Properties of determinants

On this post we are going to see all the properties of determinants. We explain each property with an example so that you understand them perfectly. In addition, you will find solved exercises of the properties of determinants.

All properties of determinants

Determinant of the transpose of a matrix

The determinant of a matrix is equivalent to the determinant of its transpose.

\lvert A \rvert = \lvert A^T\rvert

See: how to find the transpose of a matrix

Example

Let |A| be the determinant of a 2×2 square matrix:

 \lvert A \rvert = \begin{vmatrix} 2 & 3 \\[1.1ex] 1 & 5 \end{vmatrix} = 2 \cdot 5 - 1 \cdot 3 = 10 - 3 = \bm{7}

If we transpose the 2×2 matrix and solve the determinant, we get the same result as before:

 \lvert A^T\rvert = \begin{vmatrix} 2 & 1 \\[1.1ex] 3 & 5 \end{vmatrix} = 2 \cdot 5 - 3 \cdot 1 = 10 - 3 = \bm{7}

Determinant with a row or column of zeros

If a determinant has a row or column that is all zeros, the determinant results in 0.

\displaystyle \begin{vmatrix} a_{11} & 0 & a_{13} \\[1.1ex] a_{21} & 0 & a_{23} \\[1.1ex] a_{31} & 0 & a_{33}\end{vmatrix}=0

Example

In these two examples the determinants are 0. Because the second row of the first determinant is all zeros and the third column of the second determinant is also all zeros.

 \begin{vmatrix} 5 & 6 & 2 \\[1.1ex] 0 & 0 & 0 \\[1.1ex] -3 & 1 & 4 \end{vmatrix} = \bm{0} \qquad \qquad \begin{vmatrix} 1 & -5 & 0 \\[1.1ex] 6 & 2 & 0 \\[1.1ex] 1 & 3 & 0 \end{vmatrix} = \bm{0}

Determinant with two identical rows or columns

Any determinant that has two equal or multiple rows (or columns) is equal to zero.

Similarly, if there is any linear combination between rows or columns, that is, they are linearly dependent, the determinant gives 0.

Example

The following 3×3 determinant gives 0 because columns 2 and 3 are identical.

\begin{vmatrix} 3 & 4 & 4 \\[1.1ex] -1 & 5 & 5 \\[1.1ex] 6 & 2 & 2 \end{vmatrix} = 0

Changing rows or columns of a determinant

If two rows or two columns of a determinant are interchanged, the determinant gives the same result but changed sign.

\displaystyle\begin{vmatrix} a & b & c \\[1.1ex] d & e & f \\[1.1ex] g & h & i \end{vmatrix}= - \begin{vmatrix} a & c & b \\[1.1ex] d & f & e \\[1.1ex] g & i & h \end{vmatrix}

Example

To prove this property of determinants, we compute the determinant of a 3×3 dimension matrix:

 \begin{vmatrix} 3 & 2 & -4 \\[1.1ex] 1 & 5 & 6 \\[1.1ex] 1 & 0 & -3 \end{vmatrix} = \displaystyle  \bm{-7}

Now we change the order of columns 2 and 3 with each other and calculate the 3×3 determinant again. Note that the result is the same but changed sign:

\begin{vmatrix} 3 & -4 & 2 \\[1.1ex] 1 & 6 & 5 \\[1.1ex] 1 & -3 & 0 \end{vmatrix} = \displaystyle \bm{+7}

Multiplying a row or column of a determinant by a scalar

Multiplying all the elements of a row or a column by a real number is the same as multiplying the result of the determinant by that number.

\displaystyle\displaystyle \begin{vmatrix} k \cdot a_{11} &  k \cdot a_{12} & k \cdot a_{13} \\[1.1ex] a_{21} &  a_{22} & a_{23} \\[1.1ex] a_{31} &  a_{32} & a_{33} \end{vmatrix} =k \cdot \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[1.1ex] a_{21} &  a_{22} & a_{23} \\[1.1ex] a_{31} &  a_{32} & a_{33} \end{vmatrix}

