Examples and questions with their solutions on how to find the determinant of a square matrix using the row echelon form are presented. The main idea is to row reduce the given matrix to triangular form, then calculate its determinant. The determinant of the given matrix is calculated from the determinant of the triangular one by taking into account the linear properties listed below.
We reduce a given matrix into row echelon form (upper triangular or lower triangular) taking into account the following properties of determinants:
Combine rows and use the properties above to rewrite the 3 × 3 matrix given below into triangular form, and calculate its determinant.
$$ A = \begin{bmatrix} 2 & -1 & 3 \\ -2 & 5 & 6 \\ 4 & 6 & 7 \end{bmatrix} $$Let \( D \) be the determinant of the given matrix \( A \).
Step 1: Apply the row operation \( R_2 \leftarrow R_2 + R_1 \). According to Property 1, the determinant does not change.
$$ \begin{bmatrix} 2 & -1 & 3 \\ 0 & 4 & 9 \\ 4 & 6 & 7 \end{bmatrix} $$Step 2: Apply the row operation \( R_3 \leftarrow R_3 - 2R_1 \). According to Property 1, the determinant does not change.
$$ \begin{bmatrix} 2 & -1 & 3 \\ 0 & 4 & 9 \\ 0 & 8 & 1 \end{bmatrix} $$Step 3: Apply the row operation \( R_3 \leftarrow R_3 - 2R_2 \). According to Property 1, the determinant does not change.
$$ \begin{bmatrix} 2 & -1 & 3 \\ 0 & 4 & 9 \\ 0 & 0 & -17 \end{bmatrix} $$Now that the matrix is in triangular form, its determinant is calculated simply as the product of the elements on the main diagonal (upper left to lower right).
Determinant of the triangular matrix = \( (2)(4)(-17) = -136 \)
Since none of our row operations changed the value of the determinant: Det(A) = -136
Combine rows and use the above properties to rewrite the 5 × 5 matrix given below in triangular form and calculate its determinant.
$$ A = \begin{bmatrix} -1 & 0 & -1 & 3 & 6 \\ 1 & 1 & -1 & 0 & 4 \\ 1 & -3 & 0 & -2 & 2 \\ -1 & 2 & 2 & 1 & -3 \\ 0 & -1 & 2 & 0 & 2 \end{bmatrix} $$Let \( D \) be the determinant of matrix \( A \).
Step 1: Apply row operations \( R_2 \leftarrow R_2 + R_1 \), \( R_3 \leftarrow R_3 + R_1 \), and \( R_4 \leftarrow R_4 + R_2 \) (Note: using the original row 2). By Property 1, the determinant remains \( D \).
$$ \begin{bmatrix} -1 & 0 & -1 & 3 & 6 \\ 0 & 1 & -2 & 3 & 10 \\ 0 & -3 & -1 & 1 & 8 \\ 0 & 3 & 1 & 1 & 1 \\ 0 & -1 & 2 & 0 & 2 \end{bmatrix} $$Step 2: From the new matrix, apply \( R_3 \leftarrow R_3 + 3R_2 \), \( R_4 \leftarrow R_4 - 3R_2 \), and \( R_5 \leftarrow R_5 + R_2 \). The determinant remains \( D \).
$$ \begin{bmatrix} -1 & 0 & -1 & 3 & 6 \\ 0 & 1 & -2 & 3 & 10 \\ 0 & 0 & -7 & 10 & 38 \\ 0 & 0 & 7 & -8 & -29 \\ 0 & 0 & 0 & 3 & 12 \end{bmatrix} $$Step 3: Apply \( R_4 \leftarrow R_4 + R_3 \). The determinant remains \( D \).
$$ \begin{bmatrix} -1 & 0 & -1 & 3 & 6 \\ 0 & 1 & -2 & 3 & 10 \\ 0 & 0 & -7 & 10 & 38 \\ 0 & 0 & 0 & 2 & 9 \\ 0 & 0 & 0 & 3 & 12 \end{bmatrix} $$Step 4: Apply \( R_5 \leftarrow R_5 - \frac{3}{2}R_4 \). The determinant remains \( D \).
$$ \begin{bmatrix} -1 & 0 & -1 & 3 & 6 \\ 0 & 1 & -2 & 3 & 10 \\ 0 & 0 & -7 & 10 & 38 \\ 0 & 0 & 0 & 2 & 9 \\ 0 & 0 & 0 & 0 & -\frac{3}{2} \end{bmatrix} $$The matrix is now in triangular form. We find its determinant by multiplying the diagonal entries:
Det(A) = \( (-1)(1)(-7)(2)(-\frac{3}{2}) = -21 \)
Note: Compare this method with calculating the determinant using cofactors in determinant of a square matrix. Which method is more efficient for large matrices?
