After this lecture you will be able to:
Recall from Section 2.2: For \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the determinant is
\[\det A = ad - bc.\]
The matrix is invertible exactly when \(\det A \neq 0\).
For a \(1\times 1\) matrix \(A = [a]\), we define \(\det A = a\).
When we row reduce an invertible \(3\times 3\) matrix, we arrive at the expression
\[\Delta = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}.\]
This number \(\Delta\) is the determinant of the \(3\times 3\) matrix.
For the \(3 \times 3\) determinant \(\Delta\) described above, since the terms in \(\Delta\) can be grouped as \[(a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32}) - (a_{12}a_{21}a_{33} - a_{12}a_{23}a_{31}) + (a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31}),\] we have \[\Delta = a_{11} \cdot \det \begin{bmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{bmatrix} - a_{12} \cdot \det \begin{bmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{bmatrix} + a_{13} \cdot \det \begin{bmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}.\]
For brevity, it can be written as a cofactor expansion (also called Laplace expansion or expansion by minors) across the first row:
\[\det A = a_{11}\cdot\det A_{11} - a_{12}\cdot\det A_{12} + a_{13}\cdot\det A_{13},\]
where \(A_{ij}\) is the submatrix obtained by deleting row \(i\) and column \(j\) from \(A\).
For example, given the matrix
\[A = \begin{bmatrix} 1 & -2 & 5 & 0 \\ 2 & 0 & 4 & -1 \\ 3 & 1 & 0 & 7 \\ 0 & 4 & -2 & 0 \end{bmatrix},\]
the submatrix \(A_{32}\) is obtained by crossing out row 3 and column 2.
Removing those entries leaves:
\[A_{32} = \begin{bmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}.\]
Alternative formula of the determinant of the \(3\times 3\) matrix - a visual mnemonic called Sarrus’s Rule. To use it:
Positive terms (down‑right):
\[a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32}\]
Negative terms (down‑left):
\[a_{13}a_{22}a_{31} + a_{11}a_{23}a_{32} + a_{12}a_{21}a_{33}\]
This gives exactly the expanded formula above.
For \(n \ge 2\), the determinant of an \(n\times n\) matrix \(A = [a_{ij}]\) is defined by cofactor expansion across the first row: \[\begin{align}\det A &= a_{11}\det A_{11} - a_{12}\det A_{12} + \cdots + (-1)^{1+n}a_{1n}\det A_{1n}\\ &= \sum_{j=1}^{n} (-1)^{1+j} a_{1j} \det A_{1j}\end{align}.\]
The \((i,j)\)-cofactor of \(A\) is
\[C_{ij} = (-1)^{i+j} \det A_{ij}.\]
With this notation, the first‑row expansion becomes
\[\det A = a_{11}C_{11} + a_{12}C_{12} + \cdots + a_{1n}C_{1n}.\]
Example 1 (textbook). Compute \(\det A\) for
\[A = \begin{bmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}.\]
Solution. Expand across the first row:
\[\begin{aligned} \det A &= 1\cdot C_{11} + 5\cdot C_{12} + 0\cdot C_{13} \\ &= 1\cdot (-1)^{1+1}\det\begin{bmatrix}4 & -1\\ -2 & 0\end{bmatrix} \;+\; 5\cdot (-1)^{1+2}\det\begin{bmatrix}2 & -1\\ 0 & 0\end{bmatrix} \\ &= 1\cdot(4\cdot0 - (-1)(-2)) \;-\; 5\cdot(2\cdot0 - (-1)\cdot0) \\ &= (0 - 2) - 5\cdot0 = -2. \end{aligned}\]
So \(\det A = -2\).
It turns out that you can expand along any row or any column.
The determinant of an \(n\times n\) matrix \(A\) can be computed by cofactor expansion across any row or down any column.
Expansion across row \(i\):
\[\det A = a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in}.\] Expansion down column \(j\):
\[\det A = a_{1j}C_{1j} + a_{2j}C_{2j} + \cdots + a_{nj}C_{nj}.\]
The sign of the cofactor depends on the position in the matrix, following the checkerboard pattern:
\[\begin{bmatrix} + & - & + & \cdots \\ - & + & - & \cdots \\ + & - & + & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}\]
Example 2 (textbook). Use a cofactor expansion across the third row and down the third column to compute \(\det A\) for
\[A = \begin{bmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}.\]
Solution.
