MathAlgebraLinear algebraMatrices

Matrix determinant properties

9 minutes read

The determinant has a lot of applications and special qualities that will not only help you to simplify many calculations but also to analyze some relationships between matrices.

In this topic, you will take a slightly more geometric and abstract approach to understand how the properties of the determinant arise. Also, you will learn two really important applications: the explicit calculation of the unique solution of a system of linear equations and a general expression for the inverse of a matrix!

A measure of distortion

You already know that a square matrix AA of size 22 has associated a linear operator LAL_A in R2\mathbb{R}^2. This operator converts the unit square whose sides are e1e_1 and e2e_2 into a parallelogram with sides LA(e1)=Ae1=A1L_A(\mathbf{e_1})= A \, \mathbf{e_1} = A_1 and LA(e2)=Ae2=A2L_A(e_2)= A \, e_2 = A_2:

The operator of A converts the unit square into a parallelogram

This means that the area of the resulting parallelogram is precisely the determinant of AA. So, the larger the area of this parallelogram, the greater the distortion made by LAL_A. For this reason, you can intuitively think of the determinant of a matrix as a rough measure of the size of the distortion generated by it.

The determinant as a function

This section is optional, as are the demos in the following sections. Feel free to skip them!

The only goal is to understand why the properties of the determinant are true, you won't use the definition anywhere else.

In the previous section, you thought of AA in terms of its columns – both for its value in the canonical base and for the area of the parallelogram. This is why it's so natural to think of the determinant in terms of the columns rather than the matrix itself. In general, the determinant is a function that takes nn vectors in Rn\mathbb{R}^n and returns a real number. It's defined by:

det:Rn××RnntimesR given by det(v1,,vn)=A where A=[v1vn]\det: \underbrace{\mathbb{R}^n \times \dots \times \mathbb{R}^n}_{n-\text{times}} \rightarrow \mathbb{R} \text{ given by } \det(v_1, \dots, v_n) = |A| \text{ where } A=[v_1 | \dots | v_n]Clearly, the set Rn××Rnntimes\underbrace{\mathbb{R}^n \times \dots \times \mathbb{R}^n}_{n-\text{times}} is the set of square matrices of size nn. Making an analogy with the power of real numbers, this set is sometimes denoted by the symbol Rn×n\mathbb{R}^{n \times n}. But what makes this function special, and what is the use of thinking about it in terms of vectors instead of matrices as you have done so far? Well, the determinant has a lot of special properties that you will learn about in this topic, but you can only establish and understand them with this new approach, so here you go!

First of all, the determinant is completely described as the only function with the following 33 fundamental properties:

  1. det(e1,,en)=1\det(e_1, \dots, e_n) = 1. Here, by definition, det(e1,,en)=I\det(e_1, \dots, e_n) = |I| where II is the identity matrix.
  2. If among the vectors v1,,vnv_1, \dots, v_n there are two equal, then det(v1,,vn)=0\det(v_1, \dots, v_n) = 0. This property is called alternation.
  3. If j{1,,n}\textcolor{teal}{j} \in \{1,\dots, n \}, then fixing the vectors v1,,vj1,vj+1,,vnv_1, \dots, v_{j-1}, v_{j+1}, \dots, v_{n}, the function is linear with respect to column j\textcolor{teal}{j}. That is, for every u,wRnu, w \in \mathbb{R}^n and λR\lambda \in \mathbb{R}:det(v1,,vj1,u+w,vj+1,,vn)=det(v1,,vj1,u,vj+1,,vn)+det(v1,,vj1,w,vj+1,,vn)\det(v_1, \dots, v_{j-1}, \textcolor{teal}{u+w}, v_{j+1}, \dots, v_{n}) = \det(v_1, \dots, v_{j-1}, \textcolor{teal}{u}, v_{j+1}, \dots, v_{n}) + \det(v_1, \dots, v_{j-1}, \textcolor{teal}{w}, v_{j+1}, \dots, v_{n})det(v1,,vj1,λw,vj+1,,vn)=λdet(v1,,vj1,w,vj+1,,vn)\det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda w}, v_{j+1}, \dots, v_{n}) = \textcolor{teal}{\lambda }\det(v_1, \dots, v_{j-1}, \textcolor{teal}{w}, v_{j+1}, \dots, v_{n})

The last property is known as multilinearity. Therefore, the determinant is the only alternating multilinear function on Rn×n\mathbb{R}^{n \times n} that sends the identity matrix to the number 11. Now let's get to know its most useful properties!

