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# 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,Β Β 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.

## 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:

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

### 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

$$\begin{array}{l}{{C}_{i}}\to {{C}_{i}}+\alpha {{C}_{j}}+\beta {{C}_{k}},\end{array}$$
where
$$\begin{array}{l}j,k\ne i,\end{array}$$
or an operation of the form
$$\begin{array}{l}{{R}_{i}}\to {{R}_{i}}+\alpha {{R}_{j}}+\beta {{R}_{k}},\end{array}$$
where
$$\begin{array}{l}j,k\ne i\end{array}$$

### 8. Factor Property:

If a determinant Ξ becomes zero when we put

$$\begin{array}{l}x=\alpha ,\end{array}$$
then
$$\begin{array}{l}\left( x-\alpha \right)\end{array}$$
is a factor of Ξ.

### 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:

$$\begin{array}{l}\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|\;then \;{{\Delta }_{1}}=\left| \begin{matrix} {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\ {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\ {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\ \end{matrix} \right|\; where \;{{C}_{ij}}\;denotes \;the\; cofactor \;of\; the \;element \;{{a}_{ij}}\; in \;Ξ.\end{array}$$

## 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 property of determinant we can prove the given problem.

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

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

=Β (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 property 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 reflection property and then using the switching property of 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 = β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.

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

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

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

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

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

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

$$\begin{array}{l}=\left( 1+abc \right)\left( b-c \right)\left( c-a \right)\left( c+a-b-a \right)=\left( 1+abc \right)\left( b-a \right)\left( c-a \right)\left( c-b \right)\end{array}$$

$$\begin{array}{l}\Rightarrow D=\left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right);\; But \;given \;D = 0\end{array}$$

$$\begin{array}{l}\Rightarrow \left( 1+abc \right)\left( a-b \right)\left( b-c \right)\left( c-a \right)=0 \end{array}$$

β΄ ( 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 property 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 property.

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

### What do you mean by the reflection property of determinants?

If the rows are changed into columns and the columns into rows, the determinant remains unchanged. This is known as the property of reflection.

### What happens to a determinant if 2 rows or columns are interchanged?

The determinant changes its sign when 2 rows or columns are interchanged.

### 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 repetition property.

### 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.

### Can a determinant be zero?

Yes. A determinant can be zero, negative and positive.

### Is the determinant of an identity matrix equal to 1?

Yes. The determinant of an identity matrix is always 1.

### 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.

### If A-1 is the inverse of matrix A, then what is det A-1?

IfΒ  A-1 is the inverse of matrix A, then det A-1 = 1/det A.

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