Q.Find minors and cofactors of the elements of the determinant $\begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix}$ and verify that $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} = 0$.

Concept: Minor and Cofactor definitions; property that sum of products of elements of one row with cofactors of another row equals zero.

Step 1 – Minors

For element $a_{ij}$, the minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$.

• $M_{11} = \begin{vmatrix}0 & 4\\5 & -7\end{vmatrix} = 0(-7) - 4(5) = -20$
• $M_{12} = \begin{vmatrix}6 & 4\\1 & -7\end{vmatrix} = 6(-7) - 4(1) = -46$
• $M_{13} = \begin{vmatrix}6 & 0\\1 & 5\end{vmatrix} = 6(5) - 0(1) = 30$
• $M_{21} = \begin{vmatrix}-3 & 5\\5 & -7\end{vmatrix} = (-3)(-7) - 5(5) = 21 - 25 = -4$
• $M_{22} = \begin{vmatrix}2 & 5\\1 & -7\end{vmatrix} = 2(-7) - 5(1) = -14 - 5 = -19$
• $M_{23} = \begin{vmatrix}2 & -3\\1 & 5\end{vmatrix} = 2(5) - (-3)(1) = 10 + 3 = 13$
• $M_{31} = \begin{vmatrix}-3 & 5\\0 & 4\end{vmatrix} = (-3)(4) - 5(0) = -12$
• $M_{32} = \begin{vmatrix}2 & 5\\6 & 4\end{vmatrix} = 2(4) - 5(6) = 8 - 30 = -22$
• $M_{33} = \begin{vmatrix}2 & -3\\6 & 0\end{vmatrix} = 2(0) - (-3)(6) = 18$

Step 2 – Cofactors

$A_{ij} = (-1)^{i+j} M_{ij}$

• $A_{11} = (-1)^2(-20) = -20$
• $A_{12} = (-1)^3(-46) = 46$ …

We have the determinant:

Minor $M_{ij}$ is the determinant obtained by deleting the $i$-th row and $j$-th column. Cofactor $A_{ij} = (-1)^{i+j} M_{ij}$.

Let’s compute them systematically.

1. First row minors and cofactors

   • $M_{11}$: delete row 1, column 1 → $\begin{vmatrix} 0 & 4 \\ 5 & -7 \end{vmatrix} = 0(-7) - 4(5) = -20$

     $A_{11} = (-1)^{1+1}(-20) = -20$

   • $M_{12}$: delete row 1, column 2 → $\begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = 6(-7) - 4(1) = -42 - 4 = -46$

     $A_{12} = (-1)^{1+2}(-46) = 46$

   • $M_{13}$: delete row 1, column 3 → $\begin{vmatrix} 6 & 0 \\ 1 & 5 \end{vmatrix} = 6(5) - 0(1) = 30$

     $A_{13} = (-1)^{1+3}(30) = 30$

2. Second row minors and cofactors

   • $M_{21}$: delete row 2, column 1 → $\begin{vmatrix} -3 & 5 \\ 5 & -7 \end{vmatrix} = (-3)(-7) - 5(5) = 21 - 25 = -4$

     $A_{21} = (-1)^{2+1}(-4) = 4$

   • $M_{22}$: delete row 2, column 2 → $\begin{vmatrix} 2 & 5 \\ 1 & -7 \end{vmatrix} = 2(-7) - 5(1) = -14 - 5 = -19$

     $A_{22} = (-1)^{2+2}(-19) = -19$

   • $M_{23}$: delete row 2, column 3 → $\begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = 2(5) - (-3)(1) = 10 + 3 = 13$

     $A_{23} = (-1)^{2+3}(13) = -13$

3. Third row minors and cofactors

   • $M_{31}$: delete row 3, column 1 → $\begin{vmatrix} -3 & 5 \\ 0 & 4 \end{vmatrix} = (-3)(4) - 5(0) = -12$

     $A_{31} = (-1)^{3+1}(-12) = -12$

   • $M_{32}$: delete row 3, column 2 → $\begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = 2(4) - 5(6) = 8 - 30 = -22$

     $A_{32} = (-1)^{3+2}(-22) = 22$

   • $M_{33}$: delete row 3, column 3 → $\begin{vmatrix} 2 & -3 \\ 6 & 0 \end{vmatrix} = 2(0) - (-3)(6) = 0 + 18 = 18$

     $A_{33} = (-1)^{3+3}(18) = 18$

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Method: Verifying the "Foreign Row-Cofactor" Property of a Determinant

This method verifies the property that multiplying one row's elements by a different row's cofactors and summing always gives zero — a structural fact that also explains why cofactor expansion works along any row.

Steps

Step 1: Compute every minor of the determinant

Systematically find $M_{ij}$ for all nine positions (for a $3\times3$) by deleting the appropriate row and column each time.

Step 2: Convert every minor to its cofactor

Apply $A_{ij}=(-1)^{i+j}M_{ij}$ to each, using the checkerboard sign pattern.

Step 3: Identify which row's elements and which row's cofactors the problem asks you to combine

Read the expression carefully — e.g. $a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}$ combines row-1 elements with row-3 cofactors, which are from different rows.

Step 4: Multiply and sum

Compute the three products $a_{1k}\cdot A_{3k}$ for $k=1,2,3$ and add them.

Step 5: Confirm the sum is zero …

Common Mistakes

Mistake 1: Combining a row's elements with its OWN cofactors instead of a different row's

Why it's wrong: $a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$ (same row) reproduces the determinant $\Delta$ itself, not zero — confusing this with the "foreign row" combination $a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}$ leads to expecting the wrong value and "fixing" a correct calculation into an incorrect one. Correct approach: carefully check whether the elements and cofactors in the expression to verify come from the same row (expect $\Delta$) or different rows (expect $0$).

Mistake 2: A single arithmetic slip in one of the nine cofactors going unnoticed …