Determinants

Example:

Consider \[ A = \begin{bmatrix} 3 & 2 & 5 \\ 7 & 5 & 4 \\ 0 & 1 & 0 \end{bmatrix}. \] Cofactor expansion across the first row gives \[ \begin{align*} \det A &= 3 \det \begin{bmatrix} 5 & 4 \\ 1 & 0 \end{bmatrix} - 2 \det \begin{bmatrix} 7 & 4 \\ 0 & 0 \end{bmatrix} + 5 \det \begin{bmatrix} 7 & 5 \\ 0 & 1 \end{bmatrix} \\\\ &= 3(-4) - 2(0) + 5(7) \\\\ &= 23. \end{align*} \]

Example:

Recompute \(\det A\) for the same matrix using the third row: \[ \begin{align*} \det A &= 0 - 1 \cdot \det \begin{bmatrix} 3 & 5 \\ 7 & 4 \end{bmatrix} + 0 \\\\ &= -1 (12 - 35) = 23. \end{align*} \] Or expanding down the first column: \[ \begin{align*} \det A &= 3 \det \begin{bmatrix} 5 & 4 \\ 1 & 0 \end{bmatrix} - 7 \det \begin{bmatrix} 2 & 5 \\ 1 & 0 \end{bmatrix} + 0 \\\\ &= 3(-4) - 7(-5) + 0 = 23. \end{align*} \] Choosing a row or column with many zero entries dramatically reduces the work.

Proof:

We use the fact (recorded in the Cofactor definition above) that the value of \(\det A\) is independent of which row or column is chosen for cofactor expansion. For an upper triangular matrix, expand along the first column. Every entry of the first column below the \((1,1)\) entry is zero, so every term except \(a_{11} C_{11}\) drops out and \[ \det A = a_{11} \det A_{11}, \] where \(A_{11}\) is the \((n-1) \times (n-1)\) matrix obtained by deleting row \(1\) and column \(1\) of \(A\). This matrix is again upper triangular. Iterating (formally, induction on \(n\), with base case \(\det [a_{11}] = a_{11}\)) yields \(\det A = a_{11} a_{22} \cdots a_{nn}\). The lower triangular case is analogous (expand along the first row).

Example:

\[ \det \begin{bmatrix} 1 & 7 & 5 & 4 & 2 \\ 0 & 2 & 9 & 2 & 3 \\ 0 & 0 & 3 & 5 & 7\\ 0 & 0 & 0 & 4 & 7\\ 0 & 0 & 0 & 0 & 5 \end{bmatrix} = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120. \]

Proof:

We proceed by induction on \(n\). For \(n = 1\), \(A = [a]\) is its own transpose and \(\det A = \det A^\top = a\) trivially. For \(n = 2\), \[ \begin{align*} \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} &= ad - bc, \\\\ \det \begin{bmatrix} a & c \\ b & d \end{bmatrix} &= ad - cb, \end{align*} \] which are equal.

For the inductive step, assume the result holds for all \((n-1) \times (n-1)\) matrices, and let \(A\) be \(n \times n\). Compare the cofactor expansion of \(\det A\) along the first row with that of \(\det A^\top\) along the first column: \[ \begin{align*} \det A &= \sum_{j=1}^n (-1)^{1+j} a_{1j} \det A_{1j}, \\\\ \det A^\top &= \sum_{i=1}^n (-1)^{i+1} a_{1i} \det (A^\top)_{i1}. \end{align*} \] In the second expansion the \((i,1)\) entry of \(A^\top\) has been written as \(a_{1i}\), by the definition of the transpose. The minor \((A^\top)_{i1}\) is the matrix obtained by deleting row \(i\) and column \(1\) of \(A^\top\), which is exactly the transpose of the matrix \(A_{1i}\) (delete row \(1\) and column \(i\) of \(A\)). The inductive hypothesis applied to the \((n-1) \times (n-1)\) minors therefore gives \[ \det (A^\top)_{i1} = \det (A_{1i})^\top = \det A_{1i}. \]

Renaming the summation index \(i \to j\), the two cofactor expansions match term by term, so \(\det A^\top = \det A\).