Expanding across row 3 (\(i=3\)):
\[\begin{aligned} \det A &= a_{31}C_{31} + a_{32}C_{32} + a_{33}C_{33} \\ &= 0\cdot C_{31} + (-2)\cdot C_{32} + 0\cdot C_{33} \\ &= -2\,C_{32} \\ &= -2 \cdot (-1)^{3+2}\det A_{32} \\ &= -2 \cdot (-1)^5 \cdot \det\begin{bmatrix}1 & 0\\ 2 & -1\end{bmatrix} \\ &= -2 \cdot (-1) \cdot \bigl(1\cdot(-1) - 0\cdot 2\bigr) \\ &= -2 \cdot (-1) \cdot (-1) \\ &= -2. \end{aligned}\]
Expanding down column 3 (\(j=3\)):
\[\begin{aligned} \det A &= a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} \\ &= 0\cdot C_{13} + (-1)\cdot C_{23} + 0\cdot C_{33} \\ &= -C_{23} = -(-1)^{2+3}\det A_{23} \\ &= -(-1)^5\cdot\det\begin{bmatrix}1 & 5\\ 0 & -2\end{bmatrix} \\ &= -(-1)\cdot(1\cdot(-2) - 5\cdot0) = 1\cdot(-2) = -2, \end{aligned}\] which matches the previous result.
N.B. If a matrix has many zeros, choose the row or column with the most zeros for the cofactor expansion.
N.B. If an entire row or column consists of zeros, the determinant is zero.
Example 3 (textbook). Compute \(\det A\) for
\[A = \begin{bmatrix} 3 & -7 & 8 & 9 & -6 \\ 0 & 2 & -5 & 7 & 3 \\ 0 & 0 & 1 & 5 & 0 \\ 0 & 0 & 2 & 4 & -1 \\ 0 & 0 & 0 & -2 & 0 \end{bmatrix}.\]
Expand down the first column (all zeros except the top entry):
\[\det A = 3\cdot C_{11} = 3\cdot (-1)^{2}\det A_{11} =3 \cdot \begin{vmatrix} 2 & -5 & 7 & 3 \\ 0 & 1 & 5 & 0 \\ 0 & 2 & 4 & -1 \\ 0 & 0 & -2 & 0 \end{vmatrix}.\]
Now, the remaining \(4 \times 4\) determinant also has zeros in its first column (except the first entry). Expand down its first column:
\[\det A = 3 \cdot 2 \cdot \begin{vmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{vmatrix}.\]
We are left with a \(3 \times 3\) determinant that was computed in Example 1 (or can be evaluated quickly):
\[\begin{vmatrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{vmatrix} = -2.\]
Therefore,
\[\det A = 3 \cdot 2 \cdot (-2) = -12.\]
If \(A\) is a triangular matrix (upper or lower), then \(\det A\) is the product of the entries on its main diagonal.
A cofactor expansion requires roughly \(n!\) multiplications. For \(n=25\), this is about \(1.5\times10^{25}\) operations – impossible even on the fastest computers. Later we will learn how row reduction reduces this to about \(2n^3/3\) operations, making determinant calculation feasible for large matrices.
Compute the determinant of \(\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}\) using the \(2\times2\) formula.
Use a cofactor expansion across the first row to compute \(\det A\) for
\[A = \begin{bmatrix} 3 & 0 & 4 \\ 2 & 3 & 2 \\ 0 & 5 & -1 \end{bmatrix}.\]
Repeat the calculation for Problem 2 using a cofactor expansion down the second column, and verify that the result is the same.
Find the determinant of the triangular matrix
\[B = \begin{bmatrix} -2 & 0 & 0 & 0 \\ 5 & 3 & 0 & 0 \\ 1 & 7 & -4 & 0 \\ 2 & -1 & 6 & 2 \end{bmatrix}\] by inspection.
Is the matrix \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) invertible? Use the determinant to decide. (Hint: row reduction may help, or compute directly.)
(Solutions at the end.)
# Determinant of a 2x2
A2 <- matrix(c(2,3,1,4), nrow=2)
det(A2)[1] 5# Determinant of a 3x3
A3 <- matrix(c(3,2,0,0,3,5,4,2,-1), nrow=3)
det(A3)[1] 1# Example 1 from textbook
A <- matrix(c(1,2,0,5,4,-2,0,-1,0), nrow=3)
det(A) # should be -2[1] -2# Triangular determinant: product of diagonal
B <- matrix(c(-2,5,1,2,0,3,7,-1,0,0,-4,6,0,0,0,2), nrow=4)
det(B)[1] 48prod(diag(B)) # same[1] 48\(\det = 2\cdot4 - 3\cdot1 = 8 - 3 = 5\).
