Skip to main content
Logo image

Discover Linear Algebra: A first course in linear algebra

Jeremy Sylvestre

Exercises 9.5 Exercises

Recognizing zero determinants.

For each matrix, give a reason why its determinant must be zero without doing any calculations.

1.

\(\displaystyle \begin{abmatrix}{rrrrr} -6 \amp -1 \amp -7 \amp -1 \amp 3 \\ -1 \amp -2 \amp 5 \amp -2 \amp 7 \\ -4 \amp -7 \amp 5 \amp -7 \amp 1 \\ 3 \amp -6 \amp 1 \amp -6 \amp -6 \\ -6 \amp -3 \amp -6 \amp -3 \amp 2 \end{abmatrix}\)
Answer.
Contains a pair of identical columns.

2.

\(\displaystyle \begin{abmatrix}{rrrrr} 2 \amp 0 \amp 4 \amp 6 \amp -4 \\ -8 \amp -5 \amp 4 \amp 5 \amp 6 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 6 \amp -8 \amp -8 \amp -6 \\ 5 \amp 3 \amp -6 \amp -4 \amp 5 \end{abmatrix}\)
Answer.
Contains a zero row.

3.

\(\displaystyle \begin{abmatrix}{rrrrr} 4 \amp -5 \amp 4 \amp -6 \amp 0 \\ -1 \amp -5 \amp -4 \amp -2 \amp 0 \\ 3 \amp -8 \amp -7 \amp 5 \amp 0 \\ -4 \amp 4 \amp 5 \amp 9 \amp 0 \\ -5 \amp 1 \amp -2 \amp 9 \amp 0 \end{abmatrix}\)
Answer.
Contains a zero column.

4.

\(\displaystyle \begin{abmatrix}{rrrrr} 0 \amp -6 \amp -6 \amp 2 \amp 2 \\ 6 \amp 6 \amp 2 \amp -6 \amp 1 \\ 5 \amp 6 \amp 7 \amp 8 \amp -9 \\ -7 \amp 3 \amp 4 \amp 3 \amp -6 \\ 0 \amp -6 \amp -6 \amp 2 \amp 2 \end{abmatrix}\)
Answer.
Contains a pair of identical rows.

5.

\(\displaystyle \begin{abmatrix}{rrrrr} -2 \amp -1 \amp 0 \amp -1 \amp 0 \\ 1 \amp 2 \amp 2 \amp 3 \amp 3 \\ 3 \amp 6 \amp 6 \amp 9 \amp 9 \\ 3 \amp -2 \amp -1 \amp -3 \amp 3 \\ 2 \amp -2 \amp 1 \amp 2 \amp -2 \end{abmatrix}\)
Answer.
Contains a pair of proportional rows.

6.

\(\displaystyle \begin{abmatrix}{rrrrr} 0 \amp 1 \amp -3 \amp -1 \amp 1 \\ -3 \amp 1 \amp 2 \amp 0 \amp 0 \\ 2 \amp 3 \amp -1 \amp -3 \amp 3 \\ -3 \amp 1 \amp 2 \amp -3 \amp 3 \\ 3 \amp -3 \amp 2 \amp 2 \amp -2 \end{abmatrix}\)
Answer.
Contains a pair of proportional columns.

Computing determinants using row operations.

Use the row reduction method (Procedure 9.2.2) to compute the determinant of each matrix.

7.

\(\displaystyle \begin{abmatrix}{rrr} 0 \amp -2 \amp 1 \\ -8 \amp -5 \amp 4 \\ -2 \amp 7 \amp -4 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle R_1 \leftrightarrow R_3 \)
  2. \(\displaystyle R_2 - 4 R_1 \)
  3. \(\displaystyle R_2 \leftrightarrow R_3 \)
  4. \(\displaystyle -\tfrac{1}{2} R_2 \)
  5. \(R_3 + 33 R_2 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrr} -2 \amp 7 \amp -4 \\ 0 \amp 1 \amp -\frac{1}{2} \\ 0 \amp 0 \amp \frac{7}{2} \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(-7\text{.}\) From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(-d / 2\text{.}\) From \(- d / 2 = -7\) we conclude that the original matrix has determinant \(d = 14\text{.}\)
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.

8.