\displaystyle\displaystyle\begin{vmatrix} k \cdot a_{11} & a_{12} & a_{13} \\[1.1ex] k \cdot a_{21} &  a_{22} & a_{23} \\[1.1ex] k \cdot a_{31} &  a_{32} & a_{33} \end{vmatrix} =k \cdot \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[1.1ex] a_{21} &  a_{22} & a_{23} \\[1.1ex] a_{31} &  a_{32} & a_{33} \end{vmatrix}

Example

We are going to find the determinant of a 2×2 matrix to demonstrate this property of the determinants:

\displaystyle \begin{vmatrix} 2 & 3 \\[1.1ex] 1 & 4 \end{vmatrix} = 8-3= \bm{5}

Now we evaluate the same determinant and multiply all the entries of a row by 2. You will see that the result will be that of the previous determinant but multiplied by 2, that is, 10:

 \displaystyle \begin{vmatrix} 2 \cdot 2 & 2 \cdot 3 \\[1.1ex] 1 & 4 \end{vmatrix} = \begin{vmatrix} 4 & 6 \\[1.1ex] 1 & 4 \end{vmatrix} = 16-6 =\bm{10}

Determinant of a matrix product

The determinant of the product of two matrices is equal to the product of the determinant of each matrix.

\displaystyle\lvert A \cdot B \rvert = \lvert A \rvert \cdot \lvert B \rvert

See: matrix multiplication rules

Example

Let A and B be two square matrices of order 2:

\displaystyle A= \begin{pmatrix} 1 & 3 \\[1.1ex] 2 & 5 \end{pmatrix}\quad B=\begin{pmatrix} 4 & 2 \\[1.1ex] 1 & -1 \end{pmatrix}

The proof of this property of determinants is easy to do, first we compute the matrix multiplication and then we calculate the determinant of the resulting matrix:

\displaystyle \left| A \cdot B \right| =\left| \begin{pmatrix} 1 & 3 \\[1.1ex] 2 & 5 \end{pmatrix} \cdot \begin{pmatrix} 4 & 2 \\[1.1ex] 1 & -1 \end{pmatrix}\right| = \left| \begin{pmatrix} 7 & -1 \\[1.1ex] 13 & -1 \end{pmatrix} \right| = -7 - (-13) = \bm{6}

Now we find the determinant of each matrix separately, and then we multiply the results:

\displaystyle \lvert A \rvert \cdot \lvert B \rvert = \begin{vmatrix} 1 & 3 \\[1.1ex] 2 & 5 \end{vmatrix}\cdot\begin{vmatrix} 4 & 2 \\[1.1ex] 1 & -1 \end{vmatrix} = -1\cdot (-6)= \bm{6}

As you can see, first doing the matrix product and then the determinant gives the same result as first doing the determinant of each matrix and then multiplying the results.

On the other hand, this condition does not apply to matrix addition and subtraction, that is, the determinant of the addition (or subtraction) of two matrices does not give the same result as adding (or subtracting) the determinants of two matrices separately.

Determinant of the inverse of a matrix

If a matrix is invertible, the determinant of its inverse is given by the following formula:

\displaystyle  \begin{vmatrix} A^{-1} \end{vmatrix} = \cfrac{1}{\lvert A \rvert}

Example

We are going to verify this property of determinants by first calculating the inverse of a matrix and then solving its determinant. We will see that the result is equivalent to finding the determinant of the original matrix and then inverting it.

So we find the inverse of the following 2×2 matrix and compute its determinant:

\displaystyle A= \begin{pmatrix} 4 & 2 \\[1.1ex] 7 & 4 \end{pmatrix}

\displaystyle A^{-1}= \begin{pmatrix} 2 & -1 \\[1.1ex] -\frac{7}{2} & 2 \end{pmatrix}

\displaystyle \begin{vmatrix}A^{-1} \end{vmatrix}= \begin{vmatrix} 2 & -1 \\[1.1ex] -\frac{7}{2} & 2 \end{vmatrix} = 4-\cfrac{7}{2} =\cfrac{8}{2}-\cfrac{7}{2} = \cfrac{\bm{1}}{\bm{2}}

And now we solve the determinant of the original matrix and apply the formula:

\displaystyle \begin{vmatrix}A\end{vmatrix}= \begin{pmatrix} 4 & 2 \\[1.1ex] 7 & 4 \end{pmatrix}=16-14=2

\displaystyle \begin{vmatrix}A^{-1}\end{vmatrix}= \cfrac{\bm{1}}{\bm{2}}

The results of both operations are identical, so the property is proven.

