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: Understanding the Question:
The problem asks for the value of the expression $a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$, which represents the sum of the products of the elements of the first row of matrix $A$ with their corresponding cofactors.

Step 2: Key Formula or Approach:
By the fundamental definition of determinants, the sum of the products of elements of any row (or column) with their respective cofactors is equal to the determinant of that matrix: $$ |A| = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13} $$

Step 3: Detailed Explanation:
Let's compute the determinant of matrix $A$ by expanding along the first row: $$ |A| = \begin{vmatrix} 3 & 2 & 4 \\ 1 & 2 & 1 \\ 3 & 2 & 6 \end{vmatrix} $$ $$ |A| = 3 \cdot [2(6) - 1(2)] - 2 \cdot [1(6) - 1(3)] + 4 \cdot [1(2) - 2(3)] $$ Simplify the terms inside the brackets: $$ |A| = 3 \cdot (12 - 2) - 2 \cdot (6 - 3) + 4 \cdot (2 - 6) $$ $$ |A| = 3 \cdot (10) - 2 \cdot (3) + 4 \cdot (-4) $$ $$ |A| = 30 - 6 - 16 $$ $$ |A| = 30 - 22 = 8 $$

Step 4: Final Answer:
The value of the expression is 8, which corresponds to option (A).