Determinants-Properties-Differentiation and Integration of Determinants-Class 12 Math Notes Study Material free pdf download

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.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.