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If A=[aij]3×3,aij=2i−jA = [a_{ij}]_{3\times3}, a_{ij} = 2i-j then

Match the LIST-I with LIST-II

LIST-ILIST-II
A.∣A∣\vert A \vertI.44
B.∣A−I∣\vert A - I \vert, I is identity matrixII.−7-7
C.minor of a32a_{32}III.00
D.cofactor a32a_{32}IV.−4-4

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

Here A=[10−1321543]A=\begin{bmatrix}1&0&-1\\3&2&1\\5&4&3\end{bmatrix}. Its rows are in arithmetic progression, so ∣A∣=0|A|=0 (III). Subtracting II and expanding gives ∣A−I∣=−7|A-I|=-7 (II). Deleting row 3 and column 2 leaves ∣1−131∣=4\begin{vmatrix}1&-1\\3&1\end{vmatrix}=4 (I), and the cofactor is (−1)3+2×4=−4(-1)^{3+2}\times 4=-4 (IV).

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