Linear Algebra/Laplace's Expansion

We finish by applying this result to derive a new formula for the inverse of a matrix. With Theorem 1.5, the determinant of an $n\times n$ matrix $T$ can be calculated by taking linear combinations of entries from a row and their associated cofactors.

$t_{i,1}\cdot T_{i,1}+t_{i,2}\cdot T_{i,2}+\dots +t_{i,n}\cdot T_{i,n}=\left|T\right|(*)$

Recall that a matrix with two identical rows has a zero determinant. Thus, for any matrix $T$, weighing the cofactors by entries from the "wrong" row — row $k$ with $k\ne i$ — gives zero

$t_{i,1}\cdot T_{k,1}+t_{i,2}\cdot T_{k,2}+\dots +t_{i,n}\cdot T_{k,n}=0(**)$

because it represents the expansion along the row $k$ of a matrix with row $i$ equal to row $k$. This equation summarizes ($*$) and ($**$).

$\left(\begin{array}{cccc}{t}_{1,1}& t_{1,2}& \dots & t_{1,n}\\ t_{2,1}& t_{2,2}& \dots & t_{2,n}\\ & ⋮\\ t_{n,1}& t_{n,2}& \dots & t_{n,n}\end{array}\right)\left(\begin{array}{cccc}{T}_{1,1}& T_{2,1}& \dots & T_{n,1}\\ T_{1,2}& T_{2,2}& \dots & T_{n,2}\\ & ⋮& & \\ T_{1,n}& T_{2,n}& \dots & T_{n,n}\end{array}\right)=\left(\begin{array}{cccc}|T|& 0& \dots & 0\\ 0& |T|& \dots & 0\\ & ⋮& & \\ 0& 0& \dots & |T|\end{array}\right)$

Note that the order of the subscripts in the matrix of cofactors is opposite to the order of subscripts in the other matrix; e.g., along the first row of the matrix of cofactors the subscripts are $1,1$ then $2,1$, etc.

The formulas from this section are often used for by-hand calculation and are sometimes useful with special types of matrices. However, they are not the best choice for computation with arbitrary matrices because they require more arithmetic than, for instance, the Gauss-Jordan method.

Exercises

/Solutions/

References

• Template:Citation.