\(\displaystyle \begin{abmatrix}{rrr} -6 \amp 7 \amp 4 \\ -8 \amp -4 \amp -6 \\ -6 \amp -9 \amp 1 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle -\tfrac{1}{3} R_3 \)
  2. \(\displaystyle R_1 \leftrightarrow R_3 \)
  3. \(\displaystyle R_2 + 4 R_1; R_3 + 3 R_1 \)
  4. \(R_3 - 2 R_2 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrr} 2 \amp 3 \amp -\frac{ 1}{3} \\ 0 \amp 8 \amp -\frac{22}{3} \\ 0 \amp 0 \amp \frac{53}{3} \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(848/3\text{.}\) From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(d / 3\text{.}\) From \(d / 3 = 848 / 3\) we conclude that the original matrix has determinant \(d = 848\text{.}\)
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.

9.

\(\displaystyle \begin{abmatrix}{rrrr} -3 \amp 0 \amp -4 \amp -4 \\ -4 \amp 2 \amp -3 \amp 1 \\ 1 \amp -2 \amp -5 \amp -3 \\ 5 \amp 4 \amp 2 \amp 3 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle R_1 \leftrightarrow R_3 \)
  2. \(\displaystyle R_2 + 4 R_1; R_3 + 3 R_1; R_4 - 5 R_1 \)
  3. \(\displaystyle R_4 + 2 R_2; R_3 - R_2 \)
  4. \(\displaystyle \frac{1}{2} R_4 \)
  5. \(\displaystyle R_2 \leftrightarrow R_4 \)
  6. \(\displaystyle R_4 + 6 R_2 \)
  7. \(R_4 + 20 R_3 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrrr} 1 \amp -2 \amp -5 \amp - 3 \\ 0 \amp 1 \amp -\frac{19}{2} \amp - 2 \\ 0 \amp 0 \amp 4 \amp - 2 \\ 0 \amp 0 \amp 0 \amp -63 \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(-252\text{.}\) From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(d / 2\text{.}\) From \(d / 2 = -252\) we conclude that the original matrix has determinant \(d = -504\text{.}\)
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.

10.

\(\displaystyle \begin{abmatrix}{rrrr} 4 \amp 2 \amp 7 \amp 3 \\ -8 \amp 9 \amp 6 \amp 1 \\ 0 \amp -5 \amp 3 \amp -6 \\ -8 \amp -2 \amp 0 \amp -3 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle R_2 + 2 R_1; R_4 + 2 R_1 \)
  2. \(\displaystyle \frac{1}{2} R_2 \)
  3. \(\displaystyle R_2 \leftrightarrow R_4 \)
  4. \(\displaystyle R_3 + 5 R_2; R_4 - 13 R_2 \)
  5. \(\displaystyle R_4 + 2 R_3 \)
  6. \(\displaystyle \frac{1}{5} R_4 \)
  7. \(\displaystyle R_3 \leftrightarrow R_4 \)
  8. \(R_4 - 38 R_3 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{cccr} 4 \amp 2 \amp 7 \amp 3 \\ 0 \amp 1 \amp 7 \amp \frac{ 3}{ 2} \\ 0 \amp 0 \amp 1 \amp -\frac{ 19}{10} \\ 0 \amp 0 \amp 0 \amp \frac{737}{10} \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(1474/5\text{.}\) From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(d / 10\text{.}\) From \(d / 10 = 1474 / 5\) we conclude that the original matrix has determinant \(d = 2948\text{.}\)
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.

11.

\(\displaystyle \begin{abmatrix}{rrrrr} -3 \amp 1 \amp 2 \amp -1 \amp 1 \\ 0 \amp 4 \amp 1 \amp 4 \amp 2 \\ -2 \amp 0 \amp -1 \amp -3 \amp 2 \\ -1 \amp -2 \amp 3 \amp 2 \amp 4 \\ 3 \amp 3 \amp -3 \amp 0 \amp -2 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle - R_4 \)
  2. \(\displaystyle R_1 \leftrightarrow R_4 \)
  3. \(\displaystyle R_3 + 2 R_1; R_4 + 3 R_1; R_5 - 3 R_1 \)
  4. \(\displaystyle R_2 + R_5 \)
  5. \(\displaystyle R_3 - 4 R_2; R_4 - 7 R_2; R_5 + 3 R_2 \)
  6. \(\displaystyle R_4 + 2 R_5 \)
  7. \(\displaystyle -\frac{1}{2} R_4 \)
  8. \(\displaystyle R_3 \leftrightarrow R_4 \)
  9. \(\displaystyle R_4 + 35 R_3; R_5 - 27 R_3 \)
  10. \(\displaystyle R_4 + R_5 \)
  11. \(\displaystyle 2 R_5 \)
  12. \(R_5 + 7 R_4 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrrrr} 1 \amp 2 \amp -3 \amp -2 \amp -4 \\ 0 \amp 1 \amp 7 \amp 10 \amp 12 \\ 0 \amp 0 \amp 1 \amp \frac{5}{2} \amp \frac{3}{2} \\ 0 \amp 0 \amp 0 \amp 9 \amp 4 \\ 0 \amp 0 \amp 0 \amp 0 \amp 39 \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(351\text{.}\) 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, \(351\text{.}\)
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.

