A determinant is a scalar value that can be calculated from the elements of a square matrix. \(\begin{array}{l}\text{It is an arrangement of numbers in the form}\ \left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|.\end{array} \)
Determinant for a 3Γ3 matrix is determined by; \(\begin{array}{l}\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2}\\ a_{3}& b_{3} & c_{3} \end{vmatrix}\end{array} \)
= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2). In this article, we come across properties of determinants, multiplication of determinants and determinants formula.Β
Introduction to Determinants
The development of determinants took place when mathematicians were trying to solve a system of simultaneous linear equations.
\(\begin{array}{l}E.g.\left. \begin{matrix} {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} \\ \end{matrix} \right] \Rightarrow x=\frac{{{b}_{2}}{{c}_{1}}-{{b}_{1}}{{c}_{2}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\;\;\; and \;\; y=\frac{{{a}_{1}}{{c}_{2}}-{{a}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\end{array} \)
Mathematicians defined the symbol
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\end{array} \)
as a determinant of order 2 and the four numbers arranged in row and column were called its elements. If we write the coefficients of the equations in the following form \(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\end{array} \)
, then such an arrangement is called a determinant. In a determinant, horizontal lines are known as rows and vertical lines are known as columns. The shape of every determinant is a square. If a determinant is of order n, then it contains n rows and n columns.
\(\begin{array}{l}E.g.\ \left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|,\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\end{array} \)
are determinants of second and third order, respectively.
For every square matrix A of order n x n, there exists a number associated with it called the determinant of a square matrix.
For a matrix of 1 x 1, the determinant isΒ A = [a].
For a 2 x 2 matrix,
\(\begin{array}{l}A=\begin{bmatrix} {{a}} & {{b}} \\ {{c}} & {{d}} \\ \end{bmatrix},\ \text{the determinant is ad – bc.}\end{array} \)
\(\begin{array}{l}\text{In the case of a 3 x 3 matrix}\ A=\begin{bmatrix} a &b &c \\ d &e &f \\ g&h &i \end{bmatrix},\end{array} \)
the value of the determinant is = a (ei β fh) β b (di β fg) + c (dh β eg).
Note:
(i) The number of elements in a determinant of order n is n2.
(ii) A determinant of order 1 is the number itself.
Properties of Determinants
- There will be no change in the value of the determinant if the rows and columns are interchanged.
- Suppose any two rows or columns of a determinant are interchanged, then its sign changes.
- If any two rows or columns of a determinant are the same, then the determinant is 0.
- If any row or column of the determinant is multiplied by a variable k, then its value is multiplied by k.
- Say if some or all elements of a row or column are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants.
Contents in Determinants
Illustration 1: Expand
\(\begin{array}{l}\left| \begin{matrix} 3 & 2 & 5 \\ 9 & -1 & 4 \\ 2 & 3 & -5 \\ \end{matrix} \right|\end{array} \)
by Sarrus rules.
Solution:
By using Sarrus rule, i.e.,
\(\begin{array}{l}\Delta =\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right)\end{array} \)
we can expand the given determinant.
Here,
\(\begin{array}{l}\Delta =\left| \begin{matrix} 3 & 2 & 5 \\ 9 & -1 & 4 \\ 2 & 3 & -5 \\ \end{matrix} \right|\Rightarrow \Delta =15-36+90+16+135+10=230\end{array} \)
Illustration 2: Evaluate the determinant :
\(\begin{array}{l}\left| \begin{matrix} {{x}^{2}}-x+1 & x-1 \\ x+1 & x+1 \\ \end{matrix} \right|\end{array} \)
Solution:
By using the determinant expansion formula, we can get the result.
We have
\(\begin{array}{l}\left| \begin{matrix} {{x}^{2}}-x+1 & x-1 \\ x+1 & x+1 \\ \end{matrix} \right|=\left( {{x}^{2}}-x+1 \right)\left( x+1 \right)-\left( x+1 \right)\left( x-1 \right)\\={{x}^{3}}+{{x}^{2}}-{{x}^{2}}-x+x+1-{{x}^{2}}+1\\={{x}^{3}}-{{x}^{2}}+2\end{array} \)
.
Symmetric and Skew Symmetric Determinants
Symmetric Determinant
A determinant is called a symmetric determinant if aij = aji, β i, j
\(\begin{array}{l}E.g.\; [\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|\end{array} \)
.
Skew Symmetric Determinant
A determinant is called a skew symmetric determinant if aij = -aji, β i, jΒ for every element.
