### Matrix calculus: Determinants

### Higher-dimensional determinants

The determinant of an #(n\times n)#-matrix is a number. This number depends on the matrix, and in particular on the rows of the matrix. The dependence of the rows is the basis for the definition of a so-called determinantal function.

Determinantal function

A **determinantal function** on #\mathbb{R}^n# is a function #D# that assigns a number to each #n#-tuple of vectors #\vec{a}_1,\ldots ,\vec{a}_n# such that the following properties are satisfied:

**Multilinearity**: for #i=1,\ldots,n# we have \[

\begin{array}{l}

D(\vec{a}_1, \ldots , \vec{a}_{i-1}, \sum_{k=1}^m\beta_k\vec{b}_k,\vec{a}_{i+1},

\ldots ,\vec{a}_n)

=\sum_{k=1}^m\beta_k D(\vec{a}_1, \ldots , \vec{a}_{i-1},\vec{b}_k,\vec{a}_{i+1},\ldots ,\vec{a}_n)

\end{array}

\]**Antisymmetry**: when two vectors of the arguments are interchanged, the value of the determinantal function turns into its negative.**Normalization**: #D(\vec{e}_1,\vec{e}_2,\ldots ,\vec{e}_n)=1#.

We will prove that, for every #n#, there is indeed exactly one determinantal function and we will actually provide a formula for it; it will be called the *determinant.* First we draw some conclusions from the definition.

Determinantal functions disappear on dependent vectors

If #D# is a determinantal function on #\mathbb{R}^n#, then it has the following properties.

- #D(\vec{a}_1,\ldots, \vec{a} ,\ldots, \vec{a} ,\ldots ,\vec{a}_n)=0#: if two vectors among #\vec{a}_1,\ldots ,\vec{a}_n# are the same, then the determinant is equal to #0#.
- #D (\vec{a}_1,\ldots ,\vec{a}_n)=0# if the system #\vec{a}_1,\ldots ,\vec{a}_n# is linearly dependent.

For the following characterization of determinantal functions we use some facts about *permutations*.

Characterization of the determinant Consider the function #\det# on #\mathbb{R}^n# defined by

\[

\det(\vec{a}_1,\ldots ,\vec{a}_n)=\sum_{\sigma}\text{sg}(\sigma)\cdot a_{1\sigma(1)}\cdots a_{n\sigma(n)}

\] where the sum runs over all #n!# permutations #\sigma# of #\{1,\ldots ,n\}#. Here, #\vec{a}_i = \rv{a_{i1},a_{i2},\ldots,a_{in}}#, so #\det(\vec{a}_1,\ldots ,\vec{a}_n) =\det(A)#, the **determinant of** #A#, the #(n\times n)#-matrix whose #i#-th row is equal to #\vec{a}_i#.

- The function #\det# is a determinantal function.
- The function #\det# is the only determinantal function on #\mathbb{R}^n#.
- If #E# is a function of #n# arguments from #\mathbb{R}^n# which is multilinear and antisymmetric, then \(E(A) = E(I)\cdot \det(A)\).

Instead of #\det\left(\matrix{a_{11}&\cdots&a_{1n}\\ \vdots &\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}}\right)# we also write # \left | \begin{array}{ccc}a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nn}\end {array} \right | #. We will refer to the above expression for #\det# as the **sum formula for the determinant**.

The sum formula has #n!# terms, where #0! = 1# and #n! = n\cdot (n-1)!# for #n\gt0#. This is way too much to be of use even for relatively small #n#. However, the formula implies results that we will use to calculate determinants.

For what value of #k# is the determinant of the matrix #A# below equal to zero?

\[ A = \matrix{1 & 3 & 5\\ 3 & 2 &2\\ 6& k& 4} \]

Give your answer in the form of an integer or a simplified fraction.

We calculate the determinant of #A#:

\[ \begin{array}{rcl}\det(A)&=& a_{11}\cdot a_{22}\cdot a_{33} -a_{11}\cdot a_{23}\cdot a_{32} \\

&&{}-a_{12}\cdot a_{21}\cdot a_{33}+a_{12}\cdot a_{23}\cdot a_{31} \\

&&{}+a_{13}\cdot a_{21}\cdot a_{32}-a_{13}\cdot a_{22}\cdot a_{31} \\

&=& 1\cdot 2\cdot 4 -1\cdot 2\cdot k \\

&&{}-3\cdot 3\cdot 4+3\cdot 2\cdot 6 \\

&&{}+5\cdot 3\cdot k-5\cdot 2\cdot 6 \\

&=&13 k-52

\end{array} \] Thus, the determinant is equal to zero if and only if #k = 4#.

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