22
Feb 19

## Cramer's rule and invertibility criterion

### Consequences of multilinearity

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

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

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

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

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

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

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

Let $A^{\prime }$ denote the matrix obtained from $A$ by replacing column $A^{(j)}$ with column $A^{(k)}.$ The vector $L_j$ for the matrix $A^{\prime }$ is the same as for $A$ because both vectors depend on the elements from columns other than the column numbered $j.$ Since $A^\prime$ contains linearly dependent (actually two identical) columns, $\det A^\prime=0.$ Using in (1) $A^\prime$ instead of $A$ we get $0=\det A^\prime=L_jA^{(k)},$ 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 $\det A\neq 0.$ For any $y\in R^n$ denote $B_j$ the matrix formed by replacing the $j$-th column of $A$ by the column vector $y$. Then the solution of the system $Ax=y$ exists and the components of $x$ are given by

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

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

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

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

### Invertibility criterion

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

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

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

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

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

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

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