Question

If $A = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6 \end{bmatrix}$ and $A_{ij}$ are cofactors of the elements $a_{ij}$ of $A$, then $a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$ is equal to

• A. 8
• B. 6
• C. 4
• D. 0

Hint:

Remember: multiplying elements of a row by their *own* cofactors yields the determinant ($|A|$), but multiplying elements of a row by the cofactors of a *different* row always sums to 0!

Answer 1

The Correct Option is A

Step 1: Spot the shortcut.
The sum of each element of a row times its own cofactor is just the determinant of the matrix. So $a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}=|A|$. We only need to find $|A|$.

Step 2: Expand along the first row.
\[ |A|=3\,[2\cdot 6-1\cdot 2]-2\,[1\cdot 6-1\cdot 3]+4\,[1\cdot 2-2\cdot 3] \]

Step 3: Work out each bracket.
\[ |A|=3(10)-2(3)+4(-4)=30-6-16=8 \]
\[ \boxed{8} \]