Question:

The co-factors of the elements of second column of $\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 2 & 1 \\ -1 & 3 & 4\end{array}\right]$ are

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Remember the alternating sign pattern for a $3 \times 3$ matrix: the second column signs are always $(-, +, -)$. Calculate the minors and simply apply these signs to get the cofactors quickly.
Updated On: Jun 4, 2026
  • -13, 6, 5
  • 13, 5, 6
  • 13, -6, -5
  • -13, -6, 5
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
We need to calculate the cofactors specifically for the elements located in the second column of the given $3 \times 3$ matrix.

Step 2: Key Formula or Approach:
The cofactor $C_{ij}$ of an element $a_{ij}$ in a matrix is given by:
$$C_{ij} = (-1)^{i+j} M_{ij}$$ where $M_{ij}$ is the minor of the element (the determinant of the $2 \times 2$ matrix left after removing the $i$-th row and $j$-th column).

Step 3: Detailed Explanation:
Let the given matrix be $A$. The elements of the second column are $a_{12} = -1$, $a_{22} = 2$, and $a_{32} = 3$.
6. Cofactor of $a_{12}$:
$$C_{12} = (-1)^{1+2} \begin{vmatrix} 3 & 1 \\ -1 & 4 \end{vmatrix} = -1 [(3)(4) - (1)(-1)] = -[12 + 1] = -13$$ 7. Cofactor of $a_{22}$:
$$C_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 2 \\ -1 & 4 \end{vmatrix} = +1 [(1)(4) - (2)(-1)] = +[4 + 2] = 6$$ 8. Cofactor of $a_{32}$:
$$C_{32} = (-1)^{3+2} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -1 [(1)(1) - (2)(3)] = -[1 - 6] = -[-5] = 5$$ The cofactors are $-13, 6$, and $5$.

Step 4: Final Answer:
The cofactors are $-13, 6, 5$, matching option (A).
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