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I'm wondering - Other than by using row reduction on the augmented $[A|I]$ to get $[I|A^{-1}]$, and by reducing a matrix to a product of elementary matrices, is there any other way to determine what the inverse of a matrix is? I seem to recall something about using the adjoint of a matrix but I can't recall the details.

Mirrana
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  • There's also Cramer's rule. It's a lot more work, but it's an alternative to Gauss Jordan. – rurouniwallace Aug 07 '13 at 00:01
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    also, there is guessing. Every so often I have a student guess an inverse to a reasonably simple $3 \times 3$ matrix, well, maybe 1 out of 200. – James S. Cook Aug 07 '13 at 00:01
  • A related (and very fruitful) question is, what ways are there to solve $Ax = b$ without using Gaussian elimination? Numerical linear algebra has much to say about this -- for example, QR factorization, SVD, iterative methods... – littleO Aug 07 '13 at 01:13

3 Answers3

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The other way to calculate the inverse of a matrix $A$:

$A^{-1}=\dfrac{1}{detA}adjA$

This assumes $A$ is an invertible square matrix of course.

Another method is Cramer's Rule. Cramer's Rule says the following:

If $A$ is an invertible $n \times n$ matrix, the solution to the system $Ax=b$ of $n$ equations in the variables $x_1,x_2,\ldots,x_n$ is given by $$x_1=\dfrac{detA_1}{detA}, x_2=\dfrac{detA_2}{detA},\ldots,x_n=\dfrac{detA_n}{detA}$$ where, for each $k$, $A_k$ is the matrix obtained from $A$ by replacing column $k$ by $b$.

Sujaan Kunalan
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Another way to calculate an inverse is by using the characteristic polynomial of a matrix. That is, suppose $A$ matrix satisfies the polynomial equation:

$$ A^n + c_{n-1} A^{n-1} + \ldots + c_1 A + c_0 I = 0 $$

The coefficient $c_0$ must be nonzero if and only if $A$ is invertible. Then we can calculate the inverse of $A$ as a linear combination of the powers of $A$:

$$ A^{-1} = -\frac{1}{c_0}(A^{n-1} + c_{n-1}A^{n-2} + \ldots + c_2 A + c_1 I) $$

p.s.
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A "low tech" approach is to simply solve the system of equations determined by $AX=I$. For example, when $n=2$, we have

$$ \left( \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right) \left( \begin{matrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ \end{matrix} \right) =\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right)$$ which is equivalent to the four equations \begin{align} a_{11}x_{11}+a_{12}x_{21} &=1 \\ a_{11}x_{12}+a_{12}x_{22} &=0 \\ a_{21}x_{11}+a_{22}x_{21} &=0 \\ a_{21}x_{12}+a_{22}x_{22} &=1. \\ \end{align} Solve to find $X=A^{-1}$.

Rebecca J. Stones
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