DLA Exercises

Note: If you chose a different sequence of row operations, your final upper triangular matrix could be different, with a different determinant, and the relationship between that determinant and the determinant of the original matrix could be different.

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8.

[\begin{array}{rrr} - 6 & 7 & 4 \\ - 8 & - 4 & - 6 \\ - 6 & - 9 & 1 \end{array}]

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$- \frac{1}{3} R_{3}$
$R_{1} \leftrightarrow R_{3}$
$R_{2} + 4 R_{1}; R_{3} + 3 R_{1}$
$R_{3} - 2 R_{2},$

resulting in

[\begin{array}{rrr} 2 & 3 & - \frac{1}{3} \\ 0 & 8 & - \frac{22}{3} \\ 0 & 0 & \frac{53}{3} \end{array}] .

This matrix has determinant

848 / 3 .

From the sequence of operations, if the original matrix has determinant

d

then the above upper triangular matrix has determinant

d / 3 .

From

d / 3 = 848 / 3

we conclude that the original matrix has determinant

d = 848 .

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9.

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

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$R_{1} \leftrightarrow R_{3}$
$R_{2} + 4 R_{1}; R_{3} + 3 R_{1}; R_{4} - 5 R_{1}$
$R_{4} + 2 R_{2}; R_{3} - R_{2}$
$\frac{1}{2} R_{4}$
$R_{2} \leftrightarrow R_{4}$
$R_{4} + 6 R_{2}$
$R_{4} + 20 R_{3},$

resulting in

[\begin{array}{rrrr} 1 & - 2 & - 5 & - 3 \\ 0 & 1 & - \frac{19}{2} & - 2 \\ 0 & 0 & 4 & - 2 \\ 0 & 0 & 0 & - 63 \end{array}] .

This matrix has determinant

- 252 .

From the sequence of operations, if the original matrix has determinant

d

then the above upper triangular matrix has determinant

d / 2 .

From

d / 2 = - 252

we conclude that the original matrix has determinant

d = - 504 .

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10.

[\begin{array}{rrrr} 4 & 2 & 7 & 3 \\ - 8 & 9 & 6 & 1 \\ 0 & - 5 & 3 & - 6 \\ - 8 & - 2 & 0 & - 3 \end{array}]

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$R_{2} + 2 R_{1}; R_{4} + 2 R_{1}$
$\frac{1}{2} R_{2}$
$R_{2} \leftrightarrow R_{4}$
$R_{3} + 5 R_{2}; R_{4} - 13 R_{2}$
$R_{4} + 2 R_{3}$
$\frac{1}{5} R_{4}$
$R_{3} \leftrightarrow R_{4}$
$R_{4} - 38 R_{3},$

resulting in

[\begin{array}{cccr} 4 & 2 & 7 & 3 \\ 0 & 1 & 7 & \frac{3}{2} \\ 0 & 0 & 1 & - \frac{19}{10} \\ 0 & 0 & 0 & \frac{737}{10} \end{array}] .

This matrix has determinant

1474 / 5 .

From the sequence of operations, if the original matrix has determinant

d

then the above upper triangular matrix has determinant

d / 10 .

From

d / 10 = 1474 / 5

we conclude that the original matrix has determinant

d = 2948 .

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11.

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

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$- R_{4}$
$R_{1} \leftrightarrow R_{4}$
$R_{3} + 2 R_{1}; R_{4} + 3 R_{1}; R_{5} - 3 R_{1}$
$R_{2} + R_{5}$
$R_{3} - 4 R_{2}; R_{4} - 7 R_{2}; R_{5} + 3 R_{2}$
$R_{4} + 2 R_{5}$
$- \frac{1}{2} R_{4}$
$R_{3} \leftrightarrow R_{4}$
$R_{4} + 35 R_{3}; R_{5} - 27 R_{3}$
$R_{4} + R_{5}$
$2 R_{5}$
$R_{5} + 7 R_{4},$

resulting in

[\begin{array}{rrrrr} 1 & 2 & - 3 & - 2 & - 4 \\ 0 & 1 & 7 & 10 & 12 \\ 0 & 0 & 1 & \frac{5}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 9 & 4 \\ 0 & 0 & 0 & 0 & 39 \end{array}] .

This matrix has determinant

351 .

The changes to the determinant from the sequence of operations above end up cancelling each other out, and so the original matrix has the same determinant,

351 .

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12.

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

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Solution.

