[Linear Algebra] Non Invertible Matrix

Jason Lee·2022년 8월 6일

Linear Algebra

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Non-Invertible Matrix AA for Ax=bA \textbf{x} = \textbf{b}

  • If matrix AA is invertible, the solution is uniquely obtained as x=A−1b\textbf{x} = A^{-1} \textbf{b}
  • If matrix AA is non-invertible, Ax=bA \textbf{x} = \textbf{b} will have either no solution of infinitely many solutions

Does a Matrix Have an Inverse Matrix?

  • detA\textrm{det} A determines whether matrix AA is invertible or not
    • if detA=0\textrm{det} A = 0, non-invertible
    • if detA≠0\textrm{det} A \neq 0, invertible

Properties of Determinants

Defining Properties

  1. detIn=1\textrm{det} I_n = 1

  2. Exchanging rows reverse the sign of det\textrm{det}

    e.g.
    A=[abcd]A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}, B=[cdab]B = \begin{bmatrix} c & d \\ a & b \\ \end{bmatrix} then detA=−detB\textrm{det} A = -\textrm{det} B

  3. Linearity in each rows

    e.g. detC=c∗detA\textrm{det} C = c * \textrm{det} A

    A=[abcd]A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}, C=[cacbcd]C = \begin{bmatrix} ca & cb \\ c & d \\ \end{bmatrix}

    e.g. detC=detA+detB\textrm{det} C = \textrm{det} A + \textrm{det} B

    A=[abcd]A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}, B=[a′b′cd]B = \begin{bmatrix} a' & b' \\ c & d \\ \end{bmatrix}, C=[a+a′b+b′cd]C = \begin{bmatrix} a+a' & b+b' \\ c & d \\ \end{bmatrix}

    • Caution : this does not mean det(A+B)=detA+detB\textrm{det} (A + B) = \textrm{det} A + \textrm{det} B, which is wrong

Properties that are Proved from the Defining Properties

  1. If two rows of matrix AA are equal, then detA=0\textrm{det} A = 0

    • use property 2

      e.g.

      ∣abcdefabc∣=−∣abcdefabc∣=0\begin{vmatrix} a & b & c \\ d & e & f \\ a & b & c \\ \end{vmatrix} = -\begin{vmatrix} a & b & c \\ d & e & f \\ a & b & c \\ \end{vmatrix} = 0

  2. Subtracting some rows with certain row doesn't change determinant

    • use property 3 & 4

      e.g.

      ∣a+kcb+kdcd∣=∣abcd∣+∣kckdcd∣=∣abcd∣+k∣cdcd∣=∣abcd∣\begin{vmatrix} a+kc & b+kd \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} + \begin{vmatrix} kc & kd \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} + k \begin{vmatrix} c & d \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}

  3. If matrix AA has row of zeros, then detA=0\textrm{det} A = 0

    • use property 3

      e.g.

      ∣abcdef000∣=∣abcdefk×0k×0k×0∣=k∣abcdef000∣=0\begin{vmatrix} a & b & c \\ d & e & f \\ 0 & 0 & 0 \\ \end{vmatrix} = \begin{vmatrix} a & b & c \\ d & e & f \\ k \times 0 & k \times 0 & k \times 0 \\ \end{vmatrix} = k \begin{vmatrix} a & b & c \\ d & e & f \\ 0 & 0 & 0 \\ \end{vmatrix} = 0

  4. If matrix AA is upper triangular of lower triangular, then detA=a11a22⋯ann\textrm{det} A = a_{11} a_{22} \cdots a_{nn}, product of the diagonal

    • use property 1 & 3 & 5
  5. detA=0\textrm{det} A = 0 when matrix AA is singular and detA≠0\textrm{det} A \neq 0 when matrix AA is invertible

  6. If matrix AA and BB are n×nn \times n matrices, det(AB)=(detA)(detB)\textrm{det} (AB) = (\textrm{det} A) (\textrm{det} B)

    • detA−1=1detA\textrm{det} A^{-1} = \frac{1}{\textrm{det} A}
  7. detAT=detA\textrm{det} A^T = \textrm{det} A

