Determinants

Determinants Cramer's Rule Inverse Formula Invertible Matrix Theorem

Determinants

For \(n \geq 2\), the determinant of an \(n \times n\) matrix \(A\) is denined as \[ \begin{align*} \det A &= \sum_{j=1}^n (-1)^{1+j} a_{1j} detA_{1j} \\\\ &= a_{11}\det A_{11}-a_{12}\det A_{12}+\cdots+(-1)^{1+n}a_{1n}\det A_{1n}\\\\ &= a_{11}C_{11}+a_{12}C_{12} + \cdots + a_{1n}C_{1n}. \end{align*} \] Here, \(C_{ij}=(-1)^{i+j} \det A_{ij}\) is the \((i, j)\)-cofactor of \(A\).

For example, consider the determinant of the matrix: \[ A = \begin{bmatrix} 3 & 2 & 5 \\ 7 & 5 & 4 \\ 0 & 1 & 0 \end{bmatrix}. \] The determinant of \(A\) can be computed using the cofactor expansion across the first row \[ \begin{align*} \det A &= 3\det \begin{bmatrix} 5 & 4 \\ 1 & 0 \end{bmatrix} -2\det \begin{bmatrix} 7 & 4 \\ 0 & 0 \end{bmatrix} +5\det \begin{bmatrix} 7 & 5 \\ 0 & 1 \end{bmatrix} \\\\ &= 3(-4)-2(0)+5(7) \\\\ & = 23. \end{align*} \]
The cofactor expantion can be performed along any row or column.
In general, for an entry \(a_{ij}\) in a matrix \(A\) \[ \text{Cofactor of } a_{ij} = C_{ij} = (-1)^{i+j} \det A_{ij} \] where \(A_{ij}\) is the \((n-1) \times (n-1)\) submatrix obtained by removing the \(i\)-th row and \(j\)-th column of \(A\).
In addition, the cofactor matrix of \(A\) is the \(n \times n\) matrirx where each entry is the cofactor of the corresponding element of \(A\). \[ \text{cofactor }(A) = \begin{bmatrix} C_{11} & C_{12} & \cdots & C_{1n}\\ C_{21} & C_{22} & \cdots & C_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ C_{n1} & C_{n2} & \cdots & C_{nn} \end{bmatrix} \]
For example, using the third row of \(A\): \[ \begin{align*} \det A &= 0 -1\det \begin{bmatrix} 3 & 5 \\ 7 & 4 \end{bmatrix} + 0 \\\\ &= -1(12-35) \\\\ &= 23 \end{align*} \] Alternatively, expanding down the first column of \(A\): \[ \begin{align*} \det A &= 3\det \begin{bmatrix} 5 & 4 \\ 1 & 0 \end{bmatrix} -7\det \begin{bmatrix} 2 & 5 \\ 1 & 0 \end{bmatrix} +0\det \begin{bmatrix} 2 & 5 \\ 5 & 4 \end{bmatrix} \\\\ &= 3(-4)-7(-5)+0 \\\\ &= 23. \end{align*} \]
These computations demonstrate an important property: the determinant of a triangular matrix is the product of its diagonal entries. For example: \[ \det \begin{bmatrix} 1 & 7 & 5 & 4 & 2 \\ 0 & 2 & 9 & 2 & 3 \\ 0 & 0 & 3 & 5 & 7\\ 0 & 0 & 0 & 4 & 7\\ 0 & 0 & 0 & 0 & 5 \end{bmatrix} = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120. \] Additionally, note that the transpose of a matrix does not affect its determinant. Thus, for any square matrix \(A\): \[ \det A = \det A^T. \] Consider the matrix \(A\) again. The transpose of \(A\) is: \[ A^T = \begin{bmatrix} 3 & 7 & 0 \\ 2 & 5 & 1 \\ 5 & 4 & 0 \end{bmatrix} \qquad \] Then \[ \begin{align*} \det A^T &= 3\det \begin{bmatrix} 5 & 1 \\ 4 & 0 \end{bmatrix} -7\det \begin{bmatrix} 2 & 1 \\ 5 & 0 \end{bmatrix} +0\det \begin{bmatrix} 2 & 5 \\ 5 & 4 \end{bmatrix}\\\\ &= 3(-4) -7(-5)+0 = 23 \\\\ &= \det A. \end{align*} \]
This result is supported by the relationship between cofactors in \(A\) and \(A^T\). The cofactor of \(a_{1j}\) in \(A\) is equal to the cofactor of \(a_{j1}\) in \(A^T\). Therefore, the cofactor expansion across the first row of \(A\) is the same as the cofactor expansion down the first column of \(A^T\). This relationship holds for any row or column of any square matrix.

Determinants are multiplicative. For example, given: \[ A = \begin{bmatrix} 1 & 2 \\ 8 & 9 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 & 7 \\ 4 & 6 \end{bmatrix}, \qquad AB = \begin{bmatrix} 13 & 19 \\ 76 & 110 \end{bmatrix} \] then \[ \begin{align*} &\det A = 9-16=-7, \\\\ &\det B = 30-28 =2, \\\\ &\det AB = 1430-1444=-14 = (\det A)(\det B). \end{align*} \] Note: \(\det (A+B) \neq \det A + \det B\).

