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: Idea:
We are given \(A^{-1}\), so \(AA^{-1} = I\). Taking only the first column of \(A^{-1}\), which is \((\gamma, \alpha, \beta)^T\), we must get the first column of \(I\), which is \((1,0,0)^T\).

Step 2: Multiply:
\[ \begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}\begin{bmatrix}\gamma\\\alpha\\\beta\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix} \]
This gives three equations:
\(2\gamma - \beta = 1\), \(5\gamma + \alpha = 0\), \(\alpha + 3\beta = 0\).

Step 3: Solve:
From the second, \(\alpha = -5\gamma\). From the third, \(\alpha = -3\beta\), so \(\beta = \frac{5\gamma}{3}\). Put this in the first: \(2\gamma - \frac{5\gamma}{3} = 1\), so \(\frac{\gamma}{3} = 1\) and \(\gamma = 3\). Then \(\alpha = -15\) and \(\beta = 5\).

Step 4: Check with the determinant:
\(|A| = 2(3-0) - 0 + (-1)(5-0) = 1\). The (1,1) entry of the adjoint is the cofactor \(3\), matching \(\gamma = 3\).

Step 5: Compute the product:
\[ |\alpha\beta\gamma| = |(-15)(5)(3)| = 225 \]
Option (D) is \(-225\), which is wrong because an absolute value cannot be negative. Options (A) 125 and (B) 220 do not come from these values.

Final Answer:
\(|\alpha\beta\gamma| = 225\), option (C). \[ \boxed{225} \]