Expanding across row 1 (\(a_{11}=3\), \(a_{12}=0\), \(a_{13}=4\)): \[\det A = 3\cdot C_{11} + 0\cdot C_{12} + 4\cdot C_{13}
= 3\cdot(+1)\cdot\det\begin{bmatrix}3 & 2\\5 & -1\end{bmatrix}
+ 4\cdot(+1)\cdot\det\begin{bmatrix}2 & 3\\0 & 5\end{bmatrix}.\] First cofactor: \((3\cdot(-1) - 2\cdot5) = (-3-10) = -13\), times \(3\) gives \(-39\).
Second: \((2\cdot5 - 3\cdot0) = 10\), times \(4\) gives \(40\).
Sum \(= -39 + 40 = 1\). So \(\det A = 1\).
Expanding down column 2 (\(j=2\)): \(a_{12}=0\), \(a_{22}=3\), \(a_{32}=5\). \[\det A = 0\cdot C_{12} + 3\cdot C_{22} + 5\cdot C_{32}.\] \(C_{22} = (-1)^{2+2}\det\begin{bmatrix}3&4\\0&-1\end{bmatrix} = (3\cdot(-1)-4\cdot0) = -3\).
\(C_{32} = (-1)^{3+2}\det\begin{bmatrix}3&4\\2&2\end{bmatrix} = -1\cdot(3\cdot2 - 4\cdot2) = -1\cdot(6-8)=2\).
Thus \(\det A = 3\cdot(-3) + 5\cdot2 = -9 + 10 = 1\). Same result.
\(B\) is lower triangular. Its determinant is the product of the diagonal entries: \((-2)\cdot3\cdot(-4)\cdot2 = 48\).
Compute determinant. Row reduce to echelon form or use cofactor expansion. Quick row reduction: \[\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix} \xrightarrow{R2-4R1,\; R3-7R1} \begin{bmatrix}1&2&3\\0&-3&-6\\0&-6&-12\end{bmatrix} \xrightarrow{R3-2R2} \begin{bmatrix}1&2&3\\0&-3&-6\\0&0&0\end{bmatrix}.\] Upper triangular form has a zero on the diagonal, so determinant is \(0\). Therefore the matrix is not invertible.
After this lecture you will be able to:
The key to efficient determinant computation is knowing how elementary row operations change the determinant.
Let \(A\) be a square matrix.
These properties correspond exactly to the effect of elementary matrices:
Thus for any elementary matrix \(E\), \(\det(EA) = (\det E)(\det A)\).
N.B. Scaling and scalar multiplication are different: For any \(n \times n\) matrix \(A\) and any scalar \(k\),
\[\det(kA) = k^n \det A.\]
That is, multiplying every entry of a matrix by \(k\) multiplies the determinant by \(k^n\), where \(n\) is the size of the matrix.
If \(A = \begin{bmatrix} a_{ij} \end{bmatrix}_{3 \times 3}\), then
\[\det(kA) = \begin{vmatrix} ka_{11} & ka_{12} & ka_{13} \\ ka_{21} & ka_{22} & ka_{23} \\ ka_{31} & ka_{32} & ka_{33} \end{vmatrix} = k^3 \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = k^3 \det A.\]
In general, for an \(n \times n\) matrix, the factor \(k\) can be factored out of each of the \(n\) rows (or columns), giving \(k^n\).
The idea: row reduce the matrix to an upper triangular form U using only row replacements and row interchanges (no scaling until the end, if needed). Then
\[\det A = (-1)^r \cdot (\text{product of the diagonal entries of } U),\]
where \(r\) is the number of row interchanges performed.