\(\begin{array}{l}E.g.\ \left| \begin{matrix} 0 & 3 & -1 \\ -3 & 0 & 5 \\ 1 & -5 & 0 \\ \end{matrix} \right|\end{array} \)
Note: (i) det |A| = 0 βA is singular matrix (ii) det | A | β 0 βA is non-singular matrix
Multiplication of Two Determinants
(a) Multiplication of two second order determinants is as follows: (as R to C method)
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\ {{a}_{2}} & {{b}_{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} {{l}_{1}} & {{m}_{1}} \\ {{l}_{2}} & {{m}_{2}} \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}} \\ \end{matrix} \right|\end{array} \)
(b) Multiplication of two third order determinants is defined as follows
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\times \left| \begin{matrix} {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\ \end{matrix} \right|\end{array} \)
(as R to C method)
\(\begin{array}{l}=\left| \begin{matrix} {{a}_{1}}{{l}_{1}}+{{b}_{1}}{{l}_{2}}+{{c}_{1}}{{l}_{3}} & {{a}_{1}}{{m}_{1}}+{{b}_{1}}{{m}_{2}}+{{c}_{1}}{{m}_{3}} & {{a}_{1}}{{n}_{1}}+{{b}_{1}}{{n}_{2}}+{{c}_{1}}{{n}_{3}} \\ {{a}_{2}}{{l}_{1}}+{{b}_{2}}{{l}_{2}}+{{c}_{2}}{{l}_{3}} & {{a}_{2}}{{m}_{1}}+{{b}_{2}}{{m}_{2}}+{{c}_{2}}{{m}_{3}} & {{a}_{2}}{{n}_{1}}+{{b}_{2}}{{n}_{2}}+{{c}_{2}}{{n}_{3}} \\ {{a}_{3}}{{l}_{1}}+{{b}_{3}}{{l}_{2}}+{{c}_{3}}{{l}_{3}} & {{a}_{3}}{{m}_{1}}+{{b}_{3}}{{m}_{2}}+{{c}_{3}}{{m}_{3}} & {{a}_{3}}{{n}_{1}}+{{b}_{3}}{{n}_{2}}+{{c}_{3}}{{n}_{3}} \\ \end{matrix} \right|\end{array} \)
Note:
(i) The two determinants to be multiplied must be of the same order.
(ii) To get the Tmn (term in the mth row nth column) in the product, Take the mth row of the 1st determinant and multiply it by the corresponding terms of the nth column of the 2nd determinant and add.
(iii) This method is the row-by-column multiplication rule for the product of 2 determinants of the nthΒ order determinant.
(iv) IfΞβ² is the determinant formed by replacing the elements of a Ξ of order n with their corresponding co-factors, then
Ξ’ = Ξn-1 . Ξ’Β is called the reciprocal determinant.
Illustration 3: Reduce the power of the determinant
\(\begin{array}{l}{{\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|}^{2}}\end{array} \)
to 1.
Solution:
By multiplying the given determinant two times, we get the determinant as required.
\(\begin{array}{l}{{\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|}^{2}}=\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\left| \begin{matrix} 0 & c & b \\ c & 0 & a \\ b & a & 0 \\ \end{matrix} \right|\Rightarrow \left| \begin{matrix} {{b}^{2}}+{{c}^{2}} & ab & ac \\ ba & {{c}^{2}}+{{a}^{2}} & bc \\ ca & cb & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\end{array} \)
Also, Read
Matrices and Determinants Previous Year Questions
Important Matrices and Determinants Formulas for JEE Main and Advanced
Toughest JEE Advanced Problems from Matrices and Determinants
Top JEE Advanced Questions from Matrices and Determinants
Matrices and Determinants – Important Topics
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Matrices and Determinants Revision for JEE – Part 1
Matrices and Determinants Revision for JEE – Part 2
Matrices and Determinants Revision for JEE – Part 3
Matrices and Determinants – Top 10 Most Important and Expected JEE Main Questions
Matrices and Determinants Revision for JEE – Part 4
Most Difficult Matrices and Determinants Problems for JEE Advanced 2022
Frequently Asked Questions
Q1
What is the determinant of an identity matrix?
The determinant of an identity matrix is 1.
Q2
Are determinants always positive?
No, determinants can be negative, positive or zero.
Q3
Give an application of determinants.
Determinants are used to solve linear equations in two or three variables.
Q4
What is the value of the determinant of a matrix, if any two rows or columns are identical?
If any two rows or columns of a matrix are identical, then the value of the determinant is equal to zero.
Q5
What will be the determinant of a matrix when two rows and columns are interchanged?
If two rows or columns of a determinant are interchanged, then the determinant changes its sign.
Q6
Will the determinant change if transposed?
No, the determinant of a matrix is equal to the determinant of the transpose of the matrix. |AT| = |A|.
Q7
Does a matrix have more than one determinant?
No, a matrix cannot have more than one determinant.
Q8
What happens to a determinant if all the elements of a row/column are multiplied by any scalar k?
If all elements of a row/column of a determinant are multiplied by some scalar k, the value of the new determinant is k times the given determinant. If A is a square matrix having n rows, and k is any scalar, then |kA| = kn|A|.
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