Full solution

\left[\begin{array}{lll}-2 & -4 & 1 \\ -5 & -1 & 0 \\ -4 & -5 & 0\end{array}\right]\left[\begin{array}{ccc}0 & 2 & 0 \\ 2 & -1 & -1 \\ 0 & 1 & 3\end{array}\right]

Identify Size and Check: Step $1$ : Identify the size of the matrices and check if multiplication is possible. $\newline$ Matrix $A$ is $3 \times 3$ and Matrix $B$ is $3 \times 3$ . Multiplication is possible because the number of columns in $A$ equals the number of rows in $B$ .
Multiply the Matrices: Step $2$ : Multiply the matrices. $\newline$ To multiply, take the dot product of rows of $A$ with columns of $B$ . $\newline$ First row of $A$ with first column of $B$ : $(-2\times0) + (-4\times2) + (1\times0) = -8$ $\newline$ First row of $A$ with second column of $B$ : $(-2\times2) + (-4\times-1) + (1\times1) = 0$ $\newline$ First row of $A$ with third column of $B$ : $B$ $0$ $\newline$ Second row of $A$ with first column of $B$ : $B$ $3$ $\newline$ Second row of $A$ with second column of $B$ : $B$ $6$ $\newline$ Second row of $A$ with third column of $B$ : $B$ $9$ $\newline$ Third row of $A$ with first column of $B$ : $A$ $2$ $\newline$ Third row of $A$ with second column of $B$ : $A$ $5$ $\newline$ Third row of $A$ with third column of $B$ : $A$ $8$
Write Result as Matrix: Step $3$ : Write the result as a new matrix. $\newline$ Resulting Matrix $C = \begin{bmatrix} -8 & 0 & 7 \ -2 & -9 & -1 \ -10 & -3 & -5 \end{bmatrix}$