12.

\(\displaystyle \begin{abmatrix}{rrrrr} 5 \amp 2 \amp 5 \amp -1 \amp 0 \\ 2 \amp 3 \amp 4 \amp 4 \amp 3 \\ 5 \amp 1 \amp -2 \amp 4 \amp -3 \\ 3 \amp -4 \amp 3 \amp -1 \amp 5 \\ 0 \amp -3 \amp -1 \amp 4 \amp 0 \end{abmatrix}\)
Solution.
This matrix can be reduced to almost upper triangular using the sequence of operations
  1. \(\displaystyle R_1 - 2 R_2 \)
  2. \(\displaystyle R_2 - 2 R_1; R_3 - 5 R_1; R_4 - 3 R_1 \)
  3. \(\displaystyle R_2 + 4 R_5 \)
  4. \(\displaystyle - R_2 \)
  5. \(\displaystyle R_3 - 21 R_2; R_4 - 8 R_2; R_5 + 3 R_2 \)
  6. \(\displaystyle R_4 + 3 R_5 \)
  7. \(\displaystyle R_5 + 6 R_4 \)
  8. \(\displaystyle - R_5 \)
  9. \(\displaystyle R_3 \leftrightarrow R_5 \)
  10. \(R_4 - 3 R_3; R_5 - 139 R_3 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrrrr} 1 \amp -4 \amp -3 \amp -9 \amp -6 \\ 0 \amp 1 \amp -6 \amp -38 \amp 15 \\ 0 \amp 0 \amp 1 \amp 110 \amp -3 \\ 0 \amp 0 \amp 0 \amp -330 \amp 17 \\ 0 \amp 0 \amp 0 \amp -14443 \amp 759 \end{abmatrix}\text{.} \end{equation*}
By repeatedly computing cofactor expansions along the first column, we can compute
\begin{align*} \amp \det \begin{abmatrix}{rrrrr} 1 \amp -4 \amp -3 \amp -9 \amp -6 \\ 0 \amp 1 \amp -6 \amp -38 \amp 15 \\ 0 \amp 0 \amp 1 \amp 110 \amp -3 \\ 0 \amp 0 \amp 0 \amp -330 \amp 17 \\ 0 \amp 0 \amp 0 \amp -14443 \amp 759 \end{abmatrix}\\ \amp = \begin{avmatrix}{rrrr} 1 \amp -6 \amp -38 \amp 15 \\ 0 \amp 1 \amp 110 \amp -3 \\ 0 \amp 0 \amp -330 \amp 17 \\ 0 \amp 0 \amp -14443 \amp 759 \end{avmatrix}\\ \amp = \begin{avmatrix}{rrr} 1 \amp 110 \amp -3 \\ 0 \amp -330 \amp 17 \\ 0 \amp -14443 \amp 759 \end{avmatrix}\\ \amp = \begin{avmatrix}{rr} -330 \amp 17 \\ -14443 \amp 759 \end{avmatrix}\\ \amp = -300 \cdot 759 - 17 \cdot -14443\\ \amp = -4939\text{.} \end{align*}
From the sequence of operations, if the original matrix has determinant \(d\) then our final almost upper triangular matrix has determinant \(-d\text{.}\) From \(-d = -4939\) we conclude that the original matrix has determinant \(d = 4939\text{.}\)
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.

13.

