Question

How do you multiply \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] with\[\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\]?

Answer 1

Hint: In the given question, we have been asked to multiply the two given matrices. We need to know that when we multiply two given matrices then we have to observe the number of rows and columns of each of the given matrices because it will affect will resultant matrix and verify that the number of columns of first matrix must be equal to the number of rows of second matrix.

Formula used:
If \[A={{\left( {{a}_{ij}} \right)}_{m\times n}}\] and\[B={{\left( {{b}_{ij}} \right)}_{n\times p}}\], then\[A\times B=C\], where C is the matrix equals to,
\[C={{\left( {{c}_{ij}} \right)}_{m\times p}}\].

Complete step by step solution:
We have given that,
 \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\]\[\times \]\[\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\]
If \[A={{\left( {{a}_{ij}} \right)}_{m\times n}}\] and\[B={{\left( {{b}_{ij}} \right)}_{n\times p}}\], then\[A\times B=C\], where C is the matrix equals to,
\[C={{\left( {{c}_{ij}} \right)}_{m\times p}}\].
Let the matrix A = \[{{\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)}_{2\times 2}}\]
Let the matrix B = \[{{\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)}_{2\times 4}}\]
Now, we can see that we have two matrices in which we have two rows and two columns in the first matrix and in the second matrix we have two rows and 4 columns.
So after multiplication we will get a matrix of 2 rows and 4 columns.
Now we will multiply them;
Let the multiplication \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\]\[\times \]\[\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\] gives \[\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}} \\
\end{matrix} \right)\]
Now, find each term,
\[\Rightarrow{{a}_{11}}=\left( -1\times -1 \right)+\left( 0\times 3 \right)=1+0=1\]
\[\Rightarrow {{a}_{12}}=\left( -1\times -3 \right)+\left( 0\times -1 \right)=3+0=3\]
\[\Rightarrow {{a}_{13}}=\left( -1\times -3 \right)+\left( 0\times -2 \right)=3+0=3\]
\[\Rightarrow {{a}_{14}}=\left( -1\times 2 \right)+\left( 0\times 1 \right)=-2+0=-2\]
\[\Rightarrow {{a}_{21}}=\left( 0\times -1 \right)+\left( 1\times 3 \right)=0+3=3\]
\[\Rightarrow {{a}_{22}}=\left( 0\times -3 \right)+\left( 1\times -1 \right)=0-1=-1\]
\[\Rightarrow {{a}_{23}}=\left( 0\times -3 \right)+\left( 1\times -2 \right)=0-2=-2\]
\[\Rightarrow {{a}_{24}}=\left( 0\times 2 \right)+\left( 1\times 1 \right)=0+1=1\]
Substitute these values,
We get a matrix;
\[\Rightarrow \left( \begin{matrix}
   1 & 3 & 3 & -2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\]
Thus,
\[\Rightarrow \left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\]\[\times \]\[\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\]= \[\left( \begin{matrix}
   1 & 3 & 3 & -2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\]
Hence, multiply \[\left( \begin{matrix}
   -1 & 0 \\
   0 & -1 \\
\end{matrix} \right)\] with\[\left( \begin{matrix}
   -1 & -3 & -3 & 2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\] we will get\[\left( \begin{matrix}
   1 & 3 & 3 & -2 \\
   3 & -1 & -2 & 1 \\
\end{matrix} \right)\].

Note: Students need to remember that while doing multiplication of two matrices that one matrix number of columns must be equal to the second matrix number of rows only then we can further multiply the matrices otherwise not. Students should remember that If \[A={{\left( {{a}_{ij}} \right)}_{m\times n}}\] and\[B={{\left( {{b}_{ij}} \right)}_{n\times p}}\], then\[A\times B=C\], where C is the matrix equals to, \[C={{\left( {{c}_{ij}} \right)}_{m\times p}}\]. They need to know about the way to find the each term of matrices.