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What is the product?

[tex]\[
\left[\begin{array}{cc}
4 & 3 \\
-1 & -1
\end{array}\right] \times \left[\begin{array}{cc}
-3 & 1 \\
-2 & 2
\end{array}\right]
\][/tex]

Sagot :

Let's calculate the product of the given matrices step-by-step.

We need to multiply the matrix:
[tex]\[ A = \begin{pmatrix} 4 & 3 \\ -1 & -1 \end{pmatrix} \][/tex]
by the matrix:
[tex]\[ B = \begin{pmatrix} -3 & 1 \\ -2 & 2 \end{pmatrix} \][/tex]

The result of the multiplication will be a new matrix [tex]\(C\)[/tex] of the form:
[tex]\[ C = A \times B = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix} \][/tex]

Where each element [tex]\(c_{ij}\)[/tex] in the resulting matrix [tex]\(C\)[/tex] is calculated by taking the dot product of the [tex]\(i\)[/tex]-th row of matrix [tex]\(A\)[/tex] with the [tex]\(j\)[/tex]-th column of matrix [tex]\(B\)[/tex]:

1. Calculate [tex]\(c_{11}\)[/tex]:
[tex]\[ c_{11} = 4 \cdot (-3) + 3 \cdot (-2) \][/tex]
[tex]\[ c_{11} = -12 + (-6) \][/tex]
[tex]\[ c_{11} = -18 \][/tex]

2. Calculate [tex]\(c_{12}\)[/tex]:
[tex]\[ c_{12} = 4 \cdot 1 + 3 \cdot 2 \][/tex]
[tex]\[ c_{12} = 4 + 6 \][/tex]
[tex]\[ c_{12} = 10 \][/tex]

3. Calculate [tex]\(c_{21}\)[/tex]:
[tex]\[ c_{21} = -1 \cdot (-3) + (-1) \cdot (-2) \][/tex]
[tex]\[ c_{21} = 3 + 2 \][/tex]
[tex]\[ c_{21} = 5 \][/tex]

4. Calculate [tex]\(c_{22}\)[/tex]:
[tex]\[ c_{22} = -1 \cdot 1 + (-1) \cdot 2 \][/tex]
[tex]\[ c_{22} = -1 + (-2) \][/tex]
[tex]\[ c_{22} = -3 \][/tex]

So, the resulting matrix [tex]\(C\)[/tex] is:
[tex]\[ C = \begin{pmatrix} -18 & 10 \\ 5 & -3 \end{pmatrix} \][/tex]

Thus, the product of the two matrices is:
[tex]\[ \boxed{\begin{pmatrix} -18 & 10 \\ 5 & -3 \end{pmatrix}} \][/tex]