\(\displaystyle \begin{abmatrix}{rrrrrr} 2 \amp 2 \amp -10 \amp 8 \amp -13 \amp -7 \\ 1 \amp 1 \amp -5 \amp 5 \amp -3 \amp -4 \\ -1 \amp -1 \amp 5 \amp -4 \amp 7 \amp 3 \\ 1 \amp 0 \amp -2 \amp 3 \amp 0 \amp -3 \\ 2 \amp 2 \amp -10 \amp 8 \amp -14 \amp -5 \\ 1 \amp 0 \amp -1 \amp 1 \amp 3 \amp -6 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle R_1 \leftrightarrow R_2 \)
  2. \(\displaystyle R_2 - 2 R_1; R_3 + R_1; R_4 - R_1; R_5 - 2 R_1; R_6 - R_1 \)
  3. \(\displaystyle R_2 \leftrightarrow R_4 \)
  4. \(\displaystyle - R2 \)
  5. \(\displaystyle R_6 + R_2 \)
  6. \(\displaystyle R_3 \leftrightarrow R_6 \)
  7. \(\displaystyle R_4 \leftrightarrow R_6 \)
  8. \(\displaystyle R_5 + 2 R_4; R_6 + 2 R_2 \)
  9. \(R_5 \leftrightarrow R_6 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrrrrr} 1 \amp 1 \amp -5 \amp 5 \amp -3 \amp -4 \\ 0 \amp 1 \amp -3 \amp 2 \amp -3 \amp -1 \\ 0 \amp 0 \amp 1 \amp -2 \amp 3 \amp -3 \\ 0 \amp 0 \amp 0 \amp 1 \amp 4 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 1 \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(1\text{.}\) Our sequence of operations involves an odd number of row swaps, but also a single instance of multiplying a row by \(-1\text{,}\) so the original matrix and the simplified matrix have the same determinant value.
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.

14.

\(\displaystyle \begin{abmatrix}{rrrrrr} -3 \amp -2 \amp 3 \amp 1 \amp 1 \amp 0 \\ -2 \amp 1 \amp -2 \amp 1 \amp 2 \amp 2 \\ 3 \amp 1 \amp -3 \amp 2 \amp 0 \amp 2 \\ -3 \amp 3 \amp 1 \amp 1 \amp 3 \amp 3 \\ 3 \amp -3 \amp 3 \amp -1 \amp 0 \amp -3 \\ -2 \amp 0 \amp 1 \amp 1 \amp 0 \amp 3 \end{abmatrix}\)
Solution.
This matrix can be reduced to upper triangular using the sequence of operations
  1. \(\displaystyle R_1 - 2 R_2 \)
  2. \(\displaystyle R_2 + 2 R_1; R_3 - 3 R_1; R_4 + 3 R_1; R_5 - 3 R_1; R_6 + 2 R_1 \)
  3. \(\displaystyle R_2 - R_6 \)
  4. \(\displaystyle R_3 - 13 R_2; R_4 + 9 R_2; R_5 - 9 R_2; R_6 + 8 R_2 \)
  5. \(\displaystyle R_6 - 2 R_4 \)
  6. \(\displaystyle R_3 \leftrightarrow R_6 \)
  7. \(\displaystyle R_4 + 5 R_3; R_5 - 9 R_3; R_6 - 15 R_3 \)
  8. \(\displaystyle R_5 + 2 R_4 \)
  9. \(\displaystyle R_4 \leftrightarrow R_5 \)
  10. \(\displaystyle R_5 - 13 R_4; R_6 + 40 R_4 \)
  11. \(\displaystyle R_6 + 3 R_5 \)
  12. \(\displaystyle \tfrac{1}{20} R_6 \)
  13. \(\displaystyle R_5 \leftrightarrow R_6 \)
  14. \(R_6 + 71 R_5 \text{,}\)
resulting in
\begin{equation*} \begin{abmatrix}{rrrrrr} 1 \amp -4 \amp 7 \amp -1 \amp -3 \amp -4 \\ 0 \amp 1 \amp -3 \amp 0 \amp 2 \amp -1 \\ 0 \amp 0 \amp 1 \amp 3 \amp -14 \amp 23 \\ 0 \amp 0 \amp 0 \amp 1 \amp 1 \amp 5 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \amp -\frac{ 11}{10} \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \amp -\frac{461}{10} \end{abmatrix}\text{.} \end{equation*}
This matrix has determinant \(- 461 / 10\text{.}\) From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(-d / 20\text{.}\) From \(- d / 20 = - 461 / 10\) we conclude that the original matrix has determinant \(d = 922\text{.}\)
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.