Changing the row of a determinant

You can replace the row of a determinant by the addition (or subtraction) of the same row plus (or minus) another row multiplied by a scalar.

Example

With the following example we are going to show this property of determinants. We will first calculate a determinant, then we will operate on a row of the determinant and recalculate its result. You’ll see how we get the same result in both cases.

So, we first calculate a the determinant of a 3×3 matrix using cofactor expansion:

\displaystyle\begin{aligned}\begin{vmatrix}2&1&-1\\[1.1ex]3&0&1\\[1.1ex]0&-3&6\end{vmatrix}=\bm{-3}\end{aligned}

Now we add the first row multiplied by 2 to the second row:

\displaystyle \begin{vmatrix} 2 & 1 & -1 \\[1.1ex] 3 & 0 & 1 \\[1.1ex] 0 & -3 & 6 \end{vmatrix} \begin{matrix} \\[1.1ex] \xrightarrow{R_2 + 2R_1} \\[1.1ex] \ \end{matrix} \begin{vmatrix} 2 & 1 & -1 \\[1.1ex] 7 & 2 & -1 \\[1.1ex] 0 & -3 & 6 \end{vmatrix}

And we solve the determinant obtained:

\displaystyle \begin{vmatrix} 2 & 1 & -1 \\[1.1ex] 7 & 2 & -1 \\[1.1ex] 0 & -3 & 6 \end{vmatrix}=\bm{-3}

In both cases the result was -3. So it is shown that the result of a determinant does not change if a row is replaced by the sum of the same row plus another row multiplied by a number.

Determinant of a triangular matrix

The determinant of a triangular matrix is the product of the elements on its main diagonal.

See properties of triangular matrix.

Example

We are going to evaluate the determinant of the following triangular matrix as an example:

 \begin{vmatrix} 2 & 3 & 5 \\[1.1ex] 0 & -1 & 7 \\[1.1ex] 0 & 0 & 4 \end{vmatrix} \displaystyle= 2 \cdot (-1) \cdot 4 = \bm{-8}

Determinant of a diagonal matrix

The determinant of a diagonal matrix is equal to the product of the elements on its main diagonal.

See definition of diagonal matrix.

Example

Let’s take the determinant of the following diagonal matrix as an example:

\begin{vmatrix}5 & 0 & 0 \\[1.1ex] 0 & 3 & 0 \\[1.1ex] 0 & 0 & -2 \end{vmatrix} \displaystyle= 5 \cdot 3 \cdot (-2) = \bm{-30}

Computing the determinant of a matrix with its eigenvalues

The determinant of any matrix can be computed by multiplying the eigenvalues of the matrix.

See what the eigenvalues of a matrix are and how to calculate them.

Example

Let A be a 4×4 square matrix:

\displaystyle\begin{pmatrix}1&2&0&0\\[1.1ex] 2&1&0&0\\[1.1ex] 0&0&6&0\\[1.1ex] 0&1&0&2\end{pmatrix}

The eigenvalues of the 4×4 matrix are:

\lambda_1=-1\qquad\lambda_2=2\qquad\lambda_3=3\qquad\lambda_2=6

Thus, the determinant of the matrix of order 4 can be computed as follows:

\displaystyle\begin{vmatrix}1&2&0&0\\[1.1ex] 2&1&0&0\\[1.1ex] 0&0&6&0\\[1.1ex] 0&1&0&2\end{vmatrix}=-1\cdot 2\cdot 3\cdot 6=\bm{-36}

Practice problems on properties of determinants

Problem 1

Solve the following 3×3 determinant using the properties of determinants:

 \displaystyle \begin{vmatrix} 3 & 1 & 0 \\[1.1ex] 4 & 2 & 0 \\[1.1ex] -1 & 6 & 0 \end{vmatrix}

If a determinant has a row or column full of zeros, the determinant returns 0 (property 2). Therefore, the result of the determinant is 0, because the third column is all zeros.

