(I) Row Operations and Determinants
Theorem 3: How Determinants Change Under Row Operations
Let \(A\) be a square matrix.
- (Replacement) If a multiple of one row of \(A\) is added to another row to produce a matrix \(B\), then \(\det B = \det A\).
- (Interchange) If two rows of \(A\) are interchanged to produce \(B\), then \(\det B = -\det A\).
- (Scaling) If one row of \(A\) is multiplied by \(k\) to produce \(B\), then \(\det B = k \cdot \det A\).
Application of Scaling (Property 3):
If \(A\) is an \(n \times n\) matrix, and we scale every row by \(k\), then:
\[\det(kA) = k^n \det A\]
Compute the determinant by applying row operations to reach an upper triangular form.
Solution:
(Scale Row 1 by \( \frac{1}{2} \)) → factors out a \( 2 \)
\( = 2 \begin{vmatrix} 1 & -4 & 3 & 4 \\ 3 & -9 & 5 & 10 \\ -3 & 0 & 1 & -2 \\ 1 & -4 & 0 & 6 \end{vmatrix} \)
(Replacement: \( R_2 = R_2 - 3R_1 \), \( R_3 = R_3 + 3R_1 \), \( R_4 = R_4 - R_1 \)) → determinant unchanged
\( = 2 \begin{vmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & -12 & 10 & 10 \\ 0 & 0 & -3 & 2 \end{vmatrix} \)
(Replacement: \( R_3 = R_3 + 4R_2 \)) → determinant unchanged
\( = 2 \begin{vmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & 0 & -6 & 2 \\ 0 & 0 & -3 & 2 \end{vmatrix} \)
(Interchange Row 3 and Row 4) → multiplies determinant by \( -1 \)
\( = -2 \begin{vmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & -6 & 2 \end{vmatrix} \)
(Replacement: \( R_4 = R_4 - 2R_3 \)) → determinant unchanged
\( = -2 \begin{vmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & 0 & -2 \end{vmatrix} \)
The resulting matrix is upper triangular. The determinant is the product of its diagonal entries.
Therefore, \( \det A = -2 \cdot (1 \cdot 3 \cdot (-3) \cdot (-2)) = -2 \cdot 18 = -36 \).
(II) Determinants via Echelon Form
Suppose \(A \sim U\), where \(U\) is an echelon form of \(A\) obtained purely by row replacements and interchanges.
Then:
where \(r\) is the number of row interchanges.
Because \(U\) is in echelon form, it is an upper triangular matrix. Its determinant is the product of the entries on its main diagonal.
- If \(A\) is invertible, all diagonal entries of \(U\) are pivots (non-zero). Thus, \(\det U \neq 0\), and \(\det A = (-1)^r \times (\text{product of pivots in } U)\).
- If \(A\) is not invertible, \(U\) will have a row of zeros, meaning at least one diagonal entry is zero. Thus, \(\det U = 0\), and \(\det A = 0\).
Theorem 4
A square matrix \(A\) is invertible if and only if \(\det A \neq 0\).
(This statement is added to the Invertible Matrix Theorem).
Compute the determinant of the matrix:
\[ A = \begin{bmatrix} 1 & -3 & 1 & -2 \\ 2 & -5 & -1 & -2 \\ 0 & -4 & 5 & 1 \\ -3 & 10 & -6 & 8 \end{bmatrix} \]Solution: We apply row replacement operations to introduce zeros below the first pivot.
Notice that Row 2 and Row 4 are now identical (\(0, \ 1, \ -3, \ 2\)). If we perform the replacement operation \(\text{Row } 4 = \text{Row } 4 - \text{Row } 2\), the fourth row becomes all zeros.
Because there is a row of zeros, the matrix is not invertible, and its determinant is exactly \(0\).
Theorem 5
If \(A\) is an \(n \times n\) matrix, then \(\det A^T = \det A\).
(This can be proved by mathematical induction).
(III) Multiplicative Property and Applications
Theorem 6 (Multiplicative Property)
If \(A\) and \(B\) are \(n \times n\) matrices, then:
Remarks
-
\(\det(A + B) \neq \det A + \det B\) in general.
Counterexample:
Let \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \).
\(\det A = 1\), \(\det B = 4 - 6 = -2\). So \(\det A + \det B = -1\).
However, \( A + B = \begin{bmatrix} 2 & 2 \\ 3 & 5 \end{bmatrix} \), and \(\det(A + B) = (2)(5) - (2)(3) = 10 - 6 = 4\).