Computing determinants of matrices involving variables.

For each matrix:
  1. Compute the determinant as a formula in \(x\) by performing row operations to simplify before computing a cofactor expansion.
  2. Determine the values of \(x\) for which the determinant value is \(0\text{.}\)

15.

\(\displaystyle \begin{bmatrix} x + 4 \amp 3 \amp 7 \\ -1 \amp x \amp -3 \\ -1 \amp -1 \amp x - 1 \end{bmatrix}\)
Solution.
  1. One choice of a sequence of simplifying operations is:
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle - R_1 \)
    3. \(\displaystyle R_2 - (x + 4) R_1; R_3 + R_1 \)
    4. \(\displaystyle R_2 \leftrightarrow R_3 \)
    5. \(\displaystyle - R_2 \)
    6. \(R_3 - (x + 3) R_2 \text{,}\)
    resulting in upper triangular matrix
    \begin{equation*} \begin{bmatrix} 1 \amp -x \amp 3 \\ 0 \amp x + 1 \amp -(x + 2) \\ 0 \amp 0 \amp {(x + 1)}^2 \end{bmatrix}\text{.} \end{equation*}
    This matrix has determinant
    \begin{equation*} {(x + 1)}^3 \text{.} \end{equation*}
    Our sequence of operations used two row swaps and two multiplications of a row by \(-1\text{,}\) all of which cancels out in terms of the effect on the determinant. Therefore, the original matrix has the same determinant value.
  2. The determinant calculated above is equal to \(0\) only when \(x = -1\text{.}\)

16.

\(\displaystyle \begin{bmatrix} x + 12 \amp -30 \amp 60 \\ -5 \amp x + 7 \amp -20 \\ -5 \amp 10 \amp x - 23 \end{bmatrix}\)
Solution.
  1. One choice of a sequence of simplifying operations is:
    1. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle 5 R_2 \)
    3. \(\displaystyle R_2 + (x + 12) R_1; R_3 - R_1 \)
    4. \(\displaystyle R_2 \leftrightarrow R_3 \)
    5. \(\displaystyle - R_2 \)
    6. \(R_3 - (x + 22) R_2 \text{,}\)
    resulting in upper triangular matrix
    \begin{equation*} \begin{abmatrix}{rcc} -5 \amp x + 7 \amp -20 \\ 0 \amp x - 3 \amp -(x - 3) \\ 0 \amp 0 \amp (x - 3) (x + 2) \end{abmatrix}\text{.} \end{equation*}
    This matrix has determinant
    \begin{equation*} -5 {(x - 3)}^2 (x + 2) \text{.} \end{equation*}
    From the sequence of operations, if the original matrix has determinant \(d\) then the above upper triangular matrix has determinant \(-5 d\text{,}\) from which we conclude that
    \begin{equation*} d = {(x - 3)}^2 (x + 2) \text{.} \end{equation*}
  2. The determinant calculated above is equal to \(0\) when \(x = 3\) and when \(x = -2\text{.}\)

17.

\(\displaystyle \begin{bmatrix} x + 2 \amp -2 \amp -4 \\ -4 \amp x + 3 \amp 4 \\ 5 \amp -4 \amp x - 7 \end{bmatrix}\)
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. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle 4 R_2; 4 R_3 \)
    3. \(\displaystyle R_2 + (x + 2) R_1; R_3 + 5 R_1 \)
    4. \(\displaystyle R_2 - R_3 \)
    The result of these operations is
    \begin{equation*} \begin{abmatrix}{rcc} -4 \amp x + 3 \amp 4 \\ 0 \amp x^2 - 1 \amp 0 \\ 0 \amp 5 x - 1 \amp 4 (x - 2) \end{abmatrix}\text{,} \end{equation*}
    which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:
    \begin{align*} \amp \det \begin{abmatrix}{rcc} -4 \amp x + 3 \amp 4 \\ 0 \amp x^2 - 1 \amp 0 \\ 0 \amp 5 x - 1 \amp 4 (x - 2) \end{abmatrix}\\ \amp = -4 \, \begin{vmatrix} x^2 - 1 \amp 0 \\ 5 x - 1 \amp 4 (x - 2) \end{vmatrix}\\ \amp = -4 \bbrac{ (x^2 - 1) \cdot 4 (x - 2) - 0 \cdot (5 x - 1) }\\ \amp = -16 (x^2 - 1) (x - 2)\text{.} \end{align*}
    From the sequence of operations, if the original matrix has determinant \(d\) then our simplified matrix has determinant \(-16 d\text{,}\) from which we conclude that
    \begin{equation*} d = (x^2 - 1) (x - 2) \text{.} \end{equation*}
  2. The determinant calculated above is equal to \(0\) when \(x = 1\text{,}\) when \(x = -1\text{,}\) and when \(x = 2\text{.}\)

18.