Example:

With \(A = \begin{bmatrix} 3 & 2 & 5 \\ 7 & 5 & 4 \\ 0 & 1 & 0 \end{bmatrix}\) from earlier, the transpose is \[ A^\top = \begin{bmatrix} 3 & 7 & 0 \\ 2 & 5 & 1 \\ 5 & 4 & 0 \end{bmatrix}, \] and expanding \(\det A^\top\) along the first row gives \[ \det A^\top = 3(-4) - 7(-5) + 0 = 23 = \det A. \]

Proof:

We split into two cases according to whether \(A\) is invertible.

Case 1: \(A\) is invertible.
By the Inverse via Row Reduction Theorem, some sequence of elementary row operations reduces \(A\) to \(I_n\). Each of those operations is undone by an elementary row operation of the same type, so undoing them in reverse order carries \(I_n\) back to \(A\) and exhibits \(A\) as a product of elementary matrices, \(A = F_1 F_2 \cdots F_m\). It therefore suffices to prove the multiplicative law for a single elementary matrix \(E\): \(\det(E B) = \det(E)\, \det(B)\). Then iterating gives \[ \begin{align*} \det(AB) &= \det(F_1 F_2 \cdots F_m\, B) \\\\ &= \det(F_1) \det(F_2 \cdots F_m\, B) \\\\ &= \cdots = \det(F_1) \cdots \det(F_m)\, \det(B) \\\\ &= \det(A)\, \det(B). \end{align*} \] Left multiplication by an elementary matrix performs the corresponding row operation, as recorded in Elementary Matrices. The effect of that operation on \(\det B\) then comes from the Effect of Row Operations on the Determinant, and the value of \(\det E\) is justified in each of the three cases:

• Replacement (\(E\) adds a multiple of one row of \(I\) to another):
  \(\det E = 1\) (triangular with all \(1\)'s on the diagonal), and \(EB\) replaces a row of \(B\) by itself plus a multiple of another row, leaving the determinant unchanged: \(\det(EB) = \det B = \det E \cdot \det B\).
• Interchange (\(E\) swaps two rows of \(I\)):
  \(\det E = -1\) (a single swap from \(I\)), and \(EB\) swaps the corresponding rows of \(B\), so \(\det(EB) = -\det B = \det E \cdot \det B\).
• Scale (\(E\) multiplies one row of \(I\) by \(k \neq 0\)):
  \(\det E = k\) (diagonal with one \(k\)), and \(EB\) scales the corresponding row of \(B\) by \(k\), so \(\det(EB) = k \det B = \det E \cdot \det B\).

Case 2: \(A\) is singular.
We show both sides of \(\det(AB) = \det A \cdot \det B\) vanish.

We first record two consequences of singularity, stated for an arbitrary singular \(n \times n\) matrix \(C\) so that they are available at each of the places below. By the Inverse via Row Reduction Theorem, the reduced row echelon form \(R\) of \(C\) is not \(I_n\). An \(n \times n\) matrix in reduced row echelon form that has a pivot in every row must be \(I_n\): each leading \(1\) is then the only nonzero entry of its column, and the echelon ordering puts the pivot of row \(i\) in column \(i\). So \(R\) has a row carrying no pivot, and in echelon form such a row is zero. Counting columns instead, \(R\) has at most \(n - 1\) pivot columns, so at least one of the \(n\) variables of \(R\mathbf{x} = \mathbf{0}\) is free. Setting it to \(1\) gives a nontrivial solution. Each row operation is performed by an invertible elementary matrix, so \(C\mathbf{x} = \mathbf{0}\) has the same solutions and also has a nontrivial one.

First, \(\det A = 0\). Applying the above with \(C = A\) gives a reduced row echelon form \(R\) of \(A\) with a zero row. Reducing \(A\) to \(R\) by elementary row operations gives \((F_m \cdots F_1) A = R\), and by Case 1 applied to the elementary factors, \[ \det(F_m) \cdots \det(F_1) \cdot \det A = \det R = 0 \] (the right-hand side because cofactor expansion along that zero row of \(R\) yields \(0\)). Each \(\det(F_i)\) is nonzero (it equals \(\pm 1\) or some \(k \neq 0\)), so \(\det A = 0\). Hence the right-hand side \(\det A \cdot \det B = 0\).

