If \( A=[a_{ij}]_{3\times 3},a_{ij}=2i-j \) then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( |A| \) | I. | 4 |
| B. | \( |A-\mathrm{I}| \), I is identity matrix | II. | −7 |
| C. | minor of \( a_{32} \) | III. | 0 |
| D. | cofactor \( a_{32} \) | IV. | −4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-II, C-I, D-IV
Putting i = 1, 2, 3 and j = 1, 2, 3 in \( a_{ij}=2i-j \):
\( \displaystyle A=\begin{bmatrix}1 & 0 & -1\\3 & 2 & 1\\5 & 4 & 3\end{bmatrix} \)
(A) \( |A| \)
This matches III. 0. Subtracting row 2 from row 3, and row 1 from row 2, gives two equal rows (2, 2, 2), so
\( \displaystyle |A|=0 \)
(B) \( |A-\mathrm{I}| \), I is identity matrix
This matches II. −7. Expanding along the first row:
\( \displaystyle |A-\mathrm{I}|=\begin{vmatrix}0 & 0 & -1\\3 & 1 & 1\\5 & 4 & 2\end{vmatrix} \)
\( \displaystyle =-1(12-5)=-7 \)
(C) minor of \( a_{32} \)
This matches I. 4. Delete row 3 and column 2:
\( \displaystyle M_{32}=\begin{vmatrix}1 & -1\\3 & 1\end{vmatrix} \)
\( \displaystyle =1+3=4 \)
(D) cofactor \( a_{32} \)
This matches IV. −4.
\( \displaystyle A_{32}=(-1)^{3+2}M_{32}=-4 \)