Question

If \(A = \left[ \begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array} \right]_{3\times 3}\), and \(A^{-1} = \left[ \begin{array}{ccc}γ & -1 & 1 \\ α & 6 & -5 \\ β & -2 & 2\end{array} \right]_{3\times 3}\), then \(|α\cdot β\cdot γ| =\) _____ (where \(|\cdot |\) denotes the absolute value)

• A. \(125\)
• B. \(220\)
• C. \(225\)
• D. \(-225\)

Hint:

Multiply A by the first column of A inverse and set it equal to the first column of the identity matrix.

Answer 1

The Correct Option is C

Step 1: Use the adjoint:
Here $|A| = 2(1\cdot3 - 0) - 0 + (-1)(5\cdot1 - 0) = 6 - 5 = 1$, so $A^{-1} = \operatorname{adj}A$.

Step 2: Read the first column as cofactors:
The first column of $\operatorname{adj}A$ holds the cofactors of the first row of $A$: $C_{11} = \begin{vmatrix}1 & 0\\1 & 3\end{vmatrix} = 3$, $C_{12} = -\begin{vmatrix}5 & 0\\0 & 3\end{vmatrix} = -15$, $C_{13} = \begin{vmatrix}5 & 1\\0 & 1\end{vmatrix} = 5$.

Step 3: Match:
So $\gamma = 3$, $\alpha = -15$, $\beta = 5$. Then $|\alpha\beta\gamma| = 225$.

Step 4: Why not the others:
The absolute value removes the sign, so $-225$ is impossible. 125 and 220 do not equal $15\cdot5\cdot3$.

Final Answer:
The value is 225. \[ \boxed{225} \]