In linear algebra, a determinant is a special number that can be determined from a square matrix. The determinant of a matrix, say A is denoted det(A), |A| or det A. Determinants have some useful properties as they permit us to generate the same results with different and simpler entries (elements). There are 10 main properties of determinants: reflection property, all-zero property, proportionality or repetition property, switching property, scalar multiple properties, sum property, invariance property, factor property, triangle property, and co-factor matrix property. All the properties of determinants have been covered below in a detailed way, along with solved examples.
Table of Contents
Important Properties of Determinants
1. Reflection Property
The determinant remains unaltered if its rows are changed into columns and the columns into rows. This is known as the property of reflection.
2. All-zero Property
If all the elements of a row (or column) are zero, then the determinant is zero.
3. Proportionality (Repetition) Property
If all elements of a row (or column) are proportional (identical) to the elements of some other row (or column), then the determinant is zero.
4. Switching Property
The interchange of any two rows (or columns) of the determinant changes its sign.
5. Scalar Multiple Property
If all the elements of a row (or column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.
6. Sum Property
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}}+{{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{a}_{2}}+{{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{a}_{3}}+{{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|\end{array} \)
7. Property of Invariance
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}}+\alpha {{b}_{1}}+\beta {{c}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}}+\alpha {{b}_{2}}+\beta {{c}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}}+\alpha {{b}_{3}}+\beta {{c}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\end{array} \)
That is, a determinant remains unaltered under an operation of the form
Ci → Ci + αCj + βCk, where j, k ≠i
Or
An operation of the form Ri → Ri + αRj + βRk, where j, k ≠i.
8. Factor Property
If a determinant Δ becomes zero when we put x = α, then (x – α) is a factor of Δ.
9. Triangle Property
If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements. That is,
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ 0 & {{b}_{2}} & {{b}_{3}} \\ 0 & 0 & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & 0 & 0 \\ {{a}_{2}} & {{b}_{2}} & 0 \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}{{b}_{2}}{{c}_{3}}\end{array} \)
10. Determinant of Cofactor Matrix
\(\begin{array}{l}\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\;then \;{{\Delta }_{1}}=\left| \begin{matrix} {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\ {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\ {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\ \end{matrix} \right|\end{array} \)
Where Cij denotes the cofactor of the element aij in Δ.
Example Problems on Properties of Determinants
Question 1: Using properties of determinants, prove that
\(\begin{array}{l}\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=\left( a+b+c \right)\left( ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)\end{array} \)
Solution:
By using invariance and scalar multiple properties of determinants, we can prove the given problem.
\(\begin{array}{l}\Delta =\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|\\=\left| \begin{matrix} a+b+c & b & c \\ b+c+a & c & a \\ c+a+b & a & b \\ \end{matrix} \right| [Operating {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}]\end{array} \)
\(\begin{array}{l}=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|\\=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \\ \end{matrix} \right| [Operating \left( {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,and\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right)]\end{array} \)
= (a + b + c) [(c – b) (b – c) – (a – b) (a – c)]
=
\(\begin{array}{l}\left( a+b+c \right)\left( ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)\end{array} \)
Question 2: Prove the following identity
\(\begin{array}{l}\left| \begin{matrix} -{{\alpha }^{2}} & \beta \alpha & \gamma \alpha \\ \alpha \beta & -{{\beta }^{2}} & \gamma \beta \\ \alpha \gamma & \beta \gamma & -{{\gamma }^{2}} \\ \end{matrix} \right|=4{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\end{array} \)
Solution:
Take
\(\begin{array}{l}\alpha,\beta,\gamma\end{array} \)
common from the L.H.S. and then by using scalar multiple properties and invariance property of determinant, we can prove the given problem.