Example 1 (textbook). Compute \(\det A\) for
\[A = \begin{bmatrix} 1 & -4 & 2 \\ -2 & 8 & -9 \\ -1 & 7 & 0 \end{bmatrix}.\]
Solution. First, create zeros below the first pivot using row replacements (these do not change the determinant):
\[\begin{bmatrix} 1 & -4 & 2 \\ 0 & 0 & -5 \\ 0 & 3 & 2 \end{bmatrix} \qquad (R_2 \leftarrow R_2 + 2R_1,\; R_3 \leftarrow R_3 + R_1)\]
Now interchange rows 2 and 3 to bring a nonzero into the pivot position. This changes the sign:
\[\det A = -\det \begin{bmatrix} 1 & -4 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & -5 \end{bmatrix}.\]
The matrix is now upper triangular, so its determinant is the product of the diagonal entries:
\[\det A = -\,(1\cdot 3\cdot (-5)) = -(-15) = 15.\]
Example 2 (textbook). Compute \(\det A\) for
\[A = \begin{bmatrix} 2 & -8 & 6 & 8 \\ 3 & -9 & 5 & 10 \\ -3 & 0 & 1 & -2 \\ 1 & -4 & 0 & 6 \end{bmatrix}.\]
Solution. Factor out a common multiple of one row (by theorem 3(c)). Here, we factor 2 out of the first row (this multiplies the determinant by 2):
\[\det A = 2\cdot \det \begin{bmatrix} 1 & -4 & 3 & 4 \\ 3 & -9 & 5 & 10 \\ -3 & 0 & 1 & -2 \\ 1 & -4 & 0 & 6 \end{bmatrix}.\]
Create zeros below the pivot in column 1:
\[= 2\cdot \det \begin{bmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & -12 & 10 & 10 \\ 0 & 0 & -3 & 2 \end{bmatrix} \quad (R_2-3R_1,\; R_3+3R_1,\; R_4-R_1).\]
Create a zero in column 2 by adding \(4R_2\) to \(R_3\):
\[= 2\cdot \det \begin{bmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & 0 & -6 & 2 \\ 0 & 0 & -3 & 2 \end{bmatrix}.\]
Finally, add \(-\frac12 R_3\) to \(R_4\) (row replacement, determinant unchanged):
\[= 2\cdot \det \begin{bmatrix} 1 & -4 & 3 & 4 \\ 0 & 3 & -4 & -2 \\ 0 & 0 & -6 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}.\]
Now compute the determinant as the product of the diagonal entries:
\[\det A = 2 \cdot (1\cdot 3\cdot (-6)\cdot 1) = 2 \cdot (-18) = -36.\]
Comments: Although the echelon form \(U\) described above is not unique (because it is not completely row reduced), and the pivots are not unique, the product of the pivots is unique, except for a possible minus sign. In summary,
If \(A\) is an \(n \times n\) matrix, and we reduce it to an upper triangular matrix \(U\) using row operations, then
\[\det A = \begin{cases} (-1)^r \cdot \left( \text{product of pivots in } U \right) & \text{if } A \text{ is invertible}, \\[6pt] 0 & \text{if } A \text{ is not invertible}. \end{cases}\]
Here, \(r\) is the total number of row interchanges performed during the reduction.
From the row reduction process, we see that a square matrix \(A\) can be reduced to an echelon form \(U\) with diagonal entries \(u_{11},...,u_{nn}\). If \(A\) is invertible, all \(u_{ii}\) are pivots (nonzero). If \(A\) is singular, at least one \(u_{ii}=0\), making the product zero. Thus
A square matrix \(A\) is invertible if and only if \(\det A \neq 0\).
This theorem adds the statement “\(\det A \neq 0\)” to the Invertible Matrix Theorem (IMT).
N.B. Consequently, \(\det A = 0\) exactly when the columns (or rows) are linearly dependent. If two rows or columns are identical, or a row/column is all zeros, the determinant is immediately zero.
Example 3 (textbook).
\[A = \begin{bmatrix} 3 & -1 & 2 & -5 \\ 0 & 5 & -3 & -6 \\ -6 & 7 & -7 & 4 \\ -5 & -8 & 0 & 9 \end{bmatrix}.\]
Adding 2 times row 1 to row 3 gives a matrix whose second and third rows are equal,
\[\det A = \det \begin{bmatrix} 3 & -1 & 2 & -5 \\ 0 & 5 & -3 & -6 \\ 0 & 5 & -3 & -6 \\ -5 & -8 & 0 & 9 \end{bmatrix}\]
so \(\det A = 0.\)
Example 4 (textbook). Compute \(\det A\), where
\[A = \begin{bmatrix} 0 & 1 & 2 & -1 \\ 2 & 5 & -7 & 3 \\ 0 & 3 & 6 & 2 \\ -2 & -5 & 4 & -2 \end{bmatrix}.\]
Solution.
