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Invertible Coefficient Matrix

Let \(A\) be a \(n \times n\) matrix with \(det(A) \neq 0\). The Inverse of \(A\) is \(B\) such that \(AB = BA = I\).

Example

  • \(\begin{align} A &= \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} & B &= \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 4 \end{bmatrix} \end{align}\)
  • \(A \times B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = B \times A\)

The Adjugate of a matrix

The adjugate of a matrix is the transpose of the cofactor matrix.

Adjugate of a \(2 \times 2\) matrix

  • \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
  • Calculate the cofactor matrix of \(A\).
  • \(C = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}\)
  • Now, calculate the transpose of \(C\).
  • \(C^T = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)
  • The final step is to divide the transpose of the cofactor matrix by the determinant of the original matrix.

\(\begin{align} A^* &= \frac{1}{det(A)} \times C^T \times A \\ &= \frac{1}{-2} \times \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \\ &= \frac{1}{-2} \times \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \end{align}\)

Adjugate of a \(3 \times 3\) matrix

  • \(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 8 \\ 5 & 6 & 0 \end{bmatrix}\)
  • Calculate the cofactor matrix of \(A\).
  • \(C = \begin{bmatrix} -48 & 40 & -10 \\ 18 & -15 & 4 \\ 10 & -8 & 2 \end{bmatrix}\)
  • Now, calculate the transpose of \(C\).
  • \(C^T = \begin{bmatrix} -48 & 18 & 10 \\ 40 & -15 & -8 \\ -10 & 4 & 2 \end{bmatrix}\)
  • The final step is to divide the transpose of the cofactor matrix by the determinant of the original matrix.
  • Formula for the adjugate of a matrix:
\[A^* = \frac{1}{det(A)} \times C^T \times A\]

\(\begin{align} A^* &= \frac{1}{2} \times \begin{bmatrix} -48 & 18 & 10 \\ 40 & -15 & -8 \\ -10 & 4 & 2 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 8 \\ 5 & 6 & 0 \end{bmatrix} \\ &= \begin{bmatrix} -24 & 9 & 5 \\ 20 & -7.5 & -4 \\ -5 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 8 \\ 5 & 6 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\\ \end{align}\)

Properties of the Adjugate of a matrix

  • \(adj(AB) = adj(B) \times adj(A)\)
  • \(adj(AB) = adj(B) \times adj(A)\)
  • \(adj(A+B) = adj(A) + adj(B)\)
  • \(adj(A^T) = adj(A)^T\)
  • \(adj(A^{-1}) = adj(A)^{-1}\)

Solution of a system of linear equations with invertible coefficient matrix

Consider the system of linear equations \(Ax = b\) where the coefficient matrix \(A\) is an invertible matrix.

Multiplying both sides by we obtain \(A^{-1}\) we obtain:

\[\begin{align} Ax &= b \\ A^{-1}Ax &= A^{-1}b \\ I_n x &= A^{-1}b \\ x &= A^{-1}b \\ \end{align}\]

Properties of systems of linear equations

  • A system of linear equations \(Ax = b\) is called a non-homogeneous system of linear equations if \(b \neq 0.\)
  • If \(v\) is a solution of the system of linear equations \(Ax = b\), then \(\frac{1}{c}v\) is a solutions of linear equation \(cAx = b\). where \(c \neq 0\).
  • Let \(Ax = b\) be a system of linear equations. If \(A\) is invertible, then \(adj(A)x = b\) also has a solution.

The Inverse of a matrix

  • The inverse of a matrix \(A\) is a matrix \(B\) such that \(AB = BA = I\).
  • The inverse of a matrix \(A\) is denoted by \(A^{-1}\).
  • We can find the inverse of a matrix by using the adjugate of the matrix.
\[A^{-1} = \frac{1}{det(A)} \times adj(A)\]