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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]X = [x_{i j}] is given by X=[1−1234−52−13]X = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}.
Then the matrix Y=[mij]{Y} = [m_{i j}], where mij=m_{i j} = Minor of xijx_{i j}, is
Solution:     👈: Video Solution
m11=∣4−1−53∣=12−5=7m_{11} = \begin{vmatrix} 4 & -1 \\ -5 & 3 \end{vmatrix} = 12 - 5 = 7
m12=∣3−523∣=9+10=19m_{12} = \begin{vmatrix} 3 & -5 \\ 2 & 3 \end{vmatrix} = 9 + 10 = 19
m13=∣342−1∣=−3−8=−11m_{13} = \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = -3 - 8 = -11
m21=∣−12−13∣=−3+2=−1m_{21} = \begin{vmatrix} -1 & 2 \\ -1 & 3 \end{vmatrix} = -3 + 2 = -1
m22=∣1223∣=3−4=−1m_{22} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = 3 - 4 = -1
m23=∣1−12−1∣=−1+2=1m_{23} = \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = -1 + 2 = 1
m31=∣−124−5∣=5−8=−3m_{31} = \begin{vmatrix} -1 & 2 \\ 4 & -5 \end{vmatrix} = 5 - 8 = -3
m32=∣123−5∣=−5−6=−11m_{32} = \begin{vmatrix} 1 & 2 \\ 3 & -5 \end{vmatrix} = -5 - 6 = -11
m33=∣1−134∣=4+3=7m_{33} = \begin{vmatrix} 1 & -1 \\ 3 & 4 \end{vmatrix} = 4 + 3 = 7
∴    Y=[719−11−1−11−3−117]\therefore \;\; Y = \begin{bmatrix} 7 & 19 & -11 \\ -1 & -1 & 1 \\ -3 & -11 & 7 \end{bmatrix}
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