By adding row 1 to row 4 (i.e., \(R_4 \leftarrow R_4 + R_1\)),
\[\det A = \begin{vmatrix} 0 & 1 & 2 & -1 \\ 2 & 5 & -7 & 3 \\ 0 & 3 & 6 & 2 \\ 0 & 0 & -3 & 1 \end{vmatrix}.\]
Cofactor expansion along column 1:
\[\det A = 2 \cdot (-1)^{2+1} \cdot \begin{vmatrix} 1 & 2 & -1 \\ 3 & 6 & 2 \\ 0 & -3 & 1 \end{vmatrix} = -2 \cdot \begin{vmatrix} 1 & 2 & -1 \\ 3 & 6 & 2 \\ 0 & -3 & 1 \end{vmatrix}.\]
Replace row 2 with \(R_2 - 3R_1\) (i.e., subtract 3 times row 1 from row 2),
\[\det A = -2 \cdot \begin{vmatrix} 1 & 2 & -1 \\ 0 & 0 & 5 \\ 0 & -3 & 1 \end{vmatrix}=-2\cdot 1 \cdot (-1)^{1+1} \cdot \begin{vmatrix} 0 & 5 \\ -3 & 1 \end{vmatrix} = -2\cdot\begin{vmatrix} 0 & 5 \\ -3 & 1 \end{vmatrix}.\]
Get the final result
\[\det A = -2 \cdot 15 = -30.\]
Summary of the Strategy
The approach in Example 4 combines two powerful techniques:
For any square matrix \(A\), \(\det A^T = \det A\).
The proof uses induction and the fact that the cofactor expansion along the first row of \(A\) matches the expansion down the first column of \(A^T\).
Proof Strategy: Mathematical Induction (supplementary)
The proof uses the Principle of Mathematical Induction:
Since the base case holds and the inductive step shows that truth for \(k\) implies truth for \(k+1\), the theorem is true for all \(n \ge 1\).
Step-by-Step Proof. (supplementary)
Step 1: Base Case (\(n = 1\))
If \(A = \begin{bmatrix} a \end{bmatrix}\), then \(A^T = \begin{bmatrix} a \end{bmatrix}\).
Thus, \[ \det A = a \quad \text{and} \quad \det A^T = a, \] so \(\det A = \det A^T\). The base case holds.
Step 2: Inductive Hypothesis
Assume the theorem is true for all \(k \times k\) matrices. That is, for any \(k \times k\) matrix \(M\), \[ \det M^T = \det M. \]
Step 3: Inductive Step (Proving for \(n = k+1\))
Let \(A\) be a \((k+1) \times (k+1)\) matrix. We will show \(\det A = \det A^T\) by comparing the cofactor expansions.
The determinant of \(A\) expanded along its first row is: \[ \det A = a_{11} C_{11} + a_{12} C_{12} + \cdots + a_{1,\,k+1} C_{1,\,k+1}, \] where \(C_{1j}\) is the cofactor of entry \(a_{1j}\). Recall that \[ C_{1j} = (-1)^{1+j} \det M_{1j}, \] and \(M_{1j}\) is the \(k \times k\) matrix obtained by deleting row 1 and column \(j\) from \(A\).
Let \(B = A^T\). The entry in row \(j\), column 1 of \(B\) is: \[ b_{j1} = (A^T)_{j1} = a_{1j}. \]
Expanding \(\det B = \det A^T\) down its first column gives: \[ \det A^T = b_{11} C'_{11} + b_{21} C'_{21} + \cdots + b_{k+1,\,1} C'_{k+1,\,1}, \] where \(C'_{j1}\) is the cofactor of entry \(b_{j1}\) in \(B = A^T\). By definition, \[ C'_{j1} = (-1)^{j+1} \det M'_{j1}, \] and \(M'_{j1}\) is the \(k \times k\) matrix obtained by deleting row \(j\) and column 1 from \(A^T\).
We now compare the cofactor \(C_{1j}\) of \(A\) with the cofactor \(C'_{j1}\) of \(A^T\).
Observe that deleting row \(j\) and column 1 from \(A^T\) is exactly the same as deleting column \(j\) and row 1 from \(A\), and then transposing the resulting matrix. In other words, \[ M'_{j1} = (M_{1j})^T. \]
By the induction hypothesis, the theorem holds for these \(k \times k\) matrices, so: \[ \det M'_{j1} = \det (M_{1j})^T = \det M_{1j}. \]
Furthermore, the cofactor signs are identical because: \[ (-1)^{1+j} = (-1)^{j+1} \]Therefore, \[ C'_{j1} = C_{1j}. \]
Part D: Comparing the two expansions
Now substitute back into the expansion for \(\det A^T\): \[ \det A^T = \sum_{j=1}^{k+1} b_{j1} C'_{j1} = \sum_{j=1}^{k+1} a_{1j} C_{1j} = \det A. \]
Thus, \(\det A = \det A^T\) for \(n = k+1\).