Second, \(AB\) is also singular, which we show by exhibiting a nonzero vector \(\mathbf{v}\) with \((AB)\mathbf{v} = \mathbf{0}\). Applying the above with \(C = A\) once more, the equation \(A\mathbf{x} = \mathbf{0}\) has a nontrivial solution \(\mathbf{x}_0 \neq \mathbf{0}\). We split on \(B\):

• If \(B\) is invertible, set \(\mathbf{v} = B^{-1} \mathbf{x}_0 \neq \mathbf{0}\) (nonzero because \(B^{-1}\) is invertible and \(\mathbf{x}_0 \neq \mathbf{0}\)). Then \[ (AB)\mathbf{v} = A(B B^{-1} \mathbf{x}_0) = A \mathbf{x}_0 = \mathbf{0}. \]
• If \(B\) is also singular, then \(B \mathbf{y}_0 = \mathbf{0}\) for some \(\mathbf{y}_0 \neq \mathbf{0}\). Set \(\mathbf{v} = \mathbf{y}_0\). Then \[ (AB)\mathbf{v} = A(B \mathbf{y}_0) = A \mathbf{0} = \mathbf{0}. \]

Either way, \(AB\) annihilates a nonzero vector, so \(AB\) is singular. Applying the above with \(C = AB\) produces a zero row in its reduced row echelon form, and the determinant computation just made for \(A\) gives \(\det(AB) = 0\). Both sides of the identity equal zero in this case.

The corollary follows by taking determinants of \(A A^{-1} = I\): \[ (\det A)(\det A^{-1}) = \det I = 1. \] Thus, \[ \det(A^{-1}) = (\det A)^{-1}. \]

Example:

Let \[ \begin{align*} A &= \begin{bmatrix} 1 & 2 \\ 8 & 9 \end{bmatrix}, \\\\ B &= \begin{bmatrix} 5 & 7 \\ 4 & 6 \end{bmatrix}, \\\\ AB &= \begin{bmatrix} 13 & 19 \\ 76 & 110 \end{bmatrix}. \end{align*} \] Then \[ \begin{align*} \det A &= 9 - 16 = -7, \\\\ \det B &= 30 - 28 = 2, \\\\ \det(AB) &= 1430 - 1444 = -14 = (-7)(2). \end{align*} \] Warning. The determinant is not additive: \(\det(A + B) \neq \det A + \det B\) in general.

Proof:

Throughout, we use freely that the determinant may be computed by cofactor expansion along any row (the Cofactor definition above).

Scale.
Suppose \(B\) is obtained from \(A\) by multiplying row \(r\) by \(k\). Expanding \(\det B\) along row \(r\), \[ \begin{align*} \det B &= \sum_{j=1}^n (-1)^{r+j} (k a_{rj}) \det B_{rj} \\\\ &= k \sum_{j=1}^n (-1)^{r+j} a_{rj} \det A_{rj} \\\\ &= k \det A, \end{align*} \] where \(B_{rj} = A_{rj}\) because deleting row \(r\) discards the scaled row entirely. Although elementary row operations require \(k \neq 0\), the identity \(\det B = k \det A\) as stated holds for every scalar \(k\).

Interchange.
We proceed by induction on \(n\). For \(n = 2\), direct computation gives \(\det \begin{bmatrix} c & d \\ a & b \end{bmatrix} = cb - ad = -(ad - bc)\), as in the \(2 \times 2\) verification below. Assume the result for \((n-1) \times (n-1)\) matrices, and let \(B\) be obtained from an \(n \times n\) matrix \(A\) by swapping rows \(i\) and \(j\) (with \(i \lt j\)). Choose any row \(p \notin \{i, j\}\). Such a row exists since \(n \geq 3\). Expand both \(\det A\) and \(\det B\) along row \(p\): \[ \begin{align*} \det A &= \sum_{\ell=1}^n (-1)^{p+\ell} a_{p\ell} \det A_{p\ell}, \\\\ \det B &= \sum_{\ell=1}^n (-1)^{p+\ell} a_{p\ell} \det B_{p\ell}, \end{align*} \] where we used \(b_{p\ell} = a_{p\ell}\) (row \(p\) is untouched by the swap). Deleting row \(p\) from the \(n \times n\) matrix sends row \(i\) to row \(i'\) and row \(j\) to row \(j'\) in the resulting \((n-1) \times (n-1)\) minor, where \(i' = i\) if \(i \lt p\) and \(i' = i - 1\) if \(i \gt p\) (similarly for \(j'\)). Crucially, \(i' \neq j'\), since \(i \neq j\) and the shift preserves the distinction. The minor \(B_{p\ell}\) is therefore obtained from \(A_{p\ell}\) by swapping rows \(i'\) and \(j'\), so by the inductive hypothesis \(\det B_{p\ell} = -\det A_{p\ell}\). Substituting gives \(\det B = -\det A\).

