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Adjugate matrix
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=== Column substitution and Cramer's rule === {{see also|Cramer's rule}} Partition {{math|'''A'''}} into [[column vector]]s: :<math>\mathbf{A} = \begin{bmatrix}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{bmatrix}.</math> Let {{math|'''b'''}} be a column vector of size {{math|''n''}}. Fix {{math|1 β€ ''i'' β€ ''n''}} and consider the matrix formed by replacing column {{math|''i''}} of {{math|'''A'''}} by {{math|'''b'''}}: :<math>(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\ \stackrel{\text{def}}{=}\ \begin{bmatrix} \mathbf{a}_1 & \cdots & \mathbf{a}_{i-1} & \mathbf{b} & \mathbf{a}_{i+1} & \cdots & \mathbf{a}_n \end{bmatrix}.</math> Laplace expand the determinant of this matrix along column {{mvar|i}}. The result is entry {{mvar|i}} of the product {{math|adj('''A''')'''b'''}}. Collecting these determinants for the different possible {{mvar|i}} yields an equality of column vectors :<math>\left(\det(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})\right)_{i=1}^n = \operatorname{adj}(\mathbf{A})\mathbf{b}.</math> This formula has the following concrete consequence. Consider the [[linear system of equations]] :<math>\mathbf{A}\mathbf{x} = \mathbf{b}.</math> Assume that {{math|'''A'''}} is [[singular matrix|non-singular]]. Multiplying this system on the left by {{math|adj('''A''')}} and dividing by the determinant yields :<math>\mathbf{x} = \frac{\operatorname{adj}(\mathbf{A})\mathbf{b}}{\det \mathbf{A}}.</math> Applying the previous formula to this situation yields '''Cramer's rule''', :<math>x_i = \frac{\det(\mathbf{A} \stackrel{i}{\leftarrow} \mathbf{b})}{\det \mathbf{A}},</math> where {{math|''x''<sub>''i''</sub>}} is the {{mvar|i}}th entry of {{math|'''x'''}}.
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