Raising the bar

Cramer's rule and invertibility criterion

Consequences of multilinearity

For a fixed $latex j,$ $latex \det A$ is a linear function of column $latex A^{(j)}.$ Such a linear function generates a row-vector $latex L_{j}$ by way of a formula (see Exercise 3)

(1) $latex \det A=L_jA^{(j)}.$

Exercise 1. In addition to (1), we have

(2) $latex L_jA^{(k)}=0$ for any $latex k\neq l.$

Proof. Here and in the future it is useful to introduce the coordinate representation for $latex L_j=(l_{j1},...,l_{jn})$ and put

(3) $latex L=\left(\begin{array}{c}L_1 \\... \\L_n\end{array}\right).$

Then we can write (1) as $latex \det A=\sum_{i=1}^nl_{ji}a_{ij}.$ Here the element $latex l_{ji}$ does not involve $latex a_{ij}$ and therefore by the different-columns-different-rows rule it does not involve elements of the entire column $latex A^{(j)}.$ Hence, the vector $latex L_j$ does not involve elements of the column $latex A^{(j)}.$

Let $latex A^{\prime }$ denote the matrix obtained from $latex A$ by replacing column $latex A^{(j)}$ with column $latex A^{(k)}.$ The vector $latex L_j$ for the matrix $latex A^{\prime }$ is the same as for $latex A$ because both vectors depend on the elements from columns other than the column numbered $latex j.$ Since $latex A^\prime$ contains linearly dependent (actually two identical) columns, $latex \det A^\prime=0.$ Using in (1) $latex A^\prime$ instead of $latex A$ we get $latex 0=\det A^\prime=L_jA^{(k)},$ as required.

After reading the next two sections, come back and read this statement again to appreciate its power and originality.

Cramer's rule

Exercise 2. Suppose $latex \det A\neq 0.$ For any $latex y\in R^n$ denote $latex B_j$ the matrix formed by replacing the $latex j$-th column of $latex A$ by the column vector $latex y$. Then the solution of the system $latex Ax=y$ exists and the components of $latex x$ are given by

$latex x_j=\frac{\det B_j}{\det A},\ j=1,...,n.$

Proof. Premultiply $latex Ax=y$ by $latex L_{j}:$

(4) $latex L_jy=L_jAx=\left(L_jA^{(1)},...,L_jA^{(n)}\right)x=(0,...,\det A,...,0)x.$

Here we applied (1) and (2) (the $latex j$-th component of the vector $latex (0,...,\det
A,...,0)$ is $latex \det A$ and all others are zeros). From (4) it follows that $latex (\det
A)x_{j}=L_{j}y.$ On the other hand, from (1) we have $latex \det B_j=L_jy$ (the vector $latex L_j$ for $latex B_j$ is the same as for $latex A,$ see the proof of Exercise 1). The last two equations prove the statement.

Invertibility criterion

Exercise 3. $latex A$ is invertible if and only if $latex \det A\neq 0.$

Proof. If $latex A$ is invertible, then $latex AA^{-1}=I.$ By multiplicativity of determinant and Axiom 3 this implies $latex \det A\det (A^{-1})=1.$ Thus, $latex \det
A\neq 0.$

Conversely, suppose $latex \det A\neq 0.$ (1), (2) and (3) imply

(5) $latex LA=\left(\begin{array}{c}L_1 \\... \\L_n\end{array}\right) (A^{(1)},...,A^{(n)})=\left(\begin{array}{ccc}L_1A^{(1)}&...&L_1A^{(n)} \\...&...&... \\L_nA^{(1)}&...&L_nA^{(n)}\end{array}
\right)$

$latex =\left(\begin{array}{ccc}\det A&...&0 \\...&...&... \\0&...&\det A\end{array}
\right)=\det A\times I.$

This means that the matrix $latex \frac{1}{\det A}L$ is the inverse of $latex A.$ Recall that existence of the left inverse implies that of the right inverse, so $latex A$ is invertible.

Definition 1. The matrix $latex L$ is more than a transient technical twist; it is called an adjugate matrix and property (5), correspondingly, is called an adjugate identity.

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