Calculate the determinant of the following matrix:
$$ A = \begin{bmatrix} -1 & 2 & 4 & 6 \\ 0 & 0 & 1 & 7 \\ -1 & 2 & 4 & 14 \\ 0 & 2 & 4 & 6 \end{bmatrix} $$Let \( D \) be the determinant of the given matrix \( A \).
Step 1: Apply \( R_3 \leftarrow R_3 - R_1 \). By Property 1, the determinant remains \( D \).
$$ \begin{bmatrix} -1 & 2 & 4 & 6 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 8 \\ 0 & 2 & 4 & 6 \end{bmatrix} $$Step 2: Interchange Row 3 and Row 4 (\( R_3 \leftrightarrow R_4 \)). By Property 2, the sign of the determinant changes to \( -D \).
$$ \begin{bmatrix} -1 & 2 & 4 & 6 \\ 0 & 0 & 1 & 7 \\ 0 & 2 & 4 & 6 \\ 0 & 0 & 0 & 8 \end{bmatrix} $$Step 3: Interchange Row 2 and Row 3 (\( R_2 \leftrightarrow R_3 \)). By Property 2, the sign of the determinant changes again to \( -(-D) \), returning it to \( D \).
$$ \begin{bmatrix} -1 & 2 & 4 & 6 \\ 0 & 2 & 4 & 6 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 8 \end{bmatrix} $$The matrix is now in upper triangular form. We find the determinant by multiplying the diagonal entries.
Determinant of triangular matrix = \( (-1)(2)(1)(8) = -16 \)
Since the sign inversions cancelled out (\( -(-D) = D \)): Det(A) = -16
Use the method of row echelon form to calculate the determinant of the matrix:
$$ A = \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ -1 & 0 & 0 & 1 & 5 \\ 1 & -1 & 1 & 4 & 5 \\ 0 & 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 2 & 2 \end{bmatrix} $$Suppose the determinant of matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & k \end{bmatrix} \) is equal to \( D \).
Find the determinant, in terms of \( D \), for the following matrices:
a) \( \begin{bmatrix} 2a & 2b & 2c \\ d & e & f \\ -3g & -3h & -3k \end{bmatrix} \) b) \( \begin{bmatrix} d & e & f \\ a & b & c \\ 7g & 7h & 7k \end{bmatrix} \)
Let \( D \) be the determinant of the given matrix \( A \).
Step 1: Exchange Row 4 and Row 5. By Property 2, the determinant changes sign to \( -D \).
$$ \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ -1 & 0 & 0 & 1 & 5 \\ 1 & -1 & 1 & 4 & 5 \\ 1 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 0 & -1 \end{bmatrix} $$Step 2: Apply \( R_2 \leftarrow R_2+R_1 \), \( R_3 \leftarrow R_3-R_1 \), and \( R_4 \leftarrow R_4-R_1 \). The determinant remains \( -D \).
$$ \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ 0 & -1 & -3 & 1 & 6 \\ 0 & 0 & 4 & 4 & 4 \\ 0 & 1 & 4 & 2 & 1 \\ 0 & 0 & 1 & 0 & -1 \end{bmatrix} $$Step 3: Apply \( R_4 \leftarrow R_4+R_2 \). The determinant remains \( -D \).
$$ \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ 0 & -1 & -3 & 1 & 6 \\ 0 & 0 & 4 & 4 & 4 \\ 0 & 0 & 1 & 3 & 7 \\ 0 & 0 & 1 & 0 & -1 \end{bmatrix} $$Step 4: Apply \( R_4 \leftarrow R_4 - \frac{1}{4}R_3 \) and \( R_5 \leftarrow R_5 - \frac{1}{4}R_3 \). The determinant remains \( -D \).
$$ \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ 0 & -1 & -3 & 1 & 6 \\ 0 & 0 & 4 & 4 & 4 \\ 0 & 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & -1 & -2 \end{bmatrix} $$Step 5: Apply \( R_5 \leftarrow R_5 + \frac{1}{2}R_4 \). The determinant remains \( -D \).
$$ \begin{bmatrix} 1 & -1 & -3 & 0 & 1 \\ 0 & -1 & -3 & 1 & 6 \\ 0 & 0 & 4 & 4 & 4 \\ 0 & 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$Determinant of the triangular matrix = \( (1)(-1)(4)(2)(1) = -8 \).
Since the determinant of our current triangular matrix represents \( -D \), we have \( -D = -8 \). Therefore, the determinant of the original matrix is \( D = 8 \).
a) Row 1 is multiplied by 2 and Row 3 is multiplied by -3. According to Property 3, we multiply the original determinant by those scalars. The determinant is \( (2)(-3)D \) = \( -6D \).
b) Rows 1 and 2 are interchanged (which multiplies the determinant by -1 via Property 2), and Row 3 is multiplied by 7 (Property 3). The new determinant is \( (-1)(7)D \) = \( -7D \).