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Let A=\left [ a_{ij} \right ] and B=\left [ b_{ij} \right ] be two 3\times 3 real matrices such that b_{ij}=(3)^{(i+j-2)}a_{ji}, where, i, j=1,2,3. if the determinant of B is 81, then the determinant of A is :
Option: 1 1/9
Option: 2 1/81
Option: 3  1/3
Option: 4  3
 

Answers (1)

best_answer

 

 

Matrices, order of matrices, row and column matrix -

A set of \mathrm{m\times n} numbers (real or complex) or objects or symbols arranged in form of a rectangular array having m rows and n columns and bounded by brackets [?] is called matrix of order m × n, read as m by n matrix.

E.g for m = 2 and n =3, we have \begin{bmatrix} 2 & 4 & -3\\ 5 & 4 & 6 \end{bmatrix}  order of this matrix is 2×3

 

The m by n matrix is represented as : 

\begin{bmatrix} a_{11} &a_{12} &... & a_{1n}\\ a_{21}&a_{22} &... &a_{2n} \\ ...& ...& ... & ...\\ a_{m1}&a_{m2} &... &a_{mn} \end{bmatrix}

 

This representation can be represented in a more compact form as 

\left [ a_{ij} \right ]_{m\times n}

Where a_{ij} represents element of ith row and jth column and i = 1,2,...,m; j = 1,2,...,n.

 

For example, to locate the entry in matrix A identified as aij, we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a23.

A=\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]

Matrix is only a representation of the symbol, number or object. It does not have any value. Usually, a matrix denoted by capital letters.

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Properties of Determinants - Part 2 -

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:

  1. By this property, we can take out any common factor from any one row or any one column of a given determinant.  

  2. If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then the determinant value is zero. 

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|B|=\left|\begin{array}{lll}{b_{11}} & {b_{12}} & {b_{13}} \\ {b_{21}} & {b_{22}} & {b_{23}} \\ {b_{31}} & {b_{32}} & {b_{33}}\end{array}\right|

|B|=\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{1} a_{21}} & {3^{2} a_{22}} & {3^{3} a_{23}} \\ {3^{2} a_{31}} & {3^{3} a_{32}} & {3^{4} a_{33}}\end{array}\right|

Taking Common 3^2 from R_3 and 3 from R_2

|B|=3^3\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{0} a_{21}} & {3^{1} a_{22}} & {3^{2} a_{23}} \\ {3^{0} a_{31}} & {3^{1} a_{32}} & {3^{2} a_{33}}\end{array}\right|

Taking Common 3^2 from C_3 and 3 from C_2

\Rightarrow 81=3^{3} \cdot 3 \cdot 3^{2}|\mathrm{A}| \Rightarrow 3^{4}=3^{6}|\mathrm{A}| \Rightarrow|\mathrm{A}|=\frac{1}{9}

Correct Option (A)

Posted by

Ritika Jonwal

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