This matrix can be reduced to almost upper triangular using the sequence of operations

$R_{1} - 2 R_{2}$
$R_{2} - 2 R_{1}; R_{3} - 5 R_{1}; R_{4} - 3 R_{1}$
$R_{2} + 4 R_{5}$
$- R_{2}$
$R_{3} - 21 R_{2}; R_{4} - 8 R_{2}; R_{5} + 3 R_{2}$
$R_{4} + 3 R_{5}$
$R_{5} + 6 R_{4}$
$- R_{5}$
$R_{3} \leftrightarrow R_{5}$
$R_{4} - 3 R_{3}; R_{5} - 139 R_{3},$

resulting in

[\begin{array}{rrrrr} 1 & - 4 & - 3 & - 9 & - 6 \\ 0 & 1 & - 6 & - 38 & 15 \\ 0 & 0 & 1 & 110 & - 3 \\ 0 & 0 & 0 & - 330 & 17 \\ 0 & 0 & 0 & - 14443 & 759 \end{array}] .

By repeatedly computing cofactor expansions along the first column, we can compute

\begin{aligned} det [\begin{array}{rrrrr} 1 & - 4 & - 3 & - 9 & - 6 \\ 0 & 1 & - 6 & - 38 & 15 \\ 0 & 0 & 1 & 110 & - 3 \\ 0 & 0 & 0 & - 330 & 17 \\ 0 & 0 & 0 & - 14443 & 759 \end{array}] \\ = | \begin{array}{rrrr} 1 & - 6 & - 38 & 15 \\ 0 & 1 & 110 & - 3 \\ 0 & 0 & - 330 & 17 \\ 0 & 0 & - 14443 & 759 \end{array} | \\ = | \begin{array}{rrr} 1 & 110 & - 3 \\ 0 & - 330 & 17 \\ 0 & - 14443 & 759 \end{array} | \\ = | \begin{array}{rr} - 330 & 17 \\ - 14443 & 759 \end{array} | \\ = - 300 \cdot 759 - 17 \cdot - 14443 \\ = - 4939 . \end{aligned}

From the sequence of operations, if the original matrix has determinant

d

then our final almost upper triangular matrix has determinant

- d .

From

- d = - 4939

we conclude that the original matrix has determinant

d = 4939 .

Note: If you chose a different sequence of row operations, your final reduced matrix could be different, with a different determinant, and the relationship between that determinant and the determinant of the original matrix could be different.

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13.

[\begin{array}{rrrrrr} 2 & 2 & - 10 & 8 & - 13 & - 7 \\ 1 & 1 & - 5 & 5 & - 3 & - 4 \\ - 1 & - 1 & 5 & - 4 & 7 & 3 \\ 1 & 0 & - 2 & 3 & 0 & - 3 \\ 2 & 2 & - 10 & 8 & - 14 & - 5 \\ 1 & 0 & - 1 & 1 & 3 & - 6 \end{array}]

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$R_{1} \leftrightarrow R_{2}$
$R_{2} - 2 R_{1}; R_{3} + R_{1}; R_{4} - R_{1}; R_{5} - 2 R_{1}; R_{6} - R_{1}$
$R_{2} \leftrightarrow R_{4}$
$- R 2$
$R_{6} + R_{2}$
$R_{3} \leftrightarrow R_{6}$
$R_{4} \leftrightarrow R_{6}$
$R_{5} + 2 R_{4}; R_{6} + 2 R_{2}$
$R_{5} \leftrightarrow R_{6},$

resulting in

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

This matrix has determinant

1 .

Our sequence of operations involves an odd number of row swaps, but also a single instance of multiplying a row by

- 1,

so the original matrix and the simplified matrix have the same determinant value.

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14.

[\begin{array}{rrrrrr} - 3 & - 2 & 3 & 1 & 1 & 0 \\ - 2 & 1 & - 2 & 1 & 2 & 2 \\ 3 & 1 & - 3 & 2 & 0 & 2 \\ - 3 & 3 & 1 & 1 & 3 & 3 \\ 3 & - 3 & 3 & - 1 & 0 & - 3 \\ - 2 & 0 & 1 & 1 & 0 & 3 \end{array}]

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Solution.

