Find the following matrix products. Note that each product can be found​ mentally, without the use of a calculator or​ pencil-and-paper calculations.

​(a)

Start 2 By 2 Table 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndTable Start 2 By 2 Table 1st Row 1st Column negative 4 2nd Column 3 2nd Row 1st Column negative 2 2nd Column negative 5 EndTable

1

0

0

1

−4

3

−2

−5

​(b) Start 2 By 2 Table 1st Row 1st Column negative 4 2nd Column 3 2nd Row 1st Column negative 2 2nd Column negative 5 EndTable Start 2 By 2 Table 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndTable

−4

3

−2

−5

1

0

0

1

Question content area bottom

Part 1

​(a)

Start 2 By 2 Table 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndTable Start 2 By 2 Table 1st Row 1st Column negative 4 2nd Column 3 2nd Row 1st Column negative 2 2nd Column negative 5 EndTable

1

0

0

1

−4

3

−2

−5

equals=enter your response here

Part 2

​(b)

Start 2 By 2 Table 1st Row 1st Column negative 4 2nd Column 3 2nd Row 1st Column negative 2 2nd Column negative 5 EndTable Start 2 By 2 Table 1st Row 1st Column 1 2nd Column 0 2nd Row 1st Column 0 2nd Column 1 EndTable

−4

3

−2

−5

1

0

0

1

equals=enter your response here

Let's solve the given matrix products:

### Part (a)
We are asked to find the product of the following two matrices:

\[
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\cdot
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

The first matrix is the identity matrix. When you multiply any matrix by the identity matrix, the result is the original matrix itself. Therefore:

\[
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\cdot
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
=
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

So the answer for part (a) is:

\[
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

### Part (b)
Now we need to find the product of the following two matrices:

\[
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\]

Again, the second matrix is the identity matrix. Multiplying any matrix by the identity matrix results in the original matrix itself. Therefore:

\[
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

So the answer for part (b) is:

\[
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

### Summary
Both products result in the matrix:

\[
\begin{pmatrix}
-4 & 3 \\
-2 & -5
\end{pmatrix}
\]

Do you have any questions or need further details?

Here are five related questions you might find interesting:
1. What is the result of multiplying any matrix by a zero matrix?
2. How does the multiplication of matrices differ from the multiplication of numbers?
3. How do we find the inverse of a 2x2 matrix?
4. What properties does the identity matrix have?
5. Can matrix multiplication be commutative under certain conditions?

**Tip:** Remember that the identity matrix acts as the "1" in matrix multiplication, leaving any matrix it multiplies unchanged.

Explore matrix multiplication example involving identity matrices. Understand how multiplying a matrix by an identity matrix results in the original matrix. Suitable for high school students learning matrix algebra.

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