Step 4: Conclusion by Induction
By the Principle of Mathematical Induction, the theorem holds for all square matrices of size \(n \ge 1\). Hence, \[ \boxed{\det A^T = \det A} \] for every \(n \times n\) matrix \(A\).
N.B. An important consequence: Column Operations Work Like Row Operations. That is, every statement about row operations applies equally to column operations because a column operation on \(A\) is a row operation on \(A^T\), and the determinant remains the same under transposition.
Thus, if a multiple of one column is added to another, the determinant is unchanged; swapping two columns changes the sign; scaling a column multiplies the determinant by that scalar.
N.B. Practical Note: Despite the symmetry between rows and columns, in numerical calculations (e.g., Gaussian elimination), it is standard to perform only row operations to maintain consistency and avoid confusion. However, knowing that column operations are valid gives you extra flexibility when solving theoretical problems or dealing with special matrix structures.
If \(A\) and \(B\) are \(n\times n\) matrices, then \[\det(AB) = (\det A)(\det B).\]
The result extends to products of several matrices: \[ \det(A_1 A_2 \cdots A_k) = \det(A_1) \det(A_2) \cdots \det(A_k). \]
proof. (supplementary)
The proof uses elementary matrices and the fact that any invertible matrix is a product of elementary matrices.
We split the proof into two exhaustive cases, depending on whether \(A\) is invertible or singular.
Case 1: \(A\) is Singular (Non-Invertible)
Since \(A\) is singular, by Theorem 4 we have: \[ \det A = 0. \]
As established in the prerequisites, if \(A\) is singular, then the product \(AB\) is also singular, regardless of what \(B\) is. Therefore, applying Theorem 4 to \(AB\): \[ \det(AB) = 0. \]
Now compute the right-hand side of the theorem: \[ (\det A)(\det B) = 0 \cdot \det B = 0. \]
Both sides are zero: \[ \det(AB) = 0 = (\det A)(\det B). \]
Thus, the theorem holds in the singular case.
N.B. Singularity of a Product: If \(A\) is singular (non-invertible), then the product \(AB\) is also singular for any matrix \(B\).
Why? Suppose, for contradiction, that \(AB\) were invertible. Then there exists a matrix \(C\) such that \((AB)C = I\). By associativity, \(A(BC) = I\). This would imply that \(A\) has a right inverse, and since \(A\) is square, that would make \(A\) invertible (by the Invertible Matrix Theorem). This contradicts the assumption that \(A\) is singular. Therefore, \(AB\) must also be singular.
Case 2: \(A\) is Invertible
Since \(A\) is invertible, by the Invertible Matrix Theorem, it is row equivalent to \(I_n\). Therefore, there exists a sequence of elementary row operations that reduces \(A\) to \(I_n\). Equivalently, there exist elementary matrices \(E_1, E_2, \dots, E_p\) such that: \[ A = E_p E_{p-1} \cdots E_1. \]
Now consider the product \(AB\). Substitute the above expression for \(A\): \[ AB = (E_p E_{p-1} \cdots E_1) B. \]
We will compute \(\det(AB)\) by repeatedly applying the a key lemma directly from Theorem 3 (Row Operations): For any elementary matrix \(E\) and any matrix \(M\) of compatible size,
\[\det(EM) = \det(E) \det(M).\]
Let’s denote the determinant by \(|\cdot|\) for brevity. Start with: \[ |AB| = |E_p E_{p-1} \cdots E_1 B|. \]
Apply the rule to the leftmost elementary matrix \(E_p\): \[ |E_p (E_{p-1} \cdots E_1 B)| = |E_p| \cdot |E_{p-1} \cdots E_1 B|. \]
Now apply the rule again to \(E_{p-1}\): \[ = |E_p| \cdot |E_{p-1}| \cdot |E_{p-2} \cdots E_1 B|. \]
Continuing this process for all \(p\) elementary matrices, we eventually get: \[ |AB| = |E_p| \cdot |E_{p-1}| \cdots |E_1| \cdot |B|. \]
Since scalar multiplication is commutative, we can rearrange the product of the determinants of the elementary matrices: \[ |AB| = \big( |E_p| \cdot |E_{p-1}| \cdots |E_1| \big) \cdot |B|. \]
But notice that the expression in parentheses is exactly the determinant of the product of these elementary matrices, again by the same multiplicative rule applied repeatedly: \[ |E_p E_{p-1} \cdots E_1| = |E_p| \cdot |E_{p-1}| \cdots |E_1|. \]
Since \(A = E_p E_{p-1} \cdots E_1\), we have: \[ |E_p E_{p-1} \cdots E_1| = |A|. \]
Therefore: \[ |AB| = |A| \cdot |B|. \]
Thus, the theorem holds in the invertible case.
