Solve {l}{1}{2}{3}{lll}{-1}&{2}&{0}&{1}

$$\left( \begin{array} { l } { 1 } \\ { 2 } \\ { 3 } \end{array} \right) \left( \begin{array} { l l l } { - 1 } & { 2 } & { 0 } & { 1 } \end{array} \right)$$

Evaluate

$$\left(\begin{matrix}-1&2&0&1\\-2&4&0&2\\-3&6&0&3\end{matrix}\right)$$

Steps Using Matrix Multiplication Rule

Matrix multiplication is defined if the number of columns of the first matrix is equal to the number of rows of the second matrix.

$$\left(\begin{matrix}1\\2\\3\end{matrix}\right)\left(\begin{matrix}-1&2&0&1\end{matrix}\right)$$

Multiply the first element of the first matrix by the first element of the second matrix to obtain the element in the first row, first column of the product matrix.

$$\left(\begin{matrix}-1&&&\\&&&\\&&&\end{matrix}\right)$$

The remaining elements of the product matrix are found in the same way.

$$\left(\begin{matrix}-1&2&0&1\\2\left(-1\right)&2\times 2&0&2\\3\left(-1\right)&3\times 2&0&3\end{matrix}\right)$$

Simplify each element by multiplying the individual terms.

$$\left(\begin{matrix}-1&2&0&1\\-2&4&0&2\\-3&6&0&3\end{matrix}\right)$$

Transpose Matrix

$$\left(\begin{matrix}-1&-2&-3\\2&4&6\\0&0&0\\1&2&3\end{matrix}\right)$$