 

Problem 2

Find the following 4×4 determinant applying the properties of determinants:

 \displaystyle \begin{vmatrix} 4 & 2 & -3 & 5 \\[1.1ex] 1 & 5 & 3 & 2 \\[1.1ex]4 & 2 & -3 & 5 \\[1.1ex] -2 & 0 & 4 & 3 \end{vmatrix}

If a determinant has two equal or multiple rows (or columns), the determinant results in 0. Therefore, the result of the determinant is 0, because the first row and the third row are equal.

 

Problem 3

Compute the following 4×4 determinant using properties of determinants:

 \displaystyle \begin{vmatrix} 1 & 0 & 2 & 2 \\[1.1ex] 3 & 1 & 5 & 6 \\[1.1ex] 1 & 3 & -2 & 2 \\[1.1ex] 2 & 2 & 0 & 4 \end{vmatrix}

If a determinant has two equal or multiple rows or two columns, the determinant gives 0 (property 3). Therefore, the result of the determinant is 0, because the fourth column is twice the first column.

 

Problem 4

We know the result of a determinant, although we do not know the elements of the matrix:

 \displaystyle \begin{vmatrix} a & b \\[1.1ex] c & d  \end{vmatrix} = 3

From the result of the previous determinant and the properties of determinants of matrices, calculate the result of the following determinants:

\ \displaystyle \mathbf{a} \bm{)} \ \begin{vmatrix} a & c  \\[1.1ex] b & d  \end{vmatrix} \qquad \mathbf{b} \bm{)} \ \begin{vmatrix} b & a  \\[1.1ex] d & c  \end{vmatrix} \qquad \mathbf{c} \bm{)} \ \begin{vmatrix} a & 3b  \\[1.1ex] c & 3d  \end{vmatrix}

a) \begin{pmatrix} a & c  \\ b & d  \end{pmatrix} is the transpose of matrix \begin{pmatrix} a & b  \\ c & d  \end{pmatrix}  . And the determinant of a matrix is equal to the determinant of its transposed matrix (property 1). Therefore, the result of this determinant is also 3.

\displaystyle \begin{vmatrix} a & c  \\[1.1ex] b & d  \end{vmatrix}=\begin{vmatrix} a & b \\[1.1ex] c & d  \end{vmatrix}=\bm{3}

 

b) In the determinant   \begin{vmatrix} b & a  \\ d & c  \end{vmatrix} columns 1 and 2 have been changed with respect to determinant \begin{vmatrix} a & b \\ c & d  \end{vmatrix}  . Therefore, according to property 4, the result is the same as the result of the determinant of the problem but changed sign, that is, -3.

\displaystyle \begin{vmatrix} b & a  \\[1.1ex] d & c  \end{vmatrix} = - \begin{vmatrix} a & b \\[1.1ex] c & d  \end{vmatrix}= \bm{-3}

 

c) The determinant  \begin{vmatrix} a & 3b  \\ c & 3d  \end{vmatrix} is the same as the determinant of the problem but the second row is multiplied by 3. Therefore, from property 5, we can deduce that its result will also be the result of the determinant of the problem multiplied by 3, that is, 9.

\displaystyle \begin{vmatrix} a & 3b  \\[1.1ex] c & 3d  \end{vmatrix} =3 \begin{vmatrix} a & b \\[1.1ex] c & d  \end{vmatrix} =3 \cdot 3 = \bm{9}

 

Problem 5

We know the value of these two determinants:

\displaystyle\vert A \vert = \begin{vmatrix} 1 & 2 & 0 & 1 \\[1.1ex] -2 & -1 & 1 & 0 \\[1.1ex] 1 & 3 & 3 & -1 \\[1.1ex] 3 & 4 & 1 & 1 \end{vmatrix}=8

 \displaystyle\vert B \vert = \begin{vmatrix} 0 & 1 & 3 & 2 \\[1.1ex] -1 & -2 & 0 & 0 \\[1.1ex] 3 & 1 & 1 & 2 \\[1.1ex] -1 & 2 & 3 & 1 \end{vmatrix} = - 4

Calculate:

\displaystyle \vert A \cdot B \vert

To calculate the result of the determinant, it is not necessary to solve the 4×4 matrix multiplication. Since the determinant of the product of two matrices is equal to the product of the determinant of each matrix separately (property 6). So:

\vert A \cdot B \vert  = \vert A \vert \cdot \vert B \vert = 8 \cdot (-4) = \bm{-32}

 

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