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Properties of Determinants - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Properties of Determinants - Part 2 is considered one of the most asked concept.
17 Questions around this concept.

Solve by difficulty

If $\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$ ,then $\lambda, \frac{\lambda}{3}$ are the roots of the equation

$4 x^{2}-24 x-27=0$

$4 x^{2}+24 x+27=0$

$4 x^{2}-24 x+27=0$

$4 x^{2}+24 x-27=0$

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Properties of Determinants - Part 1

Properties of Determinants

Property 1: The value of the determinant remains unchanged if its rows and columns are interchanged.

For example,

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:first\:row,\:we\:get}\\\mathrm{\Delta=a_1\begin{vmatrix} b_2&b_3 \\ c_2 & c_3 \end{vmatrix}-a_2\begin{vmatrix} b_1&b_3 \\ c_1 & c_3 \end{vmatrix}+a_3\begin{vmatrix} b_1&b_2 \\ c_1 & c_2 \end{vmatrix}}\\\\\mathrm{\Delta=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)}\\\\\mathrm{By\:interchanging\:the\:rows\:and\:columns\:of\:\Delta,\:we\:get\:the\:determinant}\\\mathrm{\Delta'=\begin{vmatrix} a_1 &b_1 &c_1 \\a_2 & b_2 & c_2\\a_3 &b_3 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:\Delta'\:along\:first\:column,\:we\:get}\\\\\mathrm{\Delta'=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)}\\\mathrm{\Delta=\Delta'}$

Property 2: If any two rows or two columns of a determinant are interchanged, then sign of determinant changes but the numerical value remains unaltered.

For example

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:first\:row,\:we\:get}\\\mathrm{\Delta=a_1\begin{vmatrix} b_2&b_3 \\ c_2 & c_3 \end{vmatrix}-a_2\begin{vmatrix} b_1&b_3 \\ c_1 & c_3 \end{vmatrix}+a_3\begin{vmatrix} b_1&b_2 \\ c_1 & c_2 \end{vmatrix}}\\\\\mathrm{\Delta=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)}\\\\\mathrm{Interchanging\:first\:and\:third\:rows,\:the\:new\:determinant\:obtained\:is\:given\:by}\\\mathrm{\Delta'=\begin{vmatrix} c_1 &c_2 &c_3 \\b_1 & b_2 & b_3\\a_1 &a_2 & a_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:third\:row,\:we\:get}\\\\\mathrm{\Delta'=a_1(c_2b_3-c_3b_2)-a_2(c_1b_3-c_3b_1)+a_3(b_2c_1-b_1c_2)}\\\mathrm{\;\;\;\;=-[a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)]}\\\mathrm{\Delta=-\Delta'}$

Property 3 If there is an interchange of rows or coloumns twice, then the value of the determinant remains the same.

If $\Delta _n$ is the determinant obtained by n such successive operations, then

$\Delta_n=\left\{\begin{matrix} -\Delta, & \;\text{if n is odd}\\ \Delta,& \;\text{if n is even} \end{matrix}\right.$

Properties of Determinants - Part 2

Property 4

If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

For Example,

If we interchange the identical rows (or columns) of the determinant Δ, then by property 2, Δ changes its sign

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ a_1 & a_2 & a_3 \end{vmatrix}=-\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 & b_2 & b_3\\a_1 &a_2 & a_3 \end{vmatrix}\,\,\,\,\text{(interchanging row 1 and row 3)}=-\Delta\;\;\;\;\;\;\;\;\;\;\;[By\;property\;2]}\\\mathrm{2\Delta=0}\\\mathrm{\;\;\Delta=0}$

Property 5

If each element of a row (or a column) of a determinant is multiplied by a constant k, then the value of the determinant is multiplied by k.

For example

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{and\:\Delta'\:be\:the\:determinant\:obtained\:by\:multiplying\:the\:elements\:of\:}\\\mathrm{the\:first\:row\:by\:k.}\\\\\mathrm{\Delta'=\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:first\:row,\:we\:get}\\\\\mathrm{\Delta'=ka_1(b_2c_3-b_3c_2)-ka_2(b_1c_3-b_3c_1)+ka_3(b_1c_2-b_2c_1)}\\\mathrm{\;\;\;\;=k[a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)]}\\\mathrm{\Delta'=k\Delta}\\\mathrm{Hence,}\\\mathrm{\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}=k\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$

Note:

By this property, we can take out any common factor from any one row or any one column of a given determinant.
If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then the determinant value is zero.

Property 6

If every element of some row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants

For example

$\\\mathrm{\begin{vmatrix} a_1+x &a_2 +y &a_3+z \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}+\begin{vmatrix} x & y &z \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$

Proof:

$\\\mathrm{LHS=\begin{vmatrix} a_1+x &a_2 +y &a_3+z \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{Expanding\:along\:first\:row,\:we\:get} \\\mathrm{\Delta=(a_1+x)(b_2c_3-b_3c_2)-(a_2+y)(b_1c_3-b_3c_1)+(a_3+z)(b_1c_2-b_2c_1)}\\\mathrm{\;\;\;\;=a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;+x(b_2c_3-b_3c_2)-y(b_1c_3-b_3c_1)+z(b_1c_2-b_2c_1)}\\ \\\mathrm{=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}+\begin{vmatrix} x & y &z \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$

Properties of Determinants - Part 3

Property 7

If to each element of any row or column of a determinant, the equimultiples of corresponding elements of other rows (or column) are added, then the value of determinant remains the same, i.e., the value of determinant remains same if we apply the operation

$\\\mathrm{R_i\rightarrow R_i+kR_j\;\;\;or\;\;\;C_i\rightarrow C_i+kC_j}$

Explanation,
$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\;\;\;and\;\;\;\Delta'=\begin{vmatrix} a_1+kc_1 &a_2 +kc_2 &a_3+kc_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{Here,\;\Delta'\;is\;obtained\;by\;R_1\rightarrow R_1+kR_3}\\\\\mathrm{we\;can\;write\;\Delta'\;\;as}\\\mathrm{\Delta'=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}+\begin{vmatrix} kc_1 &kc_2 &kc_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{\;\;\;\;=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}+k\begin{vmatrix} c_1 &c_2 &c_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{\;\;\;\;=\Delta+k\cdot0}\\\mathrm{hence,\;\Delta'=\Delta}$

Property 8

If each element of one side or the other side or both sides of the principal diagonal of a determinant is zero, then the value of the determinant is the product of the diagonal elements.

I.e.

$\mathrm{\begin{vmatrix} a &f &g \\ 0& b & h\\ 0& 0& c \end{vmatrix}=\begin{vmatrix} a &0 &0 \\ f& b & 0\\ g& h& c \end{vmatrix}=\begin{vmatrix} a &0 &0 \\ 0& b & 0\\ 0& 0& c \end{vmatrix}=abc}$

Property 9

If a determinant $\text{D}$ become 0 for x = α, then (x - α) is a factor of Δ.

For example,

$\\\mathrm{If\;\;\Delta=\begin{vmatrix} x &x^2 &x^3 \\ 4 &16 &64 \\ 5 &9 &11 \end{vmatrix}}\\\mathrm{When,\;x=4\;the\;value\;of\;\Delta\;becomes\;0}\\\mathrm{\because at\;x=4,R_1\;and\;R_2\;\;are\;identical.}\\\mathrm{and\;at\;x=0,\;\Delta=0\;,because\;all\;element\;of\;R_1\;becomes\;0}\\\mathrm{hence,\;\;(x-0)\;\;and\;\;(x-4)\;\;are\;the\;factors\;of\;\Delta.}$