To find the product of the two matrices, we first need to define the matrices clearly. Let's denote the first matrix as \( A \) and the second matrix as \( B \):
\[ A = \begin{pmatrix} 3 & 2 \ 5 & -4 \ -1 & -3 \end{pmatrix} \]
\[ B = \begin{pmatrix} 1 & 4 \ -3 & 2 \end{pmatrix} \]
Matrix \( A \) has a size of \( 3 \times 2 \) (3 rows and 2 columns), and matrix \( B \) has a size of \( 2 \times 2 \) (2 rows and 2 columns). The product \( AB \) will result in a \( 3 \times 2 \) matrix, since the number of rows comes from the first matrix \( A \) and the number of columns comes from the second matrix \( B \).
To compute the matrix product \( AB \), we use the formula: \[ (AB){i,j} = \sum{k} A_{i,k} B_{k,j} \] where \( i \) is the row index of matrix \( A \), \( j \) is the column index of matrix \( B \), and \( k \) iterates over the columns of \( A \) (which are the same as the rows of \( B \)).
Now let's compute each element of the resulting matrix \( AB \):
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Row 1, Column 1: \[ (AB)_{1,1} = 3 \cdot 1 + 2 \cdot (-3) = 3 - 6 = -3 \]
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Row 1, Column 2: \[ (AB)_{1,2} = 3 \cdot 4 + 2 \cdot 2 = 12 + 4 = 16 \]
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Row 2, Column 1: \[ (AB)_{2,1} = 5 \cdot 1 + (-4) \cdot (-3) = 5 + 12 = 17 \]
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Row 2, Column 2: \[ (AB)_{2,2} = 5 \cdot 4 + (-4) \cdot 2 = 20 - 8 = 12 \]
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Row 3, Column 1: \[ (AB)_{3,1} = -1 \cdot 1 + (-3) \cdot (-3) = -1 + 9 = 8 \]
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Row 3, Column 2: \[ (AB)_{3,2} = -1 \cdot 4 + (-3) \cdot 2 = -4 - 6 = -10 \]
Putting it all together, we have the product matrix:
\[ AB = \begin{pmatrix} -3 & 16 \ 17 & 12 \ 8 & -10 \end{pmatrix} \]
Thus, the product of the two matrices \( A \) and \( B \) is:
\[ \begin{pmatrix} -3 & 16 \ 17 & 12 \ 8 & -10 \end{pmatrix} \]
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