In conclusion, both cases—\(A\) singular and \(A\) invertible—yield the same result: \[ \boxed{\det(AB) = \det(A) \det(B)}. \]
This completes the proof.
Supplementary proof of the lemma from Theorem 3:
Therefore, the lemma holds for all three types of elementary matrices.
Example 5 (textbook). Verify Theorem 6 for \(A = \begin{bmatrix} 6 & 1 \\ 3 & 2 \end{bmatrix}\), \(B = \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\).
\[\det A = 6\cdot2 - 1\cdot3 = 9,\qquad \det B = 4\cdot2 - 3\cdot1 = 5.\]
\[AB = \begin{bmatrix} 6(4)+1(1) & 6(3)+1(2) \\ 3(4)+2(1) & 3(3)+2(2) \end{bmatrix} = \begin{bmatrix} 25 & 20 \\ 14 & 13 \end{bmatrix}.\]
\[\det(AB) = 25\cdot13 - 20\cdot14 = 325 - 280 = 45 = 9\cdot5.\]
Corollary. If \(A\) is invertible, \(\det(A^{-1}) = 1/\det A\).
Warning: In general, \(\det(A+B) \neq \det A + \det B\). There is no sum rule.
The determinant is not a linear function of the entire matrix \(A\), but it does satisfy a important multi-linearity property: if we hold all but one column fixed, the determinant behaves as a linear function of the remaining column.
N.B. If we fix all columns except the \(j\)‑th, then the determinant is a linear function of that one column vector variable.
Statement of the Property
Let \(A\) be an \(n \times n\) matrix, and consider its columns: \[ A = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n \end{bmatrix}. \]
Fix all columns except the \(j\)-th column. Allow the \(j\)-th column to vary, and denote it by \(\mathbf{x}\). Define a function \(T: \mathbb{R}^n \to \mathbb{R}\) by: \[ T(\mathbf{x}) = \det \begin{bmatrix} \mathbf{a}_1 & \cdots & \mathbf{a}_{j-1} & \mathbf{x} & \mathbf{a}_{j+1} & \cdots & \mathbf{a}_n \end{bmatrix}. \]
Then \(T\) is a linear transformation from \(\mathbb{R}^n\) to \(\mathbb{R}\). That is, it satisfies two properties:
Homogeneity (Scaling):
For any scalar \(c\) and any vector \(\mathbf{x} \in \mathbb{R}^n\), \[ T(c\mathbf{x}) = c\,T(\mathbf{x}). \]
Additivity:
For any vectors \(\mathbf{u}, \mathbf{v} \in \mathbb{R}^n\), \[ T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}). \]
That is, for any vectors \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\) and scalar \(c\),
\[\det[\dots\; c\mathbf{v}\; \dots] = c\cdot\det[\dots\; \mathbf{v}\; \dots].\]
\[\det[\mathbf{a}_1\; \dots\; \mathbf{a}_{j-1}\; (\mathbf{u}+\mathbf{v})\; \mathbf{a}_{j+1}\; \dots\; \mathbf{a}_n] = \det[\dots\;\mathbf{u}\;\dots] + \det[\dots\;\mathbf{v}\;\dots],\]
Comment: This property follows from Theorem 3(c) and a cofactor expansion down column \(j\). It is used in more advanced treatments of determinants.
N.B. This linearity property applies one column at a time. The determinant is not linear as a function of the entire matrix. For example, \(\det(A + B) \neq \det(A) + \det(B)\) in general.
Use row reduction to evaluate the determinant of
\[\begin{bmatrix} 1 & 3 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}.\] (Hint: This is similar to Example 1.)
Compute \(\det A\) for \(A = \begin{bmatrix} 2 & 5 & -3 & -2 \\ -2 & -3 & 2 & -5 \\ 1 & 3 & -2 & 0 \\ -1 & -6 & 4 & 3 \end{bmatrix}\) by row reduction. (Hint: keep track of any interchanges or scaling.)
