Properties of Determinants

There are 10 main properties of determinants which include reflection property, all-zero property,  proportionality or repetition property, switching property, scalar multiple property, sum property, invariance property, factor property, triangle property, and co-factor matrix property. All the determinant properties have been covered below in a detailed way along with solved examples.

All Topics in Determinants

Determinants have some properties that are useful as they permit us to generate the same results with different and simpler configurations of entries (elements).

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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 the 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:

\(\left| \begin{matrix} {{a}_{1}}+{{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{a}_{2}}+{{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{a}_{3}}+{{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{a}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{a}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|+\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} & {{d}_{1}} \\ {{b}_{2}} & {{c}_{2}} & {{d}_{2}} \\ {{b}_{3}} & {{c}_{3}} & {{d}_{3}} \\ \end{matrix} \right|\)

7. Property of Invariance:

\(\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}}+\alpha {{b}_{1}}+\beta {{c}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}}+\alpha {{b}_{2}}+\beta {{c}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}}+\alpha {{b}_{3}}+\beta {{c}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\)

That is, a determinant remains unaltered under an operation of the form \({{C}_{i}}\to {{C}_{i}}+\alpha {{C}_{j}}+\beta {{C}_{k}},\) where \(j,k\ne i,\) or an operation of the form \({{R}_{i}}\to {{R}_{i}}+\alpha {{R}_{j}}+\beta {{R}_{k}},\) where \(j,k\ne i\)

8. Factor Property:

If a determinant Δ becomes zero when we put \(x=\alpha ,\) then \(\left( x-\alpha \right)\) 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,

\(\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ 0 & {{b}_{2}} & {{b}_{3}} \\ 0 & 0 & {{c}_{3}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & 0 & 0 \\ {{a}_{2}} & {{b}_{2}} & 0 \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|={{a}_{1}}{{b}_{2}}{{c}_{3}}\)

10. Determinant of cofactor matrix:

\(\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\;then \;{{\Delta }_{1}}=\left| \begin{matrix} {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\ {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\ {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\ \end{matrix} \right|={{\Delta }^{2}} \;where \;{{C}_{ij}}\;denotes \;the\; cofactor \;of\; the \;element \;{{a}_{ij}}\; in \;Δ.\)

Example Problems on Properties of Determinants

Question 1: Using properties of determinants, prove that \(\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=\left( a+b+c \right)\left( ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)\)

Solution:

By using invariance and scalar multiple property of determinant we can prove the given problem.

\(\Delta =\left| \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right|=\left| \begin{matrix} a+b+c & b & c \\ b+c+a & c & a \\ c+a+b & a & b \\ \end{matrix} \right| [Operating {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}]\)

\(=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{matrix} \right|=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \\ \end{matrix} \right| [Operating \left( {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,and\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right)]\)

= (a + b + c) [(c – b) (b – c) – (a – b) (a – c)]

= \(\left( a+b+c \right)\left( ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)\)

Question 2: Prove the following identity \(\left| \begin{matrix} -{{\alpha }^{2}} & \beta \alpha & \gamma \alpha \\ \alpha \beta & -{{\beta }^{2}} & \gamma \beta \\ \alpha \gamma & \beta \gamma & -{{\gamma }^{2}} \\ \end{matrix} \right|=4{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\)

Solution:

Take \(\alpha ,\beta ,\gamma\) common from the L.H.S. and then by using scalar multiple property and invariance property of determinant we can prove the given problem.

\(\Delta =\left| \begin{matrix} -{{\alpha }^{2}} & \beta \alpha & \gamma \alpha \\ \alpha \beta & -{{\beta }^{2}} & \gamma \beta \\ \alpha \gamma & \beta \gamma & -{{\gamma }^{2}} \\ \end{matrix} \right|\)

\(Taking \;\alpha ,\beta ,\gamma \; common\; from \;{{C}_{1}},{{C}_{2}},{{C}_{3}}\; \;respectively \;\;\Delta =\alpha \beta \gamma \left| \begin{matrix} -\alpha & \alpha & \alpha \\ \beta & -\beta & \beta \\ \gamma & \gamma & -\gamma \\ \end{matrix} \right|\)

\(Now \;taking\; [\alpha ,\beta ,\gamma ] \;common\; from \;{R}_{1},{R}_{2},{R}_{3}\;respectively\)

\(\Delta ={{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\left| \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{matrix} \right|\)

Now applying and \({R}_{3}\to {R}_{3}+{R}_{1}\;we\; have \;\Delta ={\alpha }^{2}{\beta }^{2}{\gamma }^{2}\left| \begin{matrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \\ \end{matrix} \right|\)

\(Now\; expanding\; along\; {{C}_{1}},\Delta {{\alpha }^{2}}\times {{\beta }^{2}}\left( -1 \right)\times {{\gamma }^{2}}\left( -1 \right)\left| \begin{matrix} 0 & 2 \\ 2 & 0 \\ \end{matrix} \right|=\;\;{{\alpha }^{2}}{{\beta }^{2}}\left( -1 \right){{\gamma }^{2}}\left( 0-4 \right)=4{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\)

Hence proved.

Question 3: Show that \(\left| \begin{matrix} \alpha & \beta & \gamma \\ \theta & \phi & \psi \\ \lambda & \mu & v \\ \end{matrix} \right|=\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|\)

Solution:

Interchange the rows and columns across the diagonal using reflection property and then using the switching property of determinant we can obtain the required result.

