CBSE Class 12 Math 2022 Term I Solved Paper

Question : 36

Total: 50

Let matrix X=[xij] is given by X=[

].
Then the matrix Y=[mij], where mij= Minor of xij, is

Solution: 👈: Video Solution

m11=|

4	−5
−1	3

|=12−5=7
m12=|

3	−5
2	3

|=9+10=19
m13=|

3	4
2	−1

|=−3−8=−11
m21=|

−1	2
−1	3

|=−3+2=−1
m22=|

1	2
2	3

|=3−4=−1
m23=|

1	−1
2	−1

|=−1+2=1
m31=|

−1	2
4	−5

|=5−8=−3
m32=|

1	2
3	−5

|=−5−6=−11
m33=|

1	−1
3	4

|=4+3=7
∴‌‌Y=[

]

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