Finally, observe how elementary row operations affect determinants: \[ \begin{align*} &\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad-bc \\\\ &\det \begin{bmatrix} c & d \\ a & b \end{bmatrix} = cb-ad=-(ad-bc) \\\\\ &\det \begin{bmatrix} a & b \\ kc & kd \end{bmatrix} = kad-kbc = k(ad-bc) \\\\ &\det \begin{bmatrix} a & b \\ c+2k & d+2k \end{bmatrix} = a(d+2k)-b(c+2k) = ad-bc \end{align*} \]

Cramer's Rule

Theorem 1: Cramer's Rule Let \(A\) be an invertible \(n \times n\) matrix. \(\forall b \in \mathbb{R}^n \), \(Ax = b\) has a solution \(x\) where the entries of \(x\) are given by: \[ x_i = \frac{detA_i(b)}{\det A}, \qquad i = 1, 2, \cdots, n. \tag{1} \] Here, \(A_i(b)\) is the matrix obtained from \(A\) by replacing its \(i\)-th column with \(b\).
Proof: Consider the \(n \times n\) identity matrix \(I\). Replace the \(i\)-th column of \(I\) with \(x\), giving a modified identity matrix \(I_i(x)\): \[ I_i(x) = \begin{bmatrix} e_1 & e_2 & \cdots & x & \cdots & e_n \end{bmatrix}. \] Multiplying \(A\) by \(I_i(x)\), we obtain: \[ \begin{align*} AI_i(x) &= \begin{bmatrix} Ae_1 & Ae_2 & \cdots & Ax & \cdots & Ae_n \end{bmatrix} \\\\ &= \begin{bmatrix} a_1 & a_2 & \cdots & b & \cdots & a_n \end{bmatrix} \\\\ &= A_i(b) \end{align*} \] By the multiplicative property of determinants, \[ \begin{align*} &(\det A)(\det I_i(x))= \det A_i(b) \\\\ &\Longrightarrow (\det A)x_i = \det A_i(b). \end{align*} \] Since \(A\) is invertible \(\det A \neq 0\) and then dividing through by \(\det A\) yields (1).

Inverse Formula

The Cramer's rule gives us the general inverse formula. Note that both inverse formula and Cramer's rule are primarily useful in theoritical discussions.

Theorem 2: Let \(A\) be an invertible matrix. Then the inverse of \(A\) is given by \[ A^{-1} = \frac{1}{\det A} \text{adj }(A) \] where \(\text{adj }A\) is the adjugate of \(A\), which is the transpose of the cofactor matrix of \(A\). \[ \begin{align*} \text{adj }(A) &= \text{cofactor }(A)^T \\\\ &= \begin{bmatrix} C_{11} & C_{21} & \cdots & C_{n1}\\ C_{12} & C_{22} & \cdots & C_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ C_{1n} & C_{2n} & \cdots & C_{nn} \end{bmatrix}. \end{align*} \]
For example, consider \[ A = \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & -4 \\ 3 & 3 & 2 \end{bmatrix}. \] To get \(\text{adj }A\), we need the nine cofactors of \(A\): \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*} and \(\det A = -1(2+12)-2(4+12)+3(6-3) = -37\). To verify the determinant, we can compute \[ \begin{align*} (\text{adj }A)A &= \begin{bmatrix} 14 & 5 & -11 \\ -16 & -11 & 2 \\ 3 & 9 & -5 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & -4 \\ 3 & 3 & 2 \end{bmatrix} \\\\ &= -37I \end{align*} \] Thus, \(\det A = -37\), and \[ A^{-1} = \begin{bmatrix} \frac{-14}{37} & \frac{-5}{37} & \frac{11}{37} \\ \frac{16}{37} & \frac{11}{37} & \frac{-2}{37} \\ \frac{-3}{37} & \frac{-9}{37} & \frac{5}{37} \end{bmatrix}. \]

Invertible Matrix Theorem

We have seen many properties of invertible matrices. It is a good time to summarize the properties.

Theorem 3: Invertible Matrix Let \(A\) be an \(n \times n\) matrix. Then the following statemets are logically equivalent.
  1. \(A\) is invertible.
  2. There is an \(n \times n\) matrix \(B\) s.t. \(AB = I\) and \(BA = I\).
  3. \(Ax = 0\) has only trivial solution.
  4. \(A\) has \(n\) pivot positions.
  5. \(A\) is row equivalent to \(I_n\).
  6. \(\forall b \in \mathbb{R}^n , Ax = b\) has at least one solution.
  7. The column of \(A\) span \(\mathbb{R}^n\).
  8. The linear transformation \(x \mapsto Ax\) maps \(\mathbb{R}^n\) onto \(\mathbb{R}^n\).
  9. The columns of \(A\) form a linearly independent set.
  10. The linear transformation \(x \mapsto Ax\) is one-to-one.
  11. \(A^T\) is invertible.
  12. \(\det A \neq 0\).
  13. \(0\) is not an eigenvalue of \(A\).
  14. \((\text{Col }A)^{\perp} = \{0\}.\) (See: Symmetry)
  15. \((\text{Nul }A)^{\perp} = \mathbb{R}^n.\)
  16. \(\text{Row }A = \mathbb{R}^n.\)
  17. \(A\) has \(n\) nonzero singular values.