Zero consequence.
A matrix with two identical rows has determinant zero. Swapping the identical rows leaves the matrix unchanged, while by Interchange the swap negates the determinant, so \(\det A = -\det A\), forcing \(\det A = 0\).

Replacement.
Suppose \(B\) is obtained from \(A\) by adding \(k\) times row \(s\) to row \(r\) (with \(r \neq s\)). Expanding \(\det B\) along row \(r\), \[ \begin{align*} \det B &= \sum_{j=1}^n (-1)^{r+j} (a_{rj} + k a_{sj}) \det B_{rj} \\\\ &= \sum_{j=1}^n (-1)^{r+j} a_{rj} \det A_{rj} + k \sum_{j=1}^n (-1)^{r+j} a_{sj} \det A_{rj}, \end{align*} \] where again \(B_{rj} = A_{rj}\) (row \(r\) is discarded). The first sum is \(\det A\). The second is the cofactor expansion along row \(r\) of the matrix \(A'\) obtained by replacing row \(r\) of \(A\) with row \(s\). Since \(A'\) has two identical rows (rows \(r\) and \(s\) both equal to the original row \(s\) of \(A\)), its determinant vanishes by the Zero consequence above. Hence \(\det B = \det A\).

Example: \(2 \times 2\) verification.

Starting from \(\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\): \[ \begin{align*} \det \begin{bmatrix} c & d \\ a & b \end{bmatrix} &= cb - ad = -(ad - bc) \quad (\text{interchange}), \\\\ \det \begin{bmatrix} a & b \\ kc & kd \end{bmatrix} &= k(ad - bc) \quad (\text{scale row 2 by } k), \\\\ \det \begin{bmatrix} a & b \\ c+2a & d+2b \end{bmatrix} &= a(d + 2b) - b(c + 2a) = ad - bc \quad (\text{replacement}). \end{align*} \]

Proof:

Consider the \(n \times n\) identity matrix \(I\), and replace its \(i\)-th column by \(\mathbf{x}\) to form the modified matrix \[ I_i(\mathbf{x}) = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{x} & \cdots & \mathbf{e}_n \end{bmatrix}. \] Multiplying on the left by \(A\): \[ \begin{align*} A I_i(\mathbf{x}) &= \begin{bmatrix} A\mathbf{e}_1 & A\mathbf{e}_2 & \cdots & A\mathbf{x} & \cdots & A\mathbf{e}_n \end{bmatrix} \\\\ &= \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{b} & \cdots & \mathbf{a}_n \end{bmatrix} \\\\ &= A_i(\mathbf{b}). \end{align*} \]

Note that \(\det I_i(\mathbf{x}) = x_i\) (cofactor expansion down the \(i\)-th column gives \(x_i\) as the only nonzero contribution). By the multiplicativity of the determinant, \[ \begin{align*} (\det A)(\det I_i(\mathbf{x})) &= \det A_i(\mathbf{b}), \\\\ \text{hence} \quad (\det A)\, x_i &= \det A_i(\mathbf{b}). \end{align*} \] Since \(A\) is invertible, \(\det A \neq 0\), and dividing through by \(\det A\) gives (1).

Proof:

We show the stronger identity \[ A \cdot \operatorname{adj}(A) = (\det A)\, I, \] from which the formula \(A^{-1} = (\det A)^{-1} \operatorname{adj}(A)\) follows by dividing through by \(\det A \neq 0\) (using invertibility).