This matrix can be reduced to upper triangular using the sequence of operations

$R_{1} - 2 R_{2}$
$R_{2} + 2 R_{1}; R_{3} - 3 R_{1}; R_{4} + 3 R_{1}; R_{5} - 3 R_{1}; R_{6} + 2 R_{1}$
$R_{2} - R_{6}$
$R_{3} - 13 R_{2}; R_{4} + 9 R_{2}; R_{5} - 9 R_{2}; R_{6} + 8 R_{2}$
$R_{6} - 2 R_{4}$
$R_{3} \leftrightarrow R_{6}$
$R_{4} + 5 R_{3}; R_{5} - 9 R_{3}; R_{6} - 15 R_{3}$
$R_{5} + 2 R_{4}$
$R_{4} \leftrightarrow R_{5}$
$R_{5} - 13 R_{4}; R_{6} + 40 R_{4}$
$R_{6} + 3 R_{5}$
$\frac{1}{20} R_{6}$
$R_{5} \leftrightarrow R_{6}$
$R_{6} + 71 R_{5},$

resulting in

[\begin{array}{rrrrrr} 1 & - 4 & 7 & - 1 & - 3 & - 4 \\ 0 & 1 & - 3 & 0 & 2 & - 1 \\ 0 & 0 & 1 & 3 & - 14 & 23 \\ 0 & 0 & 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 0 & 1 & - \frac{11}{10} \\ 0 & 0 & 0 & 0 & 0 & - \frac{461}{10} \end{array}] .

This matrix has determinant

- 461 / 10 .

From the sequence of operations, if the original matrix has determinant

d

then the above upper triangular matrix has determinant

- d / 20 .

From

- d / 20 = - 461 / 10

we conclude that the original matrix has determinant

d = 922 .

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Computing determinants of matrices involving variables.

For each matrix:

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Compute the determinant as a formula in $x$ by performing row operations to simplify before computing a cofactor expansion.
Determine the values of $x$ for which the determinant value is $0 .$

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15.

[\begin{matrix} x + 4 & 3 & 7 \\ - 1 & x & - 3 \\ - 1 & - 1 & x - 1 \end{matrix}]

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Solution.

One choice of a sequence of simplifying operations is:
1. $R_{1} \leftrightarrow R_{2}$
2. $- R_{1}$
3. $R_{2} - (x + 4) R_{1}; R_{3} + R_{1}$
4. $R_{2} \leftrightarrow R_{3}$
5. $- R_{2}$
6. $R_{3} - (x + 3) R_{2},$
resulting in upper triangular matrix

$[\begin{matrix} 1 & - x & 3 \\ 0 & x + 1 & - (x + 2) \\ 0 & 0 & {(x + 1)}^{2} \end{matrix}] .$

This matrix has determinant

${(x + 1)}^{3} .$

Our sequence of operations used two row swaps and two multiplications of a row by $- 1,$ all of which cancels out in terms of the effect on the determinant. Therefore, the original matrix has the same determinant value.
The determinant calculated above is equal to $0$ only when $x = - 1 .$

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16.

[\begin{matrix} x + 12 & - 30 & 60 \\ - 5 & x + 7 & - 20 \\ - 5 & 10 & x - 23 \end{matrix}]

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Solution.

One choice of a sequence of simplifying operations is:
1. $R_{1} \leftrightarrow R_{2}$
2. $5 R_{2}$
3. $R_{2} + (x + 12) R_{1}; R_{3} - R_{1}$
4. $R_{2} \leftrightarrow R_{3}$
5. $- R_{2}$
6. $R_{3} - (x + 22) R_{2},$
resulting in upper triangular matrix

$[\begin{array}{rcc} - 5 & x + 7 & - 20 \\ 0 & x - 3 & - (x - 3) \\ 0 & 0 & (x - 3) (x + 2) \end{array}] .$

This matrix has determinant

$- 5 {(x - 3)}^{2} (x + 2) .$

From the sequence of operations, if the original matrix has determinant $d$ then the above upper triangular matrix has determinant $- 5 d,$ from which we conclude that

$d = {(x - 3)}^{2} (x + 2) .$
The determinant calculated above is equal to $0$ when $x = 3$ and when $x = - 2 .$

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17.

[\begin{matrix} x + 2 & - 2 & - 4 \\ - 4 & x + 3 & 4 \\ 5 & - 4 & x - 7 \end{matrix}]

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Solution.