\(\displaystyle \begin{bmatrix} x - 3 \amp 0 \amp -1 \amp 0 \\ 2 \amp x - 3 \amp -2 \amp -1 \\ 0 \amp 2 \amp x - 3 \amp -2 \\ 2 \amp 1 \amp -2 \amp x - 5 \end{bmatrix}\)
Solution 1.
  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. \(\displaystyle R_1 \leftrightarrow R_2 \)
    2. \(\displaystyle 2 R_1 \)
    3. \(\displaystyle R_2 - (x - 3) R_1; R_4 - R_1 \)
    4. \(\displaystyle R_2 \leftrightarrow R_3 \)
    5. \(\displaystyle 2 R_3; 2 R_4 \)
    6. \(\displaystyle R_3 + {(x - 3)}^2 R_2; R_4 + (x - 4) R_2 \)
    The result of these operations is
    \begin{equation*} \begin{bmatrix} 2 \amp x - 3 \amp -2 \amp -1 \\ 0 \amp 2 \amp x - 3 \amp -2 \\ 0 \amp 0 \amp {(x - 3)}^3 + 4 (x - 4) \amp -2 (x - 3) (x - 4) \\ 0 \amp 0 \amp (x - 3) (x - 4) \amp 0 \end{bmatrix}\text{,} \end{equation*}
    which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:
    \begin{align*} \amp \det \begin{bmatrix} 2 \amp x - 3 \amp -2 \amp -1 \\ 0 \amp 2 \amp x - 3 \amp -2 \\ 0 \amp 0 \amp {(x - 3)}^3 + 4 (x - 4) \amp -2 (x - 3) (x - 4) \\ 0 \amp 0 \amp (x - 3) (x - 4) \amp 0 \end{bmatrix}\\ \amp = 2 \, \begin{vmatrix} 2 \amp x - 3 \amp -2 \\ 0 \amp {(x - 3)}^3 + 4 (x - 4) \amp -2 (x - 3) (x - 4) \\ 0 \amp (x - 3) (x - 4) \amp 0 \end{vmatrix}\\ \amp = 2 \cdot 2 \, \begin{vmatrix} {(x - 3)}^3 + 4 (x - 4) \amp -2 (x - 3) (x - 4) \\ (x - 3) (x - 4) \amp 0 \end{vmatrix}\\ \amp = 4 \Bbrac{ \bbrac{{(x - 3)}^3 + 4 (x - 4)} \cdot 0\\ \amp \phantom{=} \quad - \bbrac{-2 (x - 3) (x - 4)} \cdot \bbrac{(x - 3) (x - 4)} }\\ \amp = 8 {(x - 3)}^2 {(x - 4)}^2\text{.} \end{align*}
    From the sequence of operations, if the original matrix has determinant \(d\) then our simplified matrix has determinant \(8 d\text{,}\) from which we conclude that
    \begin{equation*} d = {(x - 3)}^2 {(x - 4)}^2 \text{.} \end{equation*}
  2. The determinant calculated above is equal to \(0\) when \(x = 3\) and when \(x = 4\text{.}\)
Solution 2.
  1. Using Lemma 9.4.3, we can instead compute the determinant of the transpose matrix
    \begin{equation*} \begin{bmatrix} x - 3 \amp 2 \amp 0 \amp 2 \\ 0 \amp x - 3 \amp 2 \amp 1 \\ -1 \amp -2 \amp x - 3 \amp -2 \\ 0 \amp -1 \amp -2 \amp x - 5 \end{bmatrix}\text{.} \end{equation*}
    (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. \(\displaystyle R_1 \leftrightarrow R_3 \)
    2. \(\displaystyle - R_1 \)
    3. \(\displaystyle R_3 - (x - 3) R_1 \)
    4. \(\displaystyle R_2 \leftrightarrow R_4 \)
    5. \(\displaystyle - R_2 \)
    6. \(\displaystyle R_3 + 2 (x - 4) R_2; R_4 - (x - 3) R_2 \)
    7. \(\displaystyle R_3 + 2 R_4 \)
    The result of these operations is
    \begin{equation*} \begin{bmatrix} 1 \amp 2 \amp -x + 3 \amp 2 \\ 0 \amp 1 \amp 2 \amp -x + 5 \\ 0 \amp 0 \amp {(x - 3)}^2 \amp 0 \\ 0 \amp 0 \amp -2 (x - 4) \amp {(x - 4)}^2 \end{bmatrix}\text{,} \end{equation*}
    which is enough zero entries to be as good as upper triangular for the purpose of computing the determinant:
    \begin{align*} \amp \det \begin{bmatrix} 1 \amp 2 \amp -x + 3 \amp 2 \\ 0 \amp 1 \amp 2 \amp -x + 5 \\ 0 \amp 0 \amp {(x - 3)}^2 \amp 0 \\ 0 \amp 0 \amp -2 (x - 4) \amp {(x - 4)}^2 \end{bmatrix}\\ \amp = \begin{vmatrix} 1 \amp 2 \amp -x + 5 \\ 0 \amp {(x - 3)}^2 \amp 0 \\ 0 \amp -2 (x - 4) \amp {(x - 4)}^2 \end{vmatrix}\\ \amp = \begin{vmatrix} {(x - 3)}^2 \amp 0 \\ -2 (x - 4) \amp {(x - 4)}^2 \end{vmatrix}\\ \amp = \Bbrac{ {(x - 3)}^2 \cdot {(x - 4)}^2 - 0 \cdot \bbrac{-2 (x - 4)} }\\ \amp = {(x - 3)}^2 {(x - 4)}^2\text{.} \end{align*}
    Our sequence of operations used two row swaps and two multiplications of a row by \(-1\text{,}\) all of which cancels out in terms of the effect on the determinant. Therefore, the original matrix has the same determinant value.
  2. The determinant calculated above is equal to \(0\) when \(x = 3\) and when \(x = 4\text{.}\)

