Let matrix X= [x_{i j}] is given by X=\left[\begin{array}{ccc}1 & -1 & 2 \ 3 & 4 & -5 \ 2 & -1 & 3]\end{array}. Then the matrix {\Y}= [m_{i j}], where m_{i j}= Minor of x_{i j}, is

Correct Answer is : [719−11−1−11−3−117]

\left[\begin{array}{ccc}7 & -5 & -3 \ 19 & 1 & -11 \ -11 & 1 & 7]\end{array}

\left[\begin{array}{ccc}7 & -19 & 11 \ 5 & -1 & -1 \ 3 & 11 & 7]\end{array}

\left[\begin{array}{ccc}7 & 19 & -11 \ -3 & 11 & 7 \ -2 & -1 & -1]\end{array}

\left[\begin{array}{ccc}7 & 19 & -11 \ -1 & -1 & 1 \ -3 & -11 & 7]\end{array}

CBSE Class 12 Math 2022 Term I Solved Paper - Question 36

Question : 36 of 50

Marks: +1, -0

Let matrix

X= [x_{i j}]

is given by

X={[\table 1 , -1 , 2 ; 3 , 4 , -5 ; 2 , -1 , 3]}

.
Then the matrix

{\Y}= [m_{i j}]

, where

m_{i j}=

Minor of

x_{i j}

, is

Solution:

m_{11}={|\table 4 , -5 ; -1 , 3|}=12-5=7

m_{12}={|\table 3 , -5 ; 2 , 3|}=9+10=19

m_{13}={|\table 3 , 4 ; 2 , -1|}=-3-8=-11

m_{21}={|\table -1 , 2 ; -1 , 3|}=-3+2=-1

m_{22}={|\table 1 , 2 ; 2 , 3|}=3-4=-1

m_{23}={|\table 1 , -1 ; 2 , -1|}=-1+2=1

m_{31}={|\table -1 , 2 ; 4 , -5|}=5-8=-3

m_{32}={|\table 1 , 2 ; 3 , -5|}=-5-6=-11

m_{33}={|\table 1 , -1 ; 3 , 4|}=4+3=7

∴ \;\; Y={[\table 7 , 19 , -11 ; -1 , -1 , 1 ; -3 , -11 , 7]}

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