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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 36 of 50
Marks: +1, -0
Let matrix X=[xij] is given by X=[112345213].
Then the matrix Y=[mij], where mij= Minor of xij, is
m11=|4513|=125=7
m12=|3523|=9+10=19
m13=|3421|=38=11
m21=|1213|=3+2=1
m22=|1223|=34=1
m23=|1121|=1+2=1
m31=|1245|=58=3
m32=|1235|=56=11
m33=|1134|=4+3=7
Y=[719111113117]
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