One choice of a sequence of simplifying operations is the following. (We do not aim for upper triangular here, but instead aim for as many zero entries as is reasonably possible.)
1. $R_{1} \leftrightarrow R_{2}$
2. $4 R_{2}; 4 R_{3}$
3. $R_{2} + (x + 2) R_{1}; R_{3} + 5 R_{1}$
4. $R_{2} - R_{3}$
The result of these operations is

$[\begin{array}{rcc} - 4 & x + 3 & 4 \\ 0 & x^{2} - 1 & 0 \\ 0 & 5 x - 1 & 4 (x - 2) \end{array}],$

which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:

$\begin{aligned} det [\begin{array}{rcc} - 4 & x + 3 & 4 \\ 0 & x^{2} - 1 & 0 \\ 0 & 5 x - 1 & 4 (x - 2) \end{array}] \\ = - 4 | \begin{array}{c} x^{2} - 1 & 0 \\ 5 x - 1 & 4 (x - 2) \end{array} | \\ = - 4 ((x^{2} - 1) \cdot 4 (x - 2) - 0 \cdot (5 x - 1)) \\ = - 16 (x^{2} - 1) (x - 2) . \end{aligned}$

From the sequence of operations, if the original matrix has determinant $d$ then our simplified matrix has determinant $- 16 d,$ from which we conclude that

$d = (x^{2} - 1) (x - 2) .$
The determinant calculated above is equal to $0$ when $x = 1,$ when $x = - 1,$ and when $x = 2 .$

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18.

[\begin{matrix} x - 3 & 0 & - 1 & 0 \\ 2 & x - 3 & - 2 & - 1 \\ 0 & 2 & x - 3 & - 2 \\ 2 & 1 & - 2 & x - 5 \end{matrix}]

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Solution 1.

One choice of a sequence of simplifying operations is the following. (We do not aim for upper triangular here, but instead aim for as many zero entries as is reasonably possible.)
1. $R_{1} \leftrightarrow R_{2}$
2. $2 R_{1}$
3. $R_{2} - (x - 3) R_{1}; R_{4} - R_{1}$
4. $R_{2} \leftrightarrow R_{3}$
5. $2 R_{3}; 2 R_{4}$
6. $R_{3} + {(x - 3)}^{2} R_{2}; R_{4} + (x - 4) R_{2}$
The result of these operations is

$[\begin{matrix} 2 & x - 3 & - 2 & - 1 \\ 0 & 2 & x - 3 & - 2 \\ 0 & 0 & {(x - 3)}^{3} + 4 (x - 4) & - 2 (x - 3) (x - 4) \\ 0 & 0 & (x - 3) (x - 4) & 0 \end{matrix}],$

which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:

$\begin{aligned} det [\begin{array}{c} 2 & x - 3 & - 2 & - 1 \\ 0 & 2 & x - 3 & - 2 \\ 0 & 0 & {(x - 3)}^{3} + 4 (x - 4) & - 2 (x - 3) (x - 4) \\ 0 & 0 & (x - 3) (x - 4) & 0 \end{array}] \\ = 2 | \begin{array}{c} 2 & x - 3 & - 2 \\ 0 & {(x - 3)}^{3} + 4 (x - 4) & - 2 (x - 3) (x - 4) \\ 0 & (x - 3) (x - 4) & 0 \end{array} | \\ = 2 \cdot 2 | \begin{array}{c} {(x - 3)}^{3} + 4 (x - 4) & - 2 (x - 3) (x - 4) \\ (x - 3) (x - 4) & 0 \end{array} | \\ = 4 (({(x - 3)}^{3} + 4 (x - 4)) \cdot 0 \\ - (- 2 (x - 3) (x - 4)) \cdot ((x - 3) (x - 4))) \\ = 8 {(x - 3)}^{2} {(x - 4)}^{2} . \end{aligned}$

From the sequence of operations, if the original matrix has determinant $d$ then our simplified matrix has determinant $8 d,$ from which we conclude that

$d = {(x - 3)}^{2} {(x - 4)}^{2} .$
The determinant calculated above is equal to $0$ when $x = 3$ and when $x = 4 .$

Solution 2.