\(\begin{array}{l}\Delta =\left| \begin{matrix} -{{\alpha }^{2}} & \beta \alpha & \gamma \alpha \\ \alpha \beta & -{{\beta }^{2}} & \gamma \beta \\ \alpha \gamma & \beta \gamma & -{{\gamma }^{2}} \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}Taking \;\alpha ,\beta ,\gamma \; common\; from \;{{C}_{1}},{{C}_{2}},{{C}_{3}}\; \;respectively \;\;\Delta =\alpha \beta \gamma \left| \begin{matrix} -\alpha & \alpha & \alpha \\ \beta & -\beta & \beta \\ \gamma & \gamma & -\gamma \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}Now \;taking\; [\alpha ,\beta ,\gamma ] \;common\; from \;{R}_{1},{R}_{2},{R}_{3}\;respectively\end{array} \)
\(\begin{array}{l}\Delta ={{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\left| \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{matrix} \right|\end{array} \)
Now applying and
\(\begin{array}{l}{R}_{3}\to {R}_{3}+{R}_{1}\;we\; have \;\Delta ={\alpha }^{2}{\beta }^{2}{\gamma }^{2}\left| \begin{matrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}Now\; expanding\; along\; {{C}_{1}},\Delta {{\alpha }^{2}}\times {{\beta }^{2}}\left( -1 \right)\times {{\gamma }^{2}}\left( -1 \right)\left| \begin{matrix} 0 & 2 \\ 2 & 0 \\ \end{matrix} \right|=\;\;{{\alpha }^{2}}{{\beta }^{2}}\left( -1 \right){{\gamma }^{2}}\left( 0-4 \right)=4{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\end{array} \)
Hence proved.
Question 3: Show that
\(\begin{array}{l}\left| \begin{matrix} \alpha & \beta & \gamma \\ \theta & \phi & \psi \\ \lambda & \mu & v \\ \end{matrix} \right|=\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|\end{array} \)
Solution:
Interchange the rows and columns across the diagonal using the reflection property, and then using the switching property of the determinant, we can obtain the required result.
L.H.S. =
\(\begin{array}{l}\left| \begin{matrix} \alpha & \beta & \gamma \\ \theta & \phi & \psi \\ \lambda & \mu & v \\ \end{matrix} \right|=\left| \begin{matrix} \alpha & \theta & \lambda \\ \beta & \phi & \mu \\ \gamma & \psi & v \\ \end{matrix} \right|\end{array} \)
(Interchanging rows and columns across the diagonal)
\(\begin{array}{l}=\left( -1 \right)\left| \begin{matrix} \alpha & \lambda & \theta \\ \beta & \mu & \phi \\ \gamma & v & \psi \\ \end{matrix} \right|\\={{\left( -1 \right)}^{2}}\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|\\=\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|\end{array} \)
= R.H.S.
Question 4: If a, b, c are all different and if
\(\begin{array}{l}\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0,\end{array} \)
prove that abc = –1.
Solution:
Split the given determinant using the sum property. Then, by using scalar multiple, switching and invariance properties of determinants, we can prove the given equation.
\(\begin{array}{l}D=\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{3}} & 1+{{c}^{3}} \\ \end{matrix} \right|\\=\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|\\=\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}={{\left( -1 \right)}^{1}}\left| \begin{matrix} 1 & {{a}^{2}} & a \\ 1 & {{b}^{2}} & b \\ 1 & {{c}^{2}} & c \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\; \left[ {{C}_{1}}\leftrightarrow {{C}_{3}}\,in\,\,1st\,\,\det . \right]\end{array} \)
\(\begin{array}{l}={{\left( -1 \right)}^{2}}\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right| \;\;\left[ {{C}_{2}}\leftrightarrow {{C}_{3}}\,in\,\,1st\,\,\det . \right]\end{array} \)
\(\begin{array}{l}=\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\\=\left( 1+abc \right)\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=\left( 1+abc \right)\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 0 & b-a & {{b}^{2}}-{{a}^{2}} \\ 0 & c-a & {{c}^{2}}-{{a}^{2}} \\ \end{matrix} \right| \;\;\left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1\,\,}}and\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\end{array} \)
\(\begin{array}{l}=\left( 1+abc \right)\left| \begin{matrix} b-a & {{b}^{2}}-{{a}^{2}} \\ c-a & {{c}^{2}}-{{a}^{2}} \\ \end{matrix} \right| \;(expanding \;along \;1st \;row) \\=\left( 1+abc \right)\left( b-a \right)\left( c-a \right)\left| \begin{matrix} 1 & b+a \\ 1 & c+a \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}=\left( 1+abc \right)\left( b-c \right)\left( c-a \right)\left( c+a-b-a \right)\\=\left( 1+abc \right)\left( b-a \right)\left( c-a \right)\left( c-b \right)\end{array} \)
\(\begin{array}{l}\Rightarrow D=\left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right);\; But \;given \;D = 0\end{array} \)
\(\begin{array}{l}\Rightarrow \left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right)=0 \end{array} \)
∴ ( 1 + abc) = 0
[since a, b, c are different
\(\begin{array}{l}a\ne b,b\ne c,c\ne a \end{array} \)
Hence, abc = -1
Question 5: Prove that
\(\begin{array}{l}\left| \begin{matrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \\ \end{matrix} \right|=2{{\left( a+b+c \right)}^{3}}\end{array} \)
.
