Properties of Determinants

In linear algebra, determinant is a special number that can be determined from a square matrix. The determinant of a matrix, say P is denoted det(P), |P| or det P. Determinants have some properties that are useful as they permit us to generate the same results with different and simpler configurations of entries (elements). 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

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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:

a1+b1c1d1a2+b2c2d2a3+b3c3d3=a1c1d1a2c2d2a3c3d3+b1c1d1b2c2d2b3c3d3\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:

a1b1c1a2b2c2a3b3c3=a1+αb1+βc1b1c1a2+αb2+βc2b2c2a3+αb3+βc3b3c3\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 CiCi+αCj+βCk,{{C}_{i}}\to {{C}_{i}}+\alpha {{C}_{j}}+\beta {{C}_{k}}, where j,ki,j,k\ne i, or an operation of the form RiRi+αRj+βRk,{{R}_{i}}\to {{R}_{i}}+\alpha {{R}_{j}}+\beta {{R}_{k}}, where j,kij,k\ne i

8. Factor Property:

If a determinant Δ becomes zero when we put x=α,x=\alpha , then (xα)\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,

a1a2a30b2b300c3=a100a2b20a3b3c3=a1b2c3\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:

Δ=a11a12a13a21a22a23a31a32a33  then  Δ1=C11C12C13C21C22C23C31C32C33=Δ2  where  Cij  denotes  the  cofactor  of  the  element  aij  in  Δ.\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

abcbcacab=(a+b+c)(ab+bc+caa2b2c2)\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.

Δ=abcbcacab=a+b+cbcb+c+acac+a+bab[OperatingC1C1+C2+C3]\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}}]

=(a+b+c)1bc1ca1ab=(a+b+c)1bc0cbac0abbc[Operating(R2R2R1andR3R3R1)]=\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)]

= (a+b+c)(ab+bc+caa2b2c2)\left( a+b+c \right)\left( ab+bc+ca-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \right)

Question 2: Prove the following identity α2βαγααββ2γβαγβγγ2=4α2β2γ2\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 α,β,γ\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.

Δ=α2βαγααββ2γβαγβγγ2\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  α,β,γ  common  from  C1,C2,C3    respectively    Δ=αβγαααβββγγγTaking \;\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  [α,β,γ]  common  from  R1,R2,R3  respectivelyNow \;taking\; [\alpha ,\beta ,\gamma ] \;common\; from \;{R}_{1},{R}_{2},{R}_{3}\;respectively

Δ=α2β2γ2111111111\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 R3R3+R1  we  have  Δ=α2β2γ2111002020{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  C1,Δα2×β2(1)×γ2(1)0220=    α2β2(1)γ2(04)=4α2β2γ2Now\; 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 αβγθϕψλμv=βμϕαλθγvψ\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. = αβλθϕψλμv=αθλβϕμγψv\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)

=(1)αλθβμϕγvψ=(1)2βμϕαλθγvψ=βμϕαλθγvψ==\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 aa21+a3bb21+b3cc21+c3=0,\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=aa21+a3bb21+b3cc31+c3=aa21bb21cc21+aa2a3bb2b3cc2c3=aa21bb21cc21+abc1aa21bb21cc2D=\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|

=(1)11a2a1b2b1c2c+abc1aa21bb21cc2  [C1C3in1stdet.]={{\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]

=(1)21aa21bb21cc2+abc1aa21bb21cc2    [C2C3in1stdet.]={{\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]

=1aa21bb21cc2+abc1aa21bb21cc2=(1+abc)1aa21bb21cc2=\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|

=(1+abc)1aa20bab2a20cac2a2    [R2R2R1andR3R3R1]=\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]

=(1+abc)bab2a2cac2a2  (expanding  along  1st  row)  =(1+abc)(ba)(ca)1b+a1c+a=\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|

=(1+abc)(bc)(ca)(c+aba)=(1+abc)(ba)(ca)(cb)=\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)

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

(1+abc)(ab)(bc)(ca)=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 ab,bc,caa\ne b,b\ne c,c\ne a Hence, abc = -1

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

Applying C1C1+(C2+C3),{{C}_{1}}\to {{C}_{1}}+\left( {{C}_{2}}+{{C}_{3}} \right), we obtain

2(a+b+c)ab2(a+b+c)b+c+2ab2(a+b+c)ac+a+2b=2(a+b+c)1ab1b+c+2ab1ac+a+2b\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|

R1R2R1    and    R3R3R1    (given){R}_{1}\to {R}_{2}-{R}_{1}\;\;and\;\;{R}_{3}\to {R}_{3}-{R}_{1}\;\;(given)

2(a+b+c)1ab0b+c+a000c+a+b=2(a+b+c.1){(b+c+a)(c+a+b)(0×0)}=2(a+b+c)32\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 a2+1abacabb2+1bcacbcc2+1=1+a2+b2+c2\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 a2+1abacabb2+1bcacbcc2+1\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.= a2+1abacabb2+1bcacbcc2+1;\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

=1abca(a2+1)ab2ac2a2bb(b2+1)bc2a2cb2cc(c2+1);=\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

=abcabca2+1b2c2a2b2+1c2a2b2c2+1=1+a2+b2+c2b2c21+a2+b2+c2b2+1c21+a2+b2+c2b2c2+1      [C1C1+C2+C3]=\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]

=(1+a2+b2+c2)1b2c21b2+1c21b2c2+1=(1+a2+b2+c2)1b2c2010001    [R2R2R1andR3R3R1]=\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]

=(1+a2+b2+c2)(1.1.1)=1+a2+b2+c2=R.H.S.=\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