Verify the multiplicative property for \(A = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}\) and \(B = \begin{bmatrix} 4 & 1 \\ 0 & 2 \end{bmatrix}\).
If \(A\) is a \(4\times 4\) matrix with \(\det A = 3\), find:
True or False: If two rows of a square matrix are identical, then its determinant is zero. Explain why using row operations.
(Solutions at the end.)
# Example 1 using row reduction
A <- matrix(c(1, -2, -1, -4, 8, 7, 2, -9, 0), nrow=3)
# Elementary operations and determinant tracking
library(pracma)
det(A) # should be 15[1] 15# Example 2
B <- matrix(c(2,3,-3,1, -8,-9,0,-4, 6,5,1,0, 8,10,-2,6), nrow=4)
det(B) # should be -36[1] -36# Verify Theorem 6
A1 <- matrix(c(6,3, 1,2), nrow=2)
B1 <- matrix(c(4,1, 3,2), nrow=2)
det(A1 %*% B1)[1] 45det(A1) * det(B1)[1] 45# Effect of scalar multiplication
A2 <- matrix(rnorm(16), nrow=4)
det(2*A2)[1] 7.1226662^4 * det(A2) # scalar factor appears as 2^n when entire matrix is scaled. [1] 7.122666# (For a single row scaling, it's just that scalar factor.)Row reduce \(\begin{bmatrix} 1 & 3 & 0 \\ 2 & 4 & -1 \\ 0 & -2 & 0 \end{bmatrix}\). \(R_2 \leftarrow R_2 - 2R_1\) gives \(\begin{bmatrix} 1 & 3 & 0 \\ 0 & -2 & -1 \\ 0 & -2 & 0 \end{bmatrix}\). \(R_3 \leftarrow R_3 - R_2\) (replacement, det unchanged): \(\begin{bmatrix} 1 & 3 & 0 \\ 0 & -2 & -1 \\ 0 & 0 & 1 \end{bmatrix}\). Now triangular; determinant = \(1\cdot(-2)\cdot1 = -2\).
Swap \(R_1\) and \(R_3\):
\[ \det A = -\det \begin{bmatrix} 1&3&-2&0\\ -2&-3&2&-5\\ 2&5&-3&-2\\ -1&-6&4&3 \end{bmatrix} \]
Now eliminate below the first pivot:
\[ R_2\leftarrow R_2+2R_1,\quad R_3\leftarrow R_3-2R_1,\quad R_4\leftarrow R_4+R_1 \]
\[ = -\det \begin{bmatrix} 1&3&-2&0\\ 0&3&-2&-5\\ 0&-1&1&-2\\ 0&-3&2&3 \end{bmatrix} \]
Eliminate below the second pivot:
\[ R_3\leftarrow R_3+\frac13R_2,\quad R_4\leftarrow R_4+R_2 \]
\[ = -\det \begin{bmatrix} 1&3&-2&0\\ 0&3&-2&-5\\ 0&0&\frac13&-\frac{11}{3}\\ 0&0&0&-2 \end{bmatrix} \]
The matrix is now upper triangular, so its determinant is the product of the diagonal entries:
\[ 1\cdot 3\cdot \frac13\cdot (-2)=-2 \]
Thus,
\[ \det A = -(-2)=2 \]
\(\det A = 1\cdot3 - 0\cdot2 = 3\); \(\det B = 4\cdot2 - 1\cdot0 = 8\). \(AB = \begin{bmatrix}1\cdot4+0\cdot0 & 1\cdot1+0\cdot2 \\ 2\cdot4+3\cdot0 & 2\cdot1+3\cdot2\end{bmatrix} = \begin{bmatrix}4 & 1 \\ 8 & 8\end{bmatrix}\). \(\det(AB) = 4\cdot8 - 1\cdot8 = 32 - 8 = 24 = 3\cdot8 = \det A \det B\). Verified.
True. Suppose rows \(i\) and \(j\) are identical. Adding \(-1\) times row \(i\) to row \(j\) yields a row of zeros, which does not change the determinant (replacement). The resulting matrix has a row of zeros, so its determinant is 0. Thus the original determinant is 0. (Alternatively, swapping the identical rows leaves the matrix unchanged but flips the sign of the determinant, so \(\det A = -\det A \Rightarrow \det A = 0\).)
These notes follow Chapter 3 of Lay, Lay & McDonald, “Linear Algebra and its Applications”, 5th edition.