Solution:
Simply by using switching and scalar multiple properties, we can expand the L.H.S.
Given determinant
\(\begin{array}{l}=\left| \begin{matrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \\ \end{matrix} \right|\end{array} \)
Applying
\(\begin{array}{l}{{C}_{1}}\to {{C}_{1}}+\left( {{C}_{2}}+{{C}_{3}} \right),\end{array} \)
, we obtain
\(\begin{array}{l}\left| \begin{matrix} 2\left( a+b+c \right) & a & b \\ 2\left( a+b+c \right) & b+c+2a & b \\ 2\left( a+b+c \right) & a & c+a+2b \\ \end{matrix} \right|\\=2\left( a+b+c \right)\left| \begin{matrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \\ \end{matrix} \right|\end{array} \)
\(\begin{array}{l}{R}_{1}\to {R}_{2}-{R}_{1}\;\;and\;\;{R}_{3}\to {R}_{3}-{R}_{1}\;\;(given)\end{array} \)
\(\begin{array}{l}2\left( a+b+c \right)\left| \begin{matrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \\ \end{matrix} \right|\\=2\left( a+b+c \right)\left\{ \left( b+c+a \right)\left( c+a+b \right)-\left( 0\times 0 \right) \right\}=2{{\left( a+b+c \right)}^{3}}\end{array} \)
Hence proved.
Question 6: Prove that
\(\begin{array}{l}\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\end{array} \)
.
Solution:
Expand the determinant
\(\begin{array}{l}\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|\end{array} \)
by using scalar multiple and invariance properties.
L.H.S.=
\(\begin{array}{l}\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|;\end{array} \)
Multiplying C1,C2,C3 by a, b, c, respectively
\(\begin{array}{l}=\frac{1}{abc}\left| \begin{matrix} a\left( {{a}^{2}}+1 \right) & a{{b}^{2}} & a{{c}^{2}} \\ {{a}^{2}}b & b\left( {{b}^{2}}+1 \right) & b{{c}^{2}} \\ {{a}^{2}}c & {{b}^{2}}c & c\left( {{c}^{2}}+1 \right) \\ \end{matrix} \right|;\end{array} \)
Now taking a, b, c common from R1,R2,R3, respectively.
\(\begin{array}{l}=\frac{abc}{abc}\left| \begin{matrix} {{a}^{2}}+1 & {{b}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}}+1 & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|\\=\left| \begin{matrix} 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}} & {{c}^{2}} \\ 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}}+1 & {{c}^{2}} \\ 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right| \;\;\;\left[ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \right]\end{array} \)
\(\begin{array}{l}=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix} 1 & {{b}^{2}} & {{c}^{2}} \\ 1 & {{b}^{2}}+1 & {{c}^{2}} \\ 1 & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|\\=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix} 1 & {{b}^{2}} & {{c}^{2}} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right| \;\;\left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,and\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\end{array} \)
\(\begin{array}{l}=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( 1 \right)=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=R.H.S.\end{array} \)
Hence proved.
Frequently Asked Questions
Q1
What do you mean by the reflection property of determinants?
The determinant remains unchanged if the rows are changed into columns and the columns into rows. This is known as the property of reflection.
Q2
What happens to a determinant if 2 rows or columns are interchanged?
The determinant changes its sign when 2 rows or columns are interchanged.
Q3
State the proportionality property of determinants.
If all the elements of a row or column are proportional or identical to the elements of another row or column, then the determinant is zero. This is also called the repetition property.
Q4
What do you mean by the triangle property of determinants?
If all the elements of a determinant above or below the main diagonal are zeros, then the determinant is equal to the product of diagonal elements.
Q5
Can a determinant be zero?
Yes, a determinant can be zero, negative or positive.
Q6
Is the determinant of an identity matrix equal to 1?
Yes, the determinant of an identity matrix is always 1.
Q7
What do you mean by all zero property of determinants?
According to all zero property of determinants, if all the elements of a row/column are zero, then the determinant is equal to zero.
Q8
If A-1 is the inverse of matrix A, then what is det A-1?
If A-1 is the inverse of matrix A, then det A-1 = 1/det A.