Using Lemma 9.4.3, we can instead compute the determinant of the transpose matrix

$[\begin{matrix} x - 3 & 2 & 0 & 2 \\ 0 & x - 3 & 2 & 1 \\ - 1 & - 2 & x - 3 & - 2 \\ 0 & - 1 & - 2 & x - 5 \end{matrix}] .$

(Note that performing row operations on a transposed matrix is equivalent to performing column operations on the original matrix.)
One choice of a sequence of simplifying operations for this transpose matrix is the following. (We do not aim for upper triangular here, but instead aim for as many zero entries as is reasonably possible.)
1. $R_{1} \leftrightarrow R_{3}$
2. $- R_{1}$
3. $R_{3} - (x - 3) R_{1}$
4. $R_{2} \leftrightarrow R_{4}$
5. $- R_{2}$
6. $R_{3} + 2 (x - 4) R_{2}; R_{4} - (x - 3) R_{2}$
7. $R_{3} + 2 R_{4}$
The result of these operations is

$[\begin{matrix} 1 & 2 & - x + 3 & 2 \\ 0 & 1 & 2 & - x + 5 \\ 0 & 0 & {(x - 3)}^{2} & 0 \\ 0 & 0 & - 2 (x - 4) & {(x - 4)}^{2} \end{matrix}],$

which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:

$\begin{aligned} det [\begin{array}{c} 1 & 2 & - x + 3 & 2 \\ 0 & 1 & 2 & - x + 5 \\ 0 & 0 & {(x - 3)}^{2} & 0 \\ 0 & 0 & - 2 (x - 4) & {(x - 4)}^{2} \end{array}] \\ = | \begin{array}{c} 1 & 2 & - x + 5 \\ 0 & {(x - 3)}^{2} & 0 \\ 0 & - 2 (x - 4) & {(x - 4)}^{2} \end{array} | \\ = | \begin{array}{c} {(x - 3)}^{2} & 0 \\ - 2 (x - 4) & {(x - 4)}^{2} \end{array} | \\ = ({(x - 3)}^{2} \cdot {(x - 4)}^{2} - 0 \cdot (- 2 (x - 4))) \\ = {(x - 3)}^{2} {(x - 4)}^{2} . \end{aligned}$

Our sequence of operations used two row swaps and two multiplications of a row by $- 1,$ all of which cancels out in terms of the effect on the determinant. Therefore, the original matrix has the same determinant value.
The determinant calculated above is equal to $0$ when $x = 3$ and when $x = 4 .$

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19. Determinant of a Vandermonde matrix.

Recall that in a polynomial interpolation problem, we need

n + 1

data points to uniquely determine a degree-

n

polynomial. (See Subsection 3.2.4 and Theorem 3.4.1.) And in that case, the associated Vandermonde matrix (the coefficient matrix of the polynomial interpolation problem) is square:

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[\begin{matrix} 1 & x_{1} & x_{1}^{2} & \dots & x_{1}^{n} \\ 1 & x_{2} & x_{2}^{2} & \dots & x_{2}^{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & x_{n + 1} & x_{n + 1}^{2} & \dots & x_{n + 1}^{n} \end{matrix}] .

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(a)

Compute the determinant of the

2 \times 2

Vandermonde matrix

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[\begin{matrix} 1 & x_{1} \\ 1 & x_{2} \end{matrix}],

expressed as a formula in

x_{1}

and

x_{2} .

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(b)

Compute the determinant of the

3 \times 3

Vandermonde matrix

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[\begin{matrix} 1 & x_{1} & x_{1}^{2} \\ 1 & x_{2} & x_{2}^{2} \\ 1 & x_{3} & x_{3}^{2} \end{matrix}],

expressed as a formula in

x_{1}, x_{2}, x_{3} .

Simplify your expression by factoring instead of expanding.

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Hint.

Begin by performing two simplifying row operations, then carry out a cofactor expansion. When simplifying, recall the difference of squares factorization formula:

a^{2} - b^{2} = (a - b) (a + b) .

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(c)

Based on your answers to Task a and Task b, conjecture a formula for the determinant of the

4 \times 4

Vandermonde matrix

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[\begin{matrix} 1 & x_{1} & x_{1}^{2} & x_{1}^{3} \\ 1 & x_{2} & x_{2}^{2} & x_{2}^{3} \\ 1 & x_{3} & x_{3}^{2} & x_{3}^{3} \\ 1 & x_{4} & x_{4}^{2} & x_{4}^{3} \end{matrix}] .

Then attempt to verify your conjectured formula through calculation.

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Hint.

You may find it easier to compute the determinant of the transpose of the matrix (see Lemma 9.4.3). When choosing simplifying row operations, and when attempting to simplify your final determinant calculation into factored form, you should find the following difference of like powers factorization formulas useful:

\begin{aligned} a^{2} - b^{2} & = (a - b) (a + b) \\ a^{3} - b^{3} & = (a - b) (a^{2} + a b + b^{2}) . \end{aligned}

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