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:
\begin{equation*} \begin{bmatrix} 1 \amp x_1 \amp x_1^2 \amp \cdots \amp x_1^n \\ 1 \amp x_2 \amp x_2^2 \amp \cdots \amp x_2^n \\ \vdots \amp \vdots \amp \vdots \amp \amp \vdots \\ 1 \amp x_{n + 1} \amp x_{n + 1}^2 \amp \cdots \amp x_{n + 1}^n \end{bmatrix}\text{.} \end{equation*}

(a)

Compute the determinant of the \(2 \times 2\) Vandermonde matrix
\begin{equation*} \begin{bmatrix} 1 \amp x_1 \\ 1 \amp x_2 \end{bmatrix}\text{,} \end{equation*}
expressed as a formula in \(x_1\) and \(x_2\text{.}\)

(b)

Compute the determinant of the \(3 \times 3\) Vandermonde matrix
\begin{equation*} \begin{bmatrix} 1 \amp x_1 \amp x_1^2 \\ 1 \amp x_2 \amp x_2^2 \\ 1 \amp x_3 \amp x_3^2 \end{bmatrix}\text{,} \end{equation*}
expressed as a formula in \(x_1,x_2,x_3\text{.}\) Simplify your expression by factoring instead of expanding.
Hint.
Begin by performing two simplifying row operations, then carry out a cofactor expansion. When simplifying, recall the difference of squares factorization formula:
\begin{equation*} a^2 - b^2 = (a - b) (a + b) \text{.} \end{equation*}

(c)

Based on your answers to Task a and Task b, conjecture a formula for the determinant of the \(4 \times 4\) Vandermonde matrix
\begin{equation*} \begin{bmatrix} 1 \amp x_1 \amp x_1^2 \amp x_1^3 \\ 1 \amp x_2 \amp x_2^2 \amp x_2^3 \\ 1 \amp x_3 \amp x_3^2 \amp x_3^3 \\ 1 \amp x_4 \amp x_4^2 \amp x_4^3 \end{bmatrix}\text{.} \end{equation*}
Then attempt to verify your conjectured formula through calculation.
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{align*} a^2 - b^2 \amp = (a - b) (a + b) \\ a^3 - b^3 \amp = (a - b) (a^2 + a b + b^2) \text{.} \end{align*}
PrevTopNext