If \(y = -4\) is a root of \(\begin{vmatrix}y&2&3\\1&y&1\\3&2&y\end{vmatrix} = 0\), then the product of the other two roots is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
Expanding the determinant along the first row:
\( y\begin{vmatrix}y&1\\2&y\end{vmatrix} - 2\begin{vmatrix}1&1\\3&y\end{vmatrix} + 3\begin{vmatrix}1&y\\3&2\end{vmatrix} = 0 \)
\(y(y^2 - 2) - 2(y - 3) + 3(2 - 3y) = 0\)
\(y^3 - 13y + 12 = 0\)
Since \(y = -4\) is a root, \(y + 4\) is a factor:
\((y + 4)(y^2 - 4y + 3) = 0\)
\((y + 4)(y - 1)(y - 3) = 0\)
The other two roots are 1 and 3, and their product is \(1 \times 3 = 3\).
Shortcut: For \(y^3 + 0y^2 - 13y + 12 = 0\), the product of the roots is \(-12\). Dividing by the known root: \(\frac{-12}{-4} = 3\).