Question:
If \( A = \begin{bmatrix} 1 & -2 & 3 \\ 3 & 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 3 \\ -1 & 2 \\ 4 & -5 \end{bmatrix} \) and \( C = [C_{ij}] = AB \), then \( C_{12} \) is :
If \( A = \begin{bmatrix} 1 & -2 & 3 \\ 3 & 2 & -1 \end{bmatrix} \), \( B = \begin{bmatrix} 2 & 3 \\ -1 & 2 \\ 4 & -5 \end{bmatrix} \) and \( C = [C_{ij}] = AB \), then \( C_{12} \) is :
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C12 = (row 1 of A) times (column 2 of B).
Updated On: Oct 22, 2026
- 22
- -16
- 4
- 0
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The Correct Option is B
Solution and Explanation
Step 1: Recall what C12 means:
In the product \( C = AB \), the entry \( C_{12} \) is the sum of products of row 1 of \( A \) with column 2 of \( B \).
Step 2: Write the row and column:
Row 1 of \( A \) is \( (1, -2, 3) \). Column 2 of \( B \) is \( (3, 2, -5) \).
Step 3: Multiply and add:
\[ C_{12} = (1)(3) + (-2)(2) + (3)(-5) = 3 - 4 - 15 = -16 \]
Step 4: Check the other options:
The value 22 is not an entry of \( AB \) at all. The value 4 would come from \( 3 - 4 + 5 \), which wrongly uses \( +5 \). The value 0 is \( C_{21} \), the entry from row 2 and column 1.
Final Answer:
We get \( C_{12} = -16 \), option (B). \[ \boxed{-16} \]
In the product \( C = AB \), the entry \( C_{12} \) is the sum of products of row 1 of \( A \) with column 2 of \( B \).
Step 2: Write the row and column:
Row 1 of \( A \) is \( (1, -2, 3) \). Column 2 of \( B \) is \( (3, 2, -5) \).
Step 3: Multiply and add:
\[ C_{12} = (1)(3) + (-2)(2) + (3)(-5) = 3 - 4 - 15 = -16 \]
Step 4: Check the other options:
The value 22 is not an entry of \( AB \) at all. The value 4 would come from \( 3 - 4 + 5 \), which wrongly uses \( +5 \). The value 0 is \( C_{21} \), the entry from row 2 and column 1.
Final Answer:
We get \( C_{12} = -16 \), option (B). \[ \boxed{-16} \]
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