Clearly, \(4 \neq -1\). -
Generalizing Theorem 6 yields:
\[\det(A_1 A_2 \dots A_p) = (\det A_1)(\det A_2) \dots (\det A_p)\]
\[\det(A^p) = (\det A)^p\]
Given that \( \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = 7 \), find the following determinants:
(1)
(2)
(3)
(4)
Solution: First apply a column replacement (\( \text{Column } 2 = \text{Column } 2 - \frac{1}{3}\text{Column } 1 \)) which does not change the determinant. Then factor out \(-2\) from Row 1 and interchange Row 2 and Row 3.
Use a determinant to decide if the vectors \(\vec{v}_1, \vec{v}_2, \text{ and } \vec{v}_3\) are linearly independent when:
\[ \vec{v}_1 = \begin{bmatrix} 5 \\ -7 \\ 9 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} -3 \\ 3 \\ -5 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 2 \\ -7 \\ 5 \end{bmatrix} \]Solution: Form the matrix \(A\) using these vectors as columns, and compute the determinant using a row replacement followed by cofactor expansion.
Expand across the second row:
\[ = (-1)^{2+1}(-2) \begin{vmatrix} -3 & 2 \\ -5 & 5 \end{vmatrix} + 0 + (-1)^{2+3}(-5) \begin{vmatrix} 5 & -3 \\ 9 & -5 \end{vmatrix} \] \[ = (2)\left[ (-3)(5) - (2)(-5) \right] + (5)\left[ (5)(-5) - (-3)(9) \right] \] \[ = 2(-15 + 10) + 5(-25 + 27) = 2(-5) + 5(2) = -10 + 10 = 0 \]Because the determinant is \(0\), the matrix is not invertible. By Theorem 4, the columns \(\vec{v}_1\), \(\vec{v}_2\), and \(\vec{v}_3\) are linearly dependent.
Show that if \(A\) is invertible, then \(\det A^{-1} = \frac{1}{\det A}\).
Proof:
By definition of an inverse matrix: \(A \cdot A^{-1} = I_n\).
Taking the determinant of both sides:
\[ \det(A \cdot A^{-1}) = \det(I_n) \]Applying the Multiplicative Property (Theorem 6) to the left side, and knowing \(\det(I_n) = 1\):
\[ (\det A)(\det A^{-1}) = 1 \]Dividing by \(\det A\) (which is permitted since \(A\) is invertible, so \(\det A \neq 0\)):
\[ \det A^{-1} = \frac{1}{\det A} \]Let \(A\) and \(B\) be two \(8 \times 8\) matrices. Given that \(\det B = 4\) and \(\det(A^2 B) = 2^8\).
Find \(|\det A|\) and \(|\det(\frac{1}{2}A)|\).
Solution:
Find \(|\det A|\):
\(\det(A^2 B) = (\det A)^2 \cdot \det B = 2^8\)
\((\det A)^2 \cdot 4 = 256\)
\((\det A)^2 = \frac{256}{4} = 64\)
Therefore, \(|\det A| = 8\).
Find \(\det(\frac{1}{2}A)\):
Since \(A\) is an \(8 \times 8\) matrix, scaling the entire matrix by \(\frac{1}{2}\) scales the determinant by \((\frac{1}{2})^8\):
\(\det(\frac{1}{2}A) = (\frac{1}{2})^8 \cdot \det A = \frac{1}{256} \cdot (\pm 8) = \pm \frac{1}{32}\).
The magnitude is \(|\det(\frac{1}{2}A)| = \frac{1}{32}\).
(V) Elementary Matrices and Theorem 3 Proof Foundation
To prove Theorem 3, we classify an elementary matrix \(E\) based on the row operation used to obtain it from the identity matrix \(I\):
- Row Replacement Matrix: \(E\) is obtained by adding a multiple of one row to another row.
- Interchange Matrix: \(E\) is obtained by interchanging two rows of \(I\).
- Scale by \(r\) Matrix: \(E\) is obtained by multiplying a row of \(I\) by a nonzero scalar \(r\).
Theorem 3 (Revisited for Elementary Matrices)
If \(A\) is an \(n \times n\) matrix and \(E\) is an elementary matrix, then:
where:
- \(\det E = 1\) if \(E\) is a row replacement matrix.
- \(\det E = -1\) if \(E\) is an interchange matrix.
- \(\det E = r\) if \(E\) is a scale by \(r\) matrix.
(Theorem 3 can be rigorously proved by induction on the size of \(A\)).