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: \[ \begin{align*} \det A &= 3 \det \begin{bmatrix} 5 & 4 \\ 1 & 0 \end{bmatrix} - 7 \det \begin{bmatrix} 2 & 5 \\ 1 & 0 \end{bmatrix} + 0 \\ &= 3(-4) - 7(-5) + 0 = 23. \end{align*} \] The choice of row or column does not affect the result — but choosing one 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: only the \((1,1)\) entry is nonzero, so all cofactors except \(C_{11}\) drop 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 \(A\) — itself 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 \(n = 2\), \[ \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc, \qquad \det \begin{bmatrix} a & c \\ b & d \end{bmatrix} = ad - cb, \] 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: \[ \det A = \sum_{j=1}^n (-1)^{1+j} a_{1j} \det A_{1j}, \qquad \det A^\top = \sum_{i=1}^n (-1)^{i+1} (A^\top)_{i1} \det (A^\top)_{i1}. \] Since \((A^\top)_{i1} = a_{1i}\) (definition of transpose) and 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 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 \[ \det A^\top = 3(-4) - 7(-5) + 0 = 23 = \det A. \checkmark \]

Proof:

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

Case 1: \(A\) is invertible. By the Inverse via Row Reduction theorem, \(A\) factors as a product of elementary matrices, \(A = F_1 F_2 \cdots F_k\). It therefore suffices to prove the multiplicative law for a single elementary matrix \(E\): \(\det(E B) = \det(E)\, \det(B)\). Then iterating gives \[ \det(AB) = \det(F_1 F_2 \cdots F_k\, B) = \det(F_1) \det(F_2 \cdots F_k\, B) = \cdots = \det(F_1) \cdots \det(F_k)\, \det(B) = \det(A)\, \det(B). \] For the elementary case, we read off the determinant of \(E\) and the effect of \(E B\) directly from the Row Operations theorem above. There are 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: \(\det(EB) = \det B = \det E \cdot \det B\). \(\checkmark\)
• Interchange (\(E\) swaps two rows of \(I\)): \(\det E = -1\) (a single swap from \(I\)), and \(EB\) swaps the corresponding rows of \(B\), so \(\det(EB) = -\det B = \det E \cdot \det B\). \(\checkmark\)
• 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 \(\det(EB) = k \det B = \det E \cdot \det B\). \(\checkmark\)

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

First, \(\det A = 0\). Indeed, by the Inverse via Row Reduction theorem, the RREF of a singular \(A\) is not \(I_n\), so its RREF \(R\) has at least one zero row. Reducing \(A\) to \(R\) by elementary row operations gives \((F_k \cdots F_1) A = R\), and by Case 1 applied to the elementary factors, \[ \det(F_k) \cdots \det(F_1) \cdot \det A = \det R = 0 \] (the right-hand side because cofactor expansion along a 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 \((AB)\mathbf{v} = \mathbf{0}\). Since \(A\) is singular, 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. By the same argument as for \(A\) (reducing \(AB\) to its RREF and finding a zero row), \(\det(AB) = 0\). Both sides of the identity equal zero in this case.

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

Example:

Let \[ A = \begin{bmatrix} 1 & 2 \\ 8 & 9 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 7 \\ 4 & 6 \end{bmatrix}, \quad AB = \begin{bmatrix} 13 & 19 \\ 76 & 110 \end{bmatrix}. \] Then \[ \det A = 9 - 16 = -7, \quad \det B = 30 - 28 = 2, \quad \det(AB) = 1430 - 1444 = -14 = (-7)(2). \checkmark \] 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 \(r\), \[ \det B = \sum_{j=1}^n (-1)^{r+j} (k a_{rj}) \det B_{rj} = k \sum_{j=1}^n (-1)^{r+j} a_{rj} \det A_{rj} = k \det A, \] 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 2×2 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 \(i < j\)). Choose any row \(p \notin \{i, j\}\) — such a row exists since \(n \geq 3\) — and expand both \(\det A\) and \(\det B\) along row \(p\): \[ \det A = \sum_{k=1}^n (-1)^{p+k} a_{pk} \det A_{pk}, \qquad \det B = \sum_{k=1}^n (-1)^{p+k} a_{pk} \det B_{pk}, \] where we used \(b_{pk} = a_{pk}\) (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 \(i < p\) and \(i' = i - 1\) if \(i > p\) (similarly for \(j'\)); crucially, \(i' \neq j'\) since \(i \neq j\) and the shift preserves the distinction. The minor \(B_{pk}\) is therefore obtained from \(A_{pk}\) by swapping rows \(i'\) and \(j'\), so by the inductive hypothesis \(\det B_{pk} = -\det A_{pk}\). 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 \(r\), \[ \det B = \sum_{j=1}^n (-1)^{r+j} (a_{rj} + k a_{sj}) \det B_{rj} = \sum_{j=1}^n (-1)^{r+j} a_{rj} \det A_{rj} + k \sum_{j=1}^n (-1)^{r+j} a_{sj} \det A_{rj}, \] 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: 2×2 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, \[ (\det A)(\det I_i(\mathbf{x})) = \det A_i(\mathbf{b}) \;\Longrightarrow\; (\det A)\, x_i = \det A_i(\mathbf{b}). \] 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: \[ \bigl(A \cdot \operatorname{adj}(A)\bigr)_{ij} = \sum_{k=1}^n a_{ik}\, \bigl(\operatorname{adj}(A)\bigr)_{kj} = \sum_{k=1}^n a_{ik}\, C_{jk}, \] 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 the cofactor \(\tilde{C}_{jk}\) only depends on the matrix obtained by deleting row \(j\) — which is the same minor as \(C_{jk}\) of \(A\), 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. Swapping these two equal rows leaves \(\tilde{A}\) unchanged, while by the Effect of Row Operations theorem the swap also negates the determinant: \(\det \tilde{A} = -\det \tilde{A}\), forcing \(2 \det \tilde{A} = 0\), hence \(\det \tilde{A} = 0\). 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 \(\operatorname{adj}(A)\) by \(A\) yields a scalar multiple of the identity: \[ \operatorname{adj}(A)\, A = \begin{bmatrix} 14 & 5 & -11 \\ -16 & -11 & 2 \\ 3 & 9 & -5 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & -4 \\ 3 & 3 & 2 \end{bmatrix} = -37\, I, \] confirming \(\det A = -37\). Hence \[ A^{-1} = \frac{1}{-37}\, \operatorname{adj}(A) = \begin{bmatrix} -\tfrac{14}{37} & -\tfrac{5}{37} & \tfrac{11}{37} \\[2pt] \tfrac{16}{37} & \tfrac{11}{37} & -\tfrac{2}{37} \\[2pt] -\tfrac{3}{37} & -\tfrac{9}{37} & \tfrac{5}{37} \end{bmatrix}. \]