L.H.S. = \(\left| \begin{matrix} \alpha & \beta & \lambda \\ \theta & \phi & \psi \\ \lambda & \mu & v \\ \end{matrix} \right|=\left| \begin{matrix} \alpha & \theta & \lambda \\ \beta & \phi & \mu \\ \gamma & \psi & v \\ \end{matrix} \right|\)

(Interchanging rows and columns across the diagonal)

\(=\left( -1 \right)\left| \begin{matrix} \alpha & \lambda & \theta \\ \beta & \mu & \phi \\ \gamma & v & \psi \\ \end{matrix} \right|={{\left( -1 \right)}^{2}}\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|=\left| \begin{matrix} \beta & \mu & \phi \\ \alpha & \lambda & \theta \\ \gamma & v & \psi \\ \end{matrix} \right|=\)

R.H.S.

Question 4: If a, b, c are all different and if \(\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0,\)

prove that abc = –1.

Solution:

Split the given determinant using sum property. Then by using scalar multiple, switching and invariance properties of determinants, we can prove the given equation.

\(D=\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{3}} & 1+{{c}^{3}} \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\)

\(={{\left( -1 \right)}^{1}}\left| \begin{matrix} 1 & {{a}^{2}} & a \\ 1 & {{b}^{2}} & b \\ 1 & {{c}^{2}} & c \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\; \left[ {{C}_{1}}\leftrightarrow {{C}_{3}}\,in\,\,1st\,\,\det . \right]\)

\(={{\left( -1 \right)}^{2}}\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right| \;\;\left[ {{C}_{2}}\leftrightarrow {{C}_{3}}\,in\,\,1st\,\,\det . \right]\)

\(=\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|+abc\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=\left( 1+abc \right)\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|\)

\(=\left( 1+abc \right)\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 0 & b-a & {{b}^{2}}-{{a}^{2}} \\ 0 & c-a & {{c}^{2}}-{{a}^{2}} \\ \end{matrix} \right| \;\;\left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1\,\,}}and\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\)

\(=\left( 1+abc \right)\left| \begin{matrix} b-a & {{b}^{2}}-{{a}^{2}} \\ c-a & {{c}^{2}}-{{a}^{2}} \\ \end{matrix} \right| \;(expanding \;along \;1st \;row) \;=\left( 1+abc \right)\left( b-a \right)\left( c-a \right)\left| \begin{matrix} 1 & b+a \\ 1 & c+a \\ \end{matrix} \right|\)

\(=\left( 1+abc \right)\left( b-c \right)\left( c-a \right)\left( c+a-b-a \right)=\left( 1+abc \right)\left( b-a \right)\left( c-a \right)\left( c-b \right)\)

\(\Rightarrow D=\left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right);\; But \;given \;D = 0\)

\(\Rightarrow \left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right)=0 \)

∴ ( 1 + abc) = 0

[since a, b, c are different \(a\ne b,b\ne c,c\ne a \) Hence, abc = -1

Question 5: Prove that \(\left| \begin{matrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \\ \end{matrix} \right|=2{{\left( a+b+c \right)}^{3}}\)

Solution:

Simply by using switching and scalar multiple property we can expand the L.H.S.

Given determinant \(=\left| \begin{matrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \\ \end{matrix} \right|\)

Applying \({{C}_{1}}\to {{C}_{1}}+\left( {{C}_{2}}+{{C}_{3}} \right),\) we obtain

\(\left| \begin{matrix} 2\left( a+b+c \right) & a & b \\ 2\left( a+b+c \right) & b+c+2a & b \\ 2\left( a+b+c \right) & a & c+a+2b \\ \end{matrix} \right|=2\left( a+b+c \right)\left| \begin{matrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \\ \end{matrix} \right|\)

\({R}_{1}\to {R}_{2}-{R}_{1}\;\;and\;\;{R}_{3}\to {R}_{3}-{R}_{1}\;\;(given)\)

\(2\left( a+b+c \right)\left| \begin{matrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \\ \end{matrix} \right|=2\left( a+b+c.1 \right)\left\{ \left( b+c+a \right)\left( c+a+b \right)-\left( 0\times 0 \right) \right\}=2{{\left( a+b+c \right)}^{3}}\)

Hence proved.

Question 6: Prove that \(\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\)

Solution:

Expand the determinant \(\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|\)

by using scalar multiple and invariance property.

L.H.S.= \(\left| \begin{matrix} {{a}^{2}}+1 & ab & ac \\ ab & {{b}^{2}}+1 & bc \\ ac & bc & {{c}^{2}}+1 \\ \end{matrix} \right|;\) Multiplying C1,C2,C3 by a, b, c respectively

\(=\frac{1}{abc}\left| \begin{matrix} a\left( {{a}^{2}}+1 \right) & a{{b}^{2}} & a{{c}^{2}} \\ {{a}^{2}}b & b\left( {{b}^{2}}+1 \right) & b{{c}^{2}} \\ {{a}^{2}}c & {{b}^{2}}c & c\left( {{c}^{2}}+1 \right) \\ \end{matrix} \right|;\) Now taking a, b, c common from R1,R2,R3 respectively

\(=\frac{abc}{abc}\left| \begin{matrix} {{a}^{2}}+1 & {{b}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}}+1 & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|=\left| \begin{matrix} 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}} & {{c}^{2}} \\ 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}}+1 & {{c}^{2}} \\ 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right| \;\;\;\left[ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \right]\)

\(=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix} 1 & {{b}^{2}} & {{c}^{2}} \\ 1 & {{b}^{2}}+1 & {{c}^{2}} \\ 1 & {{b}^{2}} & {{c}^{2}}+1 \\ \end{matrix} \right|=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left| \begin{matrix} 1 & {{b}^{2}} & {{c}^{2}} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right| \;\;\left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\,and\,\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\)

\(=\left( 1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( 1.1.1 \right)=1+{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=R.H.S.\)

Hence proved


Practise This Question

Which of the following factors influence the Hardy-Weinberg equilibrium?
i. Gene migration
ii. Genetic drift
iii. Mutation
iv. Reproduction
v. Genetic recombination