The most useful properties

Hereafter let's consider the vectors v1,,vnv_1, \dots, v_n in Rn\mathbb{R}^n and its matrix A=[v1vn]Rn×nA=[v_1 | \dots | v_n] \in \mathbb{R}^{n \times n}.

The first property is a generalization of one you already know. If the vectors v1,,vnv_1, \dots, v_n are linearly dependent, then det(v1,,vn)=0\det(v_1, \dots, v_n) = 0

Proof

As always, the proofs are completely optional. Feel free to skip them all. Although they may seem long, they aren't difficult.

Suppose thatvj=λ1v1++λj1vj1+λj+1vj+1++λnvnv_j = \lambda_1 v_1 + \dots + \lambda_{j-1} v_{j-1} + \lambda_{j+1} v_{j+1} + \dots + \lambda_{n} v_{n}By multilinearity:

det(v1,,vj1,vj,vj+1,,vn)=det(v1,,vj1,λ1v1++λj1vj1+λj+1vj+1++λnvn,vj+1,,vn)=det(v1,,vj1,λ1v1,vj+1,,vn)++det(v1,,vj1,λj1vj1,vj+1,,vn)+det(v1,,vj1,λj+1vj+1,vj+1,,vn)++det(v1,,vj1,λnvn,vj+1,,vn)=λ1det(v1,,vj1,v1,vj+1,,vn)++λj1det(v1,,vj1,vj1,vj+1,,vn)+λj+1det(v1,,vj1,vj+1,vj+1,,vn)++λndet(v1,,vj1,vn,vj+1,,vn)\begin{align*} \det(v_1, \dots, v_{j-1}, \textcolor{teal}{v_j}, v_{j+1}, \dots, v_{n}) &= \det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda_1 v_1 + \dots + \lambda_{j-1} v_{j-1} + \lambda_{j+1} v_{j+1} + \dots + \lambda_{n} v_{n}}, v_{j+1}, \dots, v_{n}) \\ &= \det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda_1 v_1}, v_{j+1}, \dots, v_{n}) \\ &\quad +\dots \\ &\quad + \det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda_{j-1} v_{j-1}}, v_{j+1}, \dots, v_{n}) \\ &\quad + \det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda_{j+1} v_{j+1}}, v_{j+1}, \dots, v_{n}) \\ &\quad +\dots \\ &\quad + \det(v_1, \dots, v_{j-1}, \textcolor{teal}{\lambda_{n} v_{n}}, v_{j+1}, \dots, v_{n}) \\ &= \textcolor{teal}{\lambda_1} \det(v_1, \dots, v_{j-1}, \textcolor{teal}{v_1}, v_{j+1}, \dots, v_{n}) \\ &\quad +\dots \\ &\quad + \textcolor{teal}{\lambda_{j-1}} \det(v_1, \dots, v_{j-1}, \textcolor{teal}{v_{j-1}}, v_{j+1}, \dots, v_{n}) \\ &\quad + \textcolor{teal}{\lambda_{j+1}} \det(v_1, \dots, v_{j-1}, \textcolor{teal}{v_{j+1}}, v_{j+1}, \dots, v_{n}) \\ &\quad +\dots \\ &\quad + \textcolor{teal}{\lambda_{n}} \det(v_1, \dots, v_{j-1}, \textcolor{teal}{v_{n}}, v_{j+1}, \dots, v_{n}) \\ \end{align*}But in every determinant there are two identical vectors, so by alternation all of these determinants are 0. Thus det(v1,,vn)=0\det(v_1, \dots, v_n) = 0

\blacksquare

If you have nn vectors of size nn, calculate the determinant of their matrix: if it is different from 0 then the vectors are linearly independent, otherwise there are linearly dependent.