Determinant Formulas and Cofactors

Formula for the Determinant

  • Just using the properties of determinants, we can get the formula of determinant

detA=∑n!terms±a1,αa2,βa3,γ⋯an,ω\textrm{det} A = \sum_{n! terms} \pm a_{1, \alpha} a_{2, \beta} a_{3, \gamma} \cdots a_{n, \omega}

where (α,β,γ,⋯ ,ω)(\alpha, \beta, \gamma, \cdots, \omega) is some permutation of (1,2,3,⋯ ,n)(1, 2, 3, \cdots, n)

e.g. 2 x 2 matrix

∣abcd∣=∣a0cd∣+∣0bcd∣=∣a0c0∣+∣a00d∣+∣0bc0∣+∣0b0d∣\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & 0 \\ c & d \\ \end{vmatrix} + \begin{vmatrix} 0 & b \\ c & d \\ \end{vmatrix} = \begin{vmatrix} a & 0 \\ c & 0 \\ \end{vmatrix} + \begin{vmatrix} a & 0 \\ 0 & d \\ \end{vmatrix} + \begin{vmatrix} 0 & b \\ c & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & b \\ 0 & d \\ \end{vmatrix}

=0+ad−bc+0=ad−bc= 0 + ad - bc + 0 = ad-bc

e.g. 3 x 3 matrix

∣abcdefghi∣\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{vmatrix}

=∣a000e000i∣+∣a0000f0h0∣+∣0b0d0000i∣+∣0b000fg00∣+∣00cd000h0∣+∣00c0e0g00∣= \begin{vmatrix} a & 0 & 0 \\ 0 & e & 0 \\ 0 & 0 & i \\ \end{vmatrix} + \begin{vmatrix} a & 0 & 0 \\ 0 & 0 & f \\ 0 & h & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & b & 0 \\ d & 0 & 0 \\ 0 & 0 & i \\ \end{vmatrix} + \begin{vmatrix} 0 & b & 0 \\ 0 & 0 & f \\ g & 0 & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & c \\ d & 0 & 0 \\ 0 & h & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & c \\ 0 & e & 0 \\ g & 0 & 0 \\ \end{vmatrix}

=aei−afh−bdi+bfg+cdh−ceg= aei - afh - bdi + bfg + cdh - ceg

Cofactor Formula

  • The cofactor formula rewrites the big formula for the determinant of and n×nn \times n matrix in terms of the determinants of smaller matrices
  • Cofactor of aija_{ij} = CijC_{ij} = ±det\pm \textrm{det} of n−1n-1 matrix with row ii, column jj erased
    • if i+ji + j is even, the sign is ++
    • if i+ji + j is odd, the sign is −-

detA=a11C11+a12C12+⋯+a1nC1n\textrm{det} A = a_{11} C_{11} + a_{12} C_{12} + \cdots + a_{1n} C_{1n}

e.g. 3 x 3 matrix

∣abcdefghi∣\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{vmatrix}

=∣a000ef0hi∣+∣0b0d0fg0i∣+∣00cde0gh0∣= \begin{vmatrix} a & 0 & 0 \\ 0 & e & f \\ 0 & h & i \\ \end{vmatrix} + \begin{vmatrix} 0 & b & 0 \\ d & 0 & f \\ g & 0 & i \\ \end{vmatrix} + \begin{vmatrix} 0 & 0 & c \\ d & e & 0 \\ g & h & 0 \\ \end{vmatrix}

=a(det∣efhi∣)+b(−det∣dfgi∣)+c(det∣degh∣)= a(\textrm{det} \begin{vmatrix} e & f \\ h & i \\ \end{vmatrix}) + b(- \textrm{det} \begin{vmatrix} d & f \\ g & i \\ \end{vmatrix}) + c(\textrm{det} \begin{vmatrix} d & e \\ g & h \\ \end{vmatrix})

=a(ei−fh)+b(−di+fg)+c(dh−eg)= a(ei - fh) + b(-di + fg) + c(dh - eg)

=aei−afh−bdi+bfg+cdh−ceg= aei - afh - bdi + bfg + cdh - ceg

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