begin{aligned} x & = [\begin{array}{c} w \\ x \\ y \\ z \end{array}], & b & = [\begin{array}{r} - 2 \\ 1 \\ 3 \\ 0 \end{array}] . \end{aligned}

We can use the same inverse matrix computed in Exercise 26 to compute the unique solution

x = A^{- 1} b = [\begin{array}{rrrr} 20 & - 3 & 12 & 19 \\ - 10 & 1 & - 6 & - 10 \\ 3 & 0 & 2 & 3 \\ 1 & 0 & 1 & 1 \end{array}] [\begin{array}{r} - 2 \\ 1 \\ 3 \\ 0 \end{array}] = [\begin{array}{r} - 7 \\ 3 \\ 0 \\ 1 \end{array}] .

In this unique solution,

w = - 7,

x = 3,

y = 0,

and

z = 1 .

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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 .

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Use the ideas of Subsection 6.3.3 to help verify the following properties of the row equivalence relation.

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(a) Reflexivity.

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

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(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 .

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(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 .

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