24.4 Inverse Of Matrix
Written \(A^{-1}\) (inverse of matrix \(A\))
The inverse of a matrix \(A\) written as \(A^{-1}\) is defined as \[ A^{-1}=\frac {1}{|A|}(\text {adjoint}A)^T.\]
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Adjoint \(A=\) the matrix of the co-factors of \(A\).
Note 24.5. Books differ here, so check before using a formula from elsewhere. These notes use “adjoint \(A\)” for the matrix of cofactors, which is why the transpose appears separately in \(A^{-1}=\frac {1}{|A|}(\text {adjoint }A)^{T}\).
Many texts reserve \(\operatorname {adj}(A)\) for the transpose of the cofactor matrix — also called the adjugate — and then write \(A^{-1}=\frac {1}{|A|} \operatorname {adj}(A)\) with no transpose at all. The two conventions give the same inverse; only the name moves.
- \(A^{-1}\) exists if and only if \(\left |A\right |\neq 0\); such a matrix is called non-singular. A matrix with \(\left |A\right |=0\) is singular and has no inverse. Only square matrices can have inverses.
Note 24.6. \(A^{-1}A=I.\)
Let \(A= \begin {pmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end {pmatrix} \). Then
\[ \text {adjoint}A= \begin {pmatrix} \begin {vmatrix} e & f\\ h & i\\ \end {vmatrix} & - \begin {vmatrix} d & f\\ g & i\\ \end {vmatrix} & \begin {vmatrix} d & e\\ g & h\\ \end {vmatrix}\\\\ - \begin {vmatrix} b & c\\ h & i\\ \end {vmatrix} & \begin {vmatrix} a & c\\ g & i\\ \end {vmatrix} & - \begin {vmatrix} a & b\\ g & h\\ \end {vmatrix}\\\\ \begin {vmatrix} b & c\\ e & f\\ \end {vmatrix} & - \begin {vmatrix} a & c\\ d & f\\ \end {vmatrix} & \begin {vmatrix} a & b\\ d & e\\ \end {vmatrix}\\ \end {pmatrix} \]
\[ (\text {adjoint} A)^T= \begin {pmatrix} \begin {vmatrix} e & f\\ h & i\\ \end {vmatrix} & - \begin {vmatrix} b & c\\ h & i\\ \end {vmatrix} & \begin {vmatrix} b & c\\ e & f\\ \end {vmatrix}\\\\ - \begin {vmatrix} d & f\\ g & i\\ \end {vmatrix} & \begin {vmatrix} a & c\\ g & i\\ \end {vmatrix} & - \begin {vmatrix} a & c\\ d & f\\ \end {vmatrix}\\\\ \begin {vmatrix} d & e\\ g & h\\ \end {vmatrix} & - \begin {vmatrix} a & b\\ g & h\\ \end {vmatrix} & \begin {vmatrix} a & b\\ d & e\\ \end {vmatrix}\\ \end {pmatrix} \]
\[\text {Thus},\quad A^{-1}=\frac {1}{|A|}(\text {adjoint} A )^T\]
\(\text {Let}\quad A= \begin {pmatrix} 2 & 1 & 4\\ 0 & 3 & 1\\ -1 & 2 & -1\\ \end {pmatrix} \)
\begin {align*} |A| &= 2 \begin {vmatrix} 3 & 1\\ 2 & -1\\ \end {vmatrix} -1 \begin {vmatrix} 0 & 1\\ -1 & -1\\ \end {vmatrix} + 4 \begin {vmatrix} 0 & 3\\ -1 & 2\\ \end {vmatrix}\\\\ & = 2 (-3-2)-1(0+1)+4(0+3)\\ & = -10 -1 +12\\ & = 1\\ \end {align*}
\[\text {adjoint} A= \begin {pmatrix} \begin {vmatrix} 3 & 1\\ 2 & -1\\ \end {vmatrix} & - \begin {vmatrix} 0 & 1\\ -1 & -1\\ \end {vmatrix} & \begin {vmatrix} 0 & 3\\ -1 & 2\\ \end {vmatrix}\\\\ - \begin {vmatrix} 1 & 4\\ 2 & -1\\ \end {vmatrix} & \begin {vmatrix} 2 & 4\\ -1 & -1\\ \end {vmatrix} & - \begin {vmatrix} 2 & 1\\ -1 & 2\\ \end {vmatrix}\\\\ \begin {vmatrix} 1 & 4\\ 3 & 1\\ \end {vmatrix} & - \begin {vmatrix} 2 & 4\\ 0 & 1\\ \end {vmatrix} & \begin {vmatrix} 2 & 1\\ 0 & 3\\ \end {vmatrix} \end {pmatrix} = \begin {pmatrix} -5 & -1 & 3\\ 9 & 2 & -5\\ -11 & -2 & -6\\ \end {pmatrix} \]
\[(\text {adjoint} A)^T= \begin {pmatrix} -5 & 9 & -11\\ -1 & 2 & -2\\ 3 & -5 & 6\\ \end {pmatrix} \]
\[A^{-1}=\frac {1}{|A|}(\text {adjoint} A)^T= \frac {1}{1} \begin {pmatrix} -5 & 9 & -11\\ -1 & 2 & -2\\ 3 & -5 & 6\\ \end {pmatrix} \]
\[\therefore \quad A^{-1}= \begin {pmatrix} -5 & 9 & -11\\ -1 & 2 & -2\\ 3 & -5 & 6\\ \end {pmatrix} \]
- Let \(A\) and \(B\) be \(3\times 3\) matrices, then \(AB\neq BA\).
- Multiplication of matrices is not commutative. \(A^2=AA\).
- If \(AB=I\) (\(I\) is the identity matrix, then \(A\) is an inverse of \(B\))
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