Notice that this is a quick way to know if a set of vectors forms a basis.

You already know that A is invertible if and only if its determinant is not zero. You can go a step further and calculate the determinant of A1A^{-1}:

det(A1)=1detA\det(A^{-1}) = \frac{1}{\det A}

Proof

You already know that 1=det(I)=det(AA1)1 = \det(I) = \det(A A^{-1}). You'll see later that the determinant of the product is the product of the determinants, so we get that 1=det(AA1)=det(A)det(A1)1= \det(A A^{-1}) = \det(A) \det(A^{-1}), and then det(A1)=1detA\det(A^{-1}) = \frac{1}{\det A}.

\blacksquare

Now let's take a look into geometry. If you extend the canonical vectors a bit, say by a factor of 22, then the area of the new square doubles as well. But what about the area of the parallelogram? Does it double as well? No! Its area becomes 44 times larger!

Increasing the area of the square implies a bigger change in the parallelogram

This is the perfectly natural thanks to the following property of the determinant:

For any real number λ\lambda, it holds that det(λA)=λndet(A)\det(\lambda A) = \lambda^n \det( A)

Proof

By multilinearity:

det(λA)=det(λ[v1v2vn])=det([λv1λv2λvn])=λdet([v1λv2λvn])=λ2det([v1v2λvn])==λndet([v1vn])=λndet(A)\begin{align*} \det(\lambda A) &= \det(\lambda [v_1 |v_2 | \dots | v_n]) \\ &= \det( [\lambda v_1 | \lambda v_2 |\dots | \lambda v_n]) \\ &= \lambda \det( [v_1 | \lambda v_2| \dots | \lambda v_n]) \\ &= \lambda^2 \det( [v_1 | v_2| \dots | \lambda v_n]) \\ &= \dots \\ &= \lambda^n \det( [v_1 | \dots | v_n]) \\ &= \lambda^n \det( A) \end{align*}

\blacksquare

Some other properties that come in handy and can save you more than one time are the following:

  • The determinant of the transpose doesn't change: det(AT)=det(A)\det(A^T) = \det(A)
  • The determinant of the product is the product of the determinants: det(AB)=det(A)det(B)\det(AB) = \det(A) \det(B)

Time for putting all this theory into practice.

Cramer's rule

Let's revisit systems of linear equations. Consider the system Ax=bAx =b where AA is a square matrix of size nn and both xx and bb are vectors in Rn\mathbb{R}^n.

You already know that the determinant helps to detect if the system has a unique solution. But there is a surprising result called Cramer's Rule that uses more determinants to explicitly calculate that unique solution!

Suppose that the system Ax=bAx=b has a unique solution (that is, AA is invertible). Define A(j)A_{(j)} to be the matrix that equals AA, except its jj-th column is replaced by bb. The entries of the unique solution xx of the system are:

xj=det(A(j))det(A) for all 1jnx_j=\frac{ \det(A_{(j)})}{\det(A)} \quad \text { for all } \quad 1 \leq j \leq n

