Question:
Let
\(A = \{a_{ij}\}\)
be a \(3 \times 3\) matrix,
where
\(a_{ij} = \begin{cases} (-1)^{j-i}\ \text{if } i \lt j, \\ 2\ \text{if } i = j, \\ (-1)^{i+j}\ \text{if } i \gt j \end{cases}\)
then \(\det(3\,\mathrm{Adj}(2A^{-1}))\) is equal to ______________.
Let
\(A = \{a_{ij}\}\)
be a \(3 \times 3\) matrix,
where
\(a_{ij} = \begin{cases} (-1)^{j-i}\ \text{if } i \lt j, \\ 2\ \text{if } i = j, \\ (-1)^{i+j}\ \text{if } i \gt j \end{cases}\)
then \(\det(3\,\mathrm{Adj}(2A^{-1}))\) is equal to ______________.
\(A = \{a_{ij}\}\)
be a \(3 \times 3\) matrix,
where
\(a_{ij} = \begin{cases} (-1)^{j-i}\ \text{if } i \lt j, \\ 2\ \text{if } i = j, \\ (-1)^{i+j}\ \text{if } i \gt j \end{cases}\)
then \(\det(3\,\mathrm{Adj}(2A^{-1}))\) is equal to ______________.
Show Hint
Build A explicitly to get \(\det A = 4\), then use \(\det(kB) = k^3\det B\) and \(\det(\mathrm{adj}\,B) = (\det B)^2\) for 3 x 3 matrices.
Updated On: Nov 4, 2026
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Verified By Collegedunia
Correct Answer: 108
Solution and Explanation
A has rows (2, -1, 1), (-1, 2, -1), (1, -1, 2) (using \(a_{12} = (-1)^1 = -1\), \(a_{13} = (-1)^2 = 1\), \(a_{23} = (-1)^1 = -1\), \(a_{21} = (-1)^3 = -1\), \(a_{31} = (-1)^4 = 1\), \(a_{32} = (-1)^5 = -1\)).
\(\det A = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2) = 6 - 1 - 1 = 4\).
\(\det(2A^{-1}) = 2^3 \cdot \frac{1}{4} = 2\).
For a 3 x 3 matrix B, \(\det(\mathrm{adj}\,B) = (\det B)^2\), so \(\det(\mathrm{adj}(2A^{-1})) = 4\).
\(\det(3\,\mathrm{adj}(2A^{-1})) = 3^3 \cdot 4 = 108\).
Answer: 108
\(\det A = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2) = 6 - 1 - 1 = 4\).
\(\det(2A^{-1}) = 2^3 \cdot \frac{1}{4} = 2\).
For a 3 x 3 matrix B, \(\det(\mathrm{adj}\,B) = (\det B)^2\), so \(\det(\mathrm{adj}(2A^{-1})) = 4\).
\(\det(3\,\mathrm{adj}(2A^{-1})) = 3^3 \cdot 4 = 108\).
Answer: 108
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