Question:
If $A = \left[\begin{array}{ccc}5 & 6 & 3 \\ -4 & 3 & 2 \\ -4 & -7 & 3\end{array}\right]$, then cofactors of all elements of second row are respectively
If $A = \left[\begin{array}{ccc}5 & 6 & 3 \\ -4 & 3 & 2 \\ -4 & -7 & 3\end{array}\right]$, then cofactors of all elements of second row are respectively
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You can save a lot of time by using process of elimination! As soon as you compute the very first cofactor $C_{21} = -39$, you can immediately narrow your choices down to options (A) and (B). Then, checking just the middle sign or calculating $C_{22} = 27$ isolates the correct option without needing to calculate the final term!
Updated On: Jun 12, 2026
- $-39, 3, 11$
- $-39, 27, 11$
- $39, -3, -11$
- $39, -27, 11$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The problem presents a $3 \times 3$ square matrix $A$. We need to compute the three individual cofactors corresponding specifically to the elements situated along the second row ($a_{21}$, $a_{22}$, and $a_{23}$).
Step 2: Key Formula or Approach:
The cofactor $C_{ij}$ of an element $a_{ij}$ located in row $i$ and column $j$ of a matrix is calculated using the formula:
$$C_{ij} = (-1)^{i+j} M_{ij}$$ Where $M_{ij}$ is the minor determinant obtained by completely deleting the $i$-th row and $j$-th column of matrix $A$. For the second row ($i=2$), the sign multipliers are:
$$(-1)^{2+1} = -1, \quad (-1)^{2+2} = +1, \quad (-1)^{2+3} = -1$$
Step 3: Detailed Explanation:
Let's compute the cofactors for each element of the second row systematically:
1. For element $a_{21} = -4$ (Row 2, Column 1):
Delete row 2 and column 1 to isolate the minor matrix:
$$C_{21} = (-1)^{2+1} \begin{vmatrix} 6 & 3 \\ -7 & 3 \end{vmatrix} = -1 \times [ (6 \times 3) - (3 \times -7) ]$$ $$C_{21} = -1 \times [ 18 - (-21) ] = -1 \times (18 + 21) = -39$$ 2. For element $a_{22} = 3$ (Row 2, Column 2):
Delete row 2 and column 2 to isolate the minor matrix:
$$C_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 3 \\ -4 & 3 \end{vmatrix} = +1 \times [ (5 \times 3) - (3 \times -4) ]$$ $$C_{22} = +1 \times [ 15 - (-12) ] = 15 + 12 = 27$$ 3. For element $a_{23} = 2$ (Row 2, Column 3):
Delete row 2 and column 3 to isolate the minor matrix:
$$C_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 6 \\ -4 & -7 \end{vmatrix} = -1 \times [ (5 \times -7) - (6 \times -4) ]$$ $$C_{23} = -1 \times [ -35 - (-24) ] = -1 \times ( -35 + 24 ) = -1 \times (-11) = 11$$ Assembling our three calculated cofactors in order gives: $-39, 27, 11$. This perfectly matches option (B).
Step 4: Final Answer:
The cofactors of the elements of the second row are $-39, 27, 11$, which corresponds to option (B).
The problem presents a $3 \times 3$ square matrix $A$. We need to compute the three individual cofactors corresponding specifically to the elements situated along the second row ($a_{21}$, $a_{22}$, and $a_{23}$).
Step 2: Key Formula or Approach:
The cofactor $C_{ij}$ of an element $a_{ij}$ located in row $i$ and column $j$ of a matrix is calculated using the formula:
$$C_{ij} = (-1)^{i+j} M_{ij}$$ Where $M_{ij}$ is the minor determinant obtained by completely deleting the $i$-th row and $j$-th column of matrix $A$. For the second row ($i=2$), the sign multipliers are:
$$(-1)^{2+1} = -1, \quad (-1)^{2+2} = +1, \quad (-1)^{2+3} = -1$$
Step 3: Detailed Explanation:
Let's compute the cofactors for each element of the second row systematically:
1. For element $a_{21} = -4$ (Row 2, Column 1):
Delete row 2 and column 1 to isolate the minor matrix:
$$C_{21} = (-1)^{2+1} \begin{vmatrix} 6 & 3 \\ -7 & 3 \end{vmatrix} = -1 \times [ (6 \times 3) - (3 \times -7) ]$$ $$C_{21} = -1 \times [ 18 - (-21) ] = -1 \times (18 + 21) = -39$$ 2. For element $a_{22} = 3$ (Row 2, Column 2):
Delete row 2 and column 2 to isolate the minor matrix:
$$C_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 3 \\ -4 & 3 \end{vmatrix} = +1 \times [ (5 \times 3) - (3 \times -4) ]$$ $$C_{22} = +1 \times [ 15 - (-12) ] = 15 + 12 = 27$$ 3. For element $a_{23} = 2$ (Row 2, Column 3):
Delete row 2 and column 3 to isolate the minor matrix:
$$C_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 6 \\ -4 & -7 \end{vmatrix} = -1 \times [ (5 \times -7) - (6 \times -4) ]$$ $$C_{23} = -1 \times [ -35 - (-24) ] = -1 \times ( -35 + 24 ) = -1 \times (-11) = 11$$ Assembling our three calculated cofactors in order gives: $-39, 27, 11$. This perfectly matches option (B).
Step 4: Final Answer:
The cofactors of the elements of the second row are $-39, 27, 11$, which corresponds to option (B).
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