matrix determinant

Some sources may also define the determinant function with axioms.

Axiom 1 (Homogeneity in each row): If matrix \[B\] is obtained from matrix \[A\] by multiplying one row, say, the \[i\]th row of \[A\] by a scalar \[\alpha\], then \[\det(B)=\alpha\det(A)\], i.e. \[\det(A_{1},\dots,\alpha A_{i},\dots, A_{n})=\alpha\det(A_{1},\dots,A_{i},\dots,A_{n})\].

Axiom 2 (Invariance under row addition): If matrix \[B\] is obtained from matrix \[A\] by adding one row, say the \[k\]th row of \[A\] to the \[i\]th row then \[\det(B)=\det(A)\], i.e. \[\det(A_{1},\dots,A_{i}+A_{k},\dots,A_{k},\dots,A_{n})=\det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})\].

Axiom 3 (The determinant of the identity matrix is one): \[\det(I_{n\times n})=1\].

Based on the axioms defined, we can deduce several more properties of the determinant function as a consequence of Axiom 1 and 2.

Theorem 1:
If some row of matrix \[A\] is the zero vector, then \[\det(A)=0\].

Assume the \[i\]th row, \[A_{i}\], of \[A\] is the zero vector. Then,

\begin{align*} \det(A)&=\det(\dots,A_{i-1},0,A_{i+1},\dots)\\ &=\det(\dots,A_{i-1},(-1)0,A_{i+1},\dots)\\ &=(-1)\det(\dots,A_{i-1},0,A_{i+1},\dots)\,\text{by applying Axiom 1}\\ &=(-1)\det(A) \end{align*}

Which implies that \[\det(A)\] can only be 0 for that equality to hold true.

Theorem 2:
If matrix \[B\] is obtained from matrix \[A\] by adding a scalar multiple of one row, say, the \[k\]th row of \[A\] multiplied by a scalar \[\alpha\] to the \[i\]th row, then \[\det(B)=\det(A)\], i.e. \[\det(A_{1},\dots,A_{i}+\alpha A_{k},\dots,A_{k},\dots, A_{n})=\det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})\].

It is trivial if \[a=0\]. Assume that \[a\ne 0\]. Then,

\begin{align*} \det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})&=\frac{1}{\alpha}\cdot\det(A_{1},\dots,A_{i},\dots,\alpha A_{k},\dots,A_{n})\,\text{by applying Axiom 1}\\ &=\frac{1}{\alpha}\cdot\det(A_{1},\dots,A_{i}+\alpha A_{k},\dots,\alpha A_{k},\dots,A_{n})\,\text{by applying Axiom 2}\\ &=\frac{\alpha}{\alpha}\cdot\det(A_{1},\dots,A_{i}+\alpha A_{k},\dots,A_{k},\dots,A_{n})\\ \end{align*}

Theorem 3:
If matrix \[B\] is obtained from matrix \[A\] by interchanging two rows of \[A\] , then \[\det(B)=(-1)\det(A)\], i.e. \[\det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})=(-1)\det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})\].

\begin{align*} \det(A_{1},\dots,A_{i},\dots,A_{k},\dots,A_{n})&=\det(A_{1},\dots,A_{i}+A_{k},\dots,A_{k},\dots,A_{n})\,\text{by applying Axiom 2}\\ &=\det(A_{1},\dots,A_{i}+A_{k},\dots,A_{k}+(-1)(A_{i}+A_{k}),\dots,A_{n})\,\text{by applying Theorem 2}\\ &=\det(A_{1},\dots,A_{i}+A_{k},\dots,(-1)A_{i},\dots,A_{n})\\ &=\det(A_{1},\dots,A_{i}+A_{k}+(-1)A_{i},\dots,(-1)A_{i},\dots,A_{n})\\ &=\det(A_{1},\dots,A_{k},\dots,(-1)A_{i},\dots,A_{n})\\ &=(-1)\det(A_{1},\dots,A_{k},\dots,A_{i},\dots,A_{n}) \end{align*}

Theorem 4:
If two rows of matrix \[A\] are equal, then \[\det(A)=0\].

As a consequence of Theorem 3, if we switch the two equal rows of \[A\] to form an (identical) matrix \[B\], then \[\det(A)=(-1)\det(B)\], implying \[\det(A)=\det(B)=0\].

The formula for the area of the parallelogram, \[ad-bc\] can be derived via:

\begin{align*} \text{area of parallelogram}&=\text{area of rectangle}-\text{area of A}-\text{area of B}-\cdots-\text{area of F}\\ &=(a+c)(b+d)-cb-\frac{1}{2}ba-\frac{1}{2}cd-\frac{1}{2}cd-\frac{1}{2}ba-cb\\ &=ab+ad+bc+cd-2bc-ab-cd\\ &=ad-bc \end{align*}

The fact that the area of the parallelogram created by the column vectors of a \[2\times 2\] matrix coincides with the value of the determinant of that matrix is no coincidence.

In fact, our definition (or axioms) for determinants make quite reasonable postulates for a function that measures the area of the region enclosed by the vectors in the matrix.

For Axiom 1, i.e. multiplying a row by a scalar will result in it's determinant being multiplied by that scalar, we can understand it by taking the area enclosed by two vectors defined as \[\vec{v}\] and \[\vec{w}\], defined as \[\begin{pmatrix} v_{1}&w_{1}\\v_{2}&w_{2} \end{pmatrix}\], then scaling one of the vector by a scalar, where \[k=1.4\], will scale the area by that scalar as shown in the diagram.

Although we defined that multiplying a row by a scalar multiples the determinant by that scalar, here we are multiplying a column not a row. However, because the transpose of a matrix will still have the same determinant, i.e. \[\det(A)=\det(A^{T})\], this distinction does not really matter in this context.

We can illustrate Theorem 2, i.e. determinants being unaffected by pivoting, in a similar manner.

Although the region formed by \[\vec{v}\] and \[k\vec{v}+\vec{w}\], where \[k=0.35\], is slightly slanted than the original region, the two have the same base and height, hence the same area.

For Theorem 3, we define \[\vec{v}=\begin{pmatrix} 1\\3 \end{pmatrix}\] and \[\vec{u}=\begin{pmatrix} 4\\2 \end{pmatrix}\].

Then, \[\det\begin{pmatrix} 4&1\\2&3 \end{pmatrix}=10\] and \[\det\begin{pmatrix} 1&4\\3&2 \end{pmatrix}=-10\], which tells us that the sign returned by the determinant reflects the "orientation" or "sense" of the box.

Finally, a square defined by an identity matrix, or sides of 1, will have an size of 1, which is just a restatement that \[\det(I_{n\times n})=1\].

• rule of Sarrus
  
• linear transformations
  
• linear equation
  
• minor (matrix)
  
• determinants and volumes
  
• Gaussian elimination
  
• cross product
  
• matrix inverse