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Use the matrix multiplication below to show how to find element \( a_{2,1} \) of the product.

[tex]\[
\left[\begin{array}{rrr}
2 & 0 & -8 \\
-1 & 10 & -9
\end{array}\right] \cdot \left[\begin{array}{rrr}
-2 & -7 & 3 \\
-5 & 4 & -8 \\
2 & -4 & -6
\end{array}\right]
\][/tex]

Note: Use parentheses around all numbers to show multiplication!


Sagot :

Sure! Let's find the element at position (2,1) of the product matrix resulting from the multiplication of the given matrices.

To perform matrix multiplication, the element at the position (i,j) in the resulting matrix is calculated by taking the dot product of the i-th row of the first matrix and the j-th column of the second matrix. In our case, we need to find the element in the 2nd row and 1st column.

Given matrices:
[tex]\[ \text{Matrix A} = \left[\begin{array}{rrr} 2 & 0 & -8 \\ -1 & 10 & -9 \end{array}\right] \][/tex]

[tex]\[ \text{Matrix B} = \left[\begin{array}{rrr} -2 & -7 & 3 \\ -5 & 4 & -8 \\ 2 & -4 & -6 \end{array}\right] \][/tex]

We are searching for the element at position (2,1) in the product matrix.

The formula for finding this element is as follows:

[tex]\[ \text{Element (2,1)} = (\text{Element (2,1)} \text{ of Matrix A} \times \text{Element (1,1)} \text{ of Matrix B}) \\ + (\text{Element (2,2)} \text{ of Matrix A} \times \text{Element (2,1)} \text{ of Matrix B}) \\ + (\text{Element (2,3)} \text{ of Matrix A} \times \text{Element (3,1)} \text{ of Matrix B}) \][/tex]

Substituting the values:

[tex]\[ \text{Element (2,1)} = ((-1) \times (-2)) + (10 \times (-5)) + ((-9) \times 2) \][/tex]

Calculate each term:

1. \( (-1) \times (-2) = 2 \)
2. \( 10 \times (-5) = -50 \)
3. \( (-9) \times 2 = -18 \)

Now, add these results together:

[tex]\[ 2 + (-50) + (-18) = 2 - 50 - 18 = -66 \][/tex]

So, the element at position (2,1) of the product matrix is

[tex]\[ \boxed{-66} \][/tex]