Compute the \((i, j)\)-entry of \(A \cdot \operatorname{adj}(A)\) directly: \[ \begin{align*} \bigl(A \cdot \operatorname{adj}(A)\bigr)_{ij} &= \sum_{k=1}^n a_{ik}\, \bigl(\operatorname{adj}(A)\bigr)_{kj} \\\\ &= \sum_{k=1}^n a_{ik}\, C_{jk}, \end{align*} \] using \(\bigl(\operatorname{adj}(A)\bigr)_{kj} = C_{jk}\) (the index swap from the transpose).

Diagonal entries (\(i = j\)).
The sum becomes \(\sum_k a_{ik} C_{ik}\), which is exactly the cofactor expansion of \(\det A\) along row \(i\). Hence \[ \bigl(A \cdot \operatorname{adj}(A)\bigr)_{ii} = \det A. \]

Off-diagonal entries (\(i \neq j\)).
The sum becomes \(\sum_k a_{ik} C_{jk}\). We claim this equals zero. Define an auxiliary matrix \(\tilde{A}\) obtained from \(A\) by replacing row \(j\) with a copy of row \(i\) (so \(\tilde{A}\) has two identical rows: rows \(i\) and \(j\) both equal to row \(i\) of \(A\)). The cofactor expansion of \(\det \tilde{A}\) along row \(j\) reads \[ \det \tilde{A} = \sum_k \tilde{a}_{jk}\, \tilde{C}_{jk} = \sum_k a_{ik}\, C_{jk}, \] because \(\tilde{a}_{jk} = a_{ik}\) (we copied row \(i\) into row \(j\)), and because the cofactor \(\tilde{C}_{jk}\) depends only on the matrix obtained by deleting row \(j\). That matrix is the same minor as the one defining \(C_{jk}\) for \(A\), since the \((n-1)\)-row matrix used to compute these cofactors does not involve row \(j\). Thus \[ \sum_k a_{ik}\, C_{jk} = \det \tilde{A}. \]

But \(\tilde{A}\) has two identical rows, so \(\det \tilde{A} = 0\) by the Zero consequence recorded with the Effect of Row Operations theorem. Therefore \(\bigl(A \cdot \operatorname{adj}(A)\bigr)_{ij} = 0\) for \(i \neq j\).

Combining diagonal and off-diagonal entries: \(A \cdot \operatorname{adj}(A) = (\det A)\, I\). Dividing by \(\det A\) gives the claimed formula.

Example:

Consider \[ A = \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & -4 \\ 3 & 3 & 2 \end{bmatrix}. \] Computing the nine cofactors: \[ \begin{align*} C_{11} &= +(2+12) = 14, & C_{12} &= -(4+12) = -16, & C_{13} &= +(6-3) = 3, \\\\ C_{21} &= -(4-9) = 5, & C_{22} &= +(-2-9) = -11, & C_{23} &= -(-3-6) = 9, \\\\ C_{31} &= +(-8-3) = -11, & C_{32} &= -(4-6) = 2, & C_{33} &= +(-1-4) = -5. \end{align*} \] Direct cofactor expansion gives \(\det A = -1(2+12) - 2(4+12) + 3(6-3) = -37\).

As a cross-check, multiplying \(A\) by \(\operatorname{adj}(A)\) yields a scalar multiple of the identity: \[ A\, \operatorname{adj}(A) = \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & -4 \\ 3 & 3 & 2 \end{bmatrix} \begin{bmatrix} 14 & 5 & -11 \\ -16 & -11 & 2 \\ 3 & 9 & -5 \end{bmatrix} = -37\, I, \] confirming \(\det A = -37\). Hence \[ A^{-1} = \frac{1}{-37}\, \operatorname{adj}(A) = \begin{bmatrix} -\tfrac{14}{37} & -\tfrac{5}{37} & \tfrac{11}{37} \\[2pt] \tfrac{16}{37} & \tfrac{11}{37} & -\tfrac{2}{37} \\[2pt] -\tfrac{3}{37} & -\tfrac{9}{37} & \tfrac{5}{37} \end{bmatrix}. \]