Proof

Since AA is invertible, its columns are linearly independent. So they are nn linearly independent vectors in Rn\mathbb{R}^n, then they form a basis. By the matrix product, you can write the system Ax=bAx=b as x1A1+x2A2++xnAn=bx_1A_1 + x_2 A_2 + \dots + x_n A_n = b . Since the columns are a basis, there are unique numbers x1,,xnRnx_1, \dots, x_n \in \mathbb{R}^n such that:x1A1+x2A2++xnAn=bx_1A_1 + x_2 A_2 + \dots + x_n A_n = bTherefore, the system has a unique solution. Then for every j{1,,n}j \in \{1, \dots, n \} by multilinearity:det(A(j))=det([A1bAn])=det([A1x1A1++xjAj++xnAnAn])=x1det([A1A1An])++xjdet([A1AjAn])++xndet([A1AnAn])=x10++xjdet([A1AjAn])++xn0=xjdet([A1AjAn])=xjdet(A)\begin{align*} \det(A_{(j)}) &= \det([A_1 | \dots | \textcolor{teal}{b} | \dots | A_n]) \\ &= \det([A_1 | \dots | \textcolor{teal}{x_1A_1 + \dots + x_j A_j + \dots + x_n A_n} | \dots | A_n]) \\ &= \textcolor{teal}{x_1} \det([A_1 | \dots | \textcolor{teal}{A_1} | \dots | A_n]) \\ &\quad + \dots \\ &\quad+ \textcolor{teal}{x_{j}} \det([A_1 | \dots | \textcolor{teal}{A_j} | \dots | A_n]) \\ &\quad + \dots \\ &\quad+ \textcolor{teal}{x_{n}} \det([A_1 | \dots | \textcolor{teal}{A_n} | \dots | A_n]) \\ &= \textcolor{teal}{x_1} 0\\ &\quad + \dots \\ &\quad+ \textcolor{teal}{x_{j}} \det([A_1 | \dots | \textcolor{teal}{A_j} | \dots | A_n])\\ &\quad + \dots \\ &\quad+ \textcolor{teal}{x_{n}} 0 \\ &= \textcolor{teal}{x_{j}} \det([A_1 | \dots | \textcolor{teal}{A_j} | \dots | A_n]) \\ &= \textcolor{teal}{x_{j}} \det(A) \\ \end{align*}That's because in nearly every determinant, there are two identical vectors. So by alternation, nearly all of the determinants are 00. Therefore xj=det(A(j))det(A)x_j=\frac{ \det(A_{(j)})}{\det(A)}

\blacksquare

Now let's see an example with the 3×33 \times 3 system Ax=bAx=b where:

A=[111124139]b=[301]A =\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{bmatrix} \quad b=\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}You can easily verify that detA=2\det A = 2 and:

  • detA(1)=311024139=16\det A_{(1)} = \begin{vmatrix} \textcolor{teal}{-3} & 1 & 1 \\ \textcolor{teal}{0} & 2 & 4 \\ \textcolor{teal}{1} & 3 & 9 \end{vmatrix} = -16
  • detA(2)=[131104119]=12\det A_{(2)} = \begin{bmatrix} 1 & \textcolor{teal}{-3} & 1 \\ 1 & \textcolor{teal}{0} & 4 \\ 1 & \textcolor{teal}{1} & 9 \end{bmatrix} = 12
  • detA(2)=[113120131]=2\det A_{(2)} = \begin{bmatrix} 1 & 1 & \textcolor{teal}{-3} \\ 1 & 2 & \textcolor{teal}{0} \\ 1 & 3 & \textcolor{teal}{1} \end{bmatrix} = -2

Thus, the unique solution is:

x=1detA[detA(1)detA(2)detA(3)]=12[16122]=[861]x = \frac{1}{\det A} \begin{bmatrix} \det A_{(1)} \\ \det A_{(2)} \\ \det A_{(3)} \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -16 \\ 12 \\ -2 \end{bmatrix} = \begin{bmatrix} -8 \\ 6 \\ -1 \end{bmatrix}

Conclusion

Let's take some vectors v1,,vnv_1, \dots, v_n in Rn\mathbb{R}^n and their matrix A=[v1vn]Rn×nA=[v_1 | \dots | v_n] \in \mathbb{R}^{n \times n}.

  • The determinant of AA is the area of the parallelogram into which LAL_A converts to the unit square.
  • The determinant is the only alternating multilinear function on Rn×n\mathbb{R}^{n \times n} that sends the identity matrix to the number 11.

  • v1,,vnv_1, \dots, v_n are linearly independent if and only if det(v1,,vn)0\det(v_1, \dots, v_n) \neq 0.

  • When detA0\det A \neq 0, det(A1)=1detA\det(A^{-1}) = \frac{1}{\det A}

  • For any real number λ\lambda, it holds that det(λA)=λndet(A)\det(\lambda A) = \lambda^n \det( A)

  • When a system of equations has a unique solution, you can use Cramer's rule to calculate it.

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