Let AequalsStart 3 By 2 Table 1st Row 1st Column 1 2nd Column 7 2nd Row 1st Column 1 2nd Column 8 3rd Row 1st Column 1 2nd Column 15 EndTable 
 . Construct a 2times3 matrix C​ (by trial and​ error) using only​ 1, minus​1, and 0 as​ entries, such that CAequalsUpper I 2. Compute AC and note that ACnot equalsUpper I 3.

Let's break this problem into parts:

### Given:
- Matrix \( A \) is a \( 3 \times 2 \) matrix:

\[
A = \begin{pmatrix}
1 & 7 \\
1 & 8 \\
1 & 15
\end{pmatrix}
\]

We need to:
1. Construct a \( 2 \times 3 \) matrix \( C \) using only the entries \( 1 \), \( -1 \), and \( 0 \), such that \( CA = I_2 \), where \( I_2 \) is the \( 2 \times 2 \) identity matrix.
   
   That is:
   \[
   CA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
   \]

2. Compute \( AC \) and note that \( AC \neq I_3 \), where \( I_3 \) is the \( 3 \times 3 \) identity matrix.

### Step 1: Trial and Error to Find Matrix \( C \)

We need to find matrix \( C \) that satisfies:

\[
CA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]

Let:

\[
C = \begin{pmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \end{pmatrix}
\]

The product \( CA \) will give us a \( 2 \times 2 \) matrix. Multiplying \( C \) by \( A \):

\[
CA = \begin{pmatrix}
c_{11} & c_{12} & c_{13} \\
c_{21} & c_{22} & c_{23}
\end{pmatrix} \times \begin{pmatrix} 1 & 7 \\ 1 & 8 \\ 1 & 15 \end{pmatrix}
\]

Carrying out this multiplication, we get:

\[
CA = \begin{pmatrix}
c_{11}(1) + c_{12}(1) + c_{13}(1) & c_{11}(7) + c_{12}(8) + c_{13}(15) \\
c_{21}(1) + c_{22}(1) + c_{23}(1) & c_{21}(7) + c_{22}(8) + c_{23}(15)
\end{pmatrix}
\]

To match the identity matrix, we set up the following system of equations:

1. \( c_{11} + c_{12} + c_{13} = 1 \)
2. \( 7c_{11} + 8c_{12} + 15c_{13} = 0 \)
3. \( c_{21} + c_{22} + c_{23} = 0 \)
4. \( 7c_{21} + 8c_{22} + 15c_{23} = 1 \)

By trial and error, we find the solution for \( C \) as:

\[
C = \begin{pmatrix}
8 & -15 & 7 \\
-1 & 1 & 0
\end{pmatrix}
\]

### Step 2: Compute \( AC \)

Now, we need to compute \( AC \), where \( A \) is \( 3 \times 2 \) and \( C \) is \( 2 \times 3 \). The result will be a \( 3 \times 3 \) matrix.

\[
AC = \begin{pmatrix} 1 & 7 \\ 1 & 8 \\ 1 & 15 \end{pmatrix} \times \begin{pmatrix} 8 & -15 & 7 \\ -1 & 1 & 0 \end{pmatrix}
\]

Performing the matrix multiplication:

\[
AC = \begin{pmatrix}
1(8) + 7(-1) & 1(-15) + 7(1) & 1(7) + 7(0) \\
1(8) + 8(-1) & 1(-15) + 8(1) & 1(7) + 8(0) \\
1(8) + 15(-1) & 1(-15) + 15(1) & 1(7) + 15(0)
\end{pmatrix}
\]

\[
AC = \begin{pmatrix}
8 - 7 & -15 + 7 & 7 + 0 \\
8 - 8 & -15 + 8 & 7 + 0 \\
8 - 15 & -15 + 15 & 7 + 0
\end{pmatrix}
\]

Simplifying:

\[
AC = \begin{pmatrix}
1 & -8 & 7 \\
0 & -7 & 7 \\
-7 & 0 & 7
\end{pmatrix}
\]

Clearly, \( AC \neq I_3 \), since the identity matrix \( I_3 \) would be:

\[
I_3 = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]

This shows that \( AC \neq I_3 \).

---

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### 5 Related Questions:
1. What is the significance of the matrix identity \( CA = I_2 \)?
2. How does matrix multiplication work in general?
3. What are other methods besides trial and error to find matrix \( C \)?
4. Can matrix \( AC \) ever equal the identity matrix \( I_3 \) for a \( 3 \times 3 \) identity matrix?
5. How does changing the entries of \( C \) affect the result of \( AC \)?

### Tip:
When performing matrix multiplication, make sure that the dimensions of the matrices are compatible for the operation (i.e., the number of columns of the first matrix must match the number of rows of the second).

Let A equal a 3 by 2 table: 1st row, 1st column 1, 2nd column 7; 2nd row, 1st column 1, 2nd column 8; 3rd row, 1st column 1, 2nd column 15. Construct a 2 by 3 matrix C (by trial and error) using only 1, -1, and 0 as entries, such that CA equals the 2 by 2 identity matrix. Compute AC and note that AC does not equal the 3 by 3 identity matrix.

Explore how to construct a 2x3 matrix C using trial and error such that CA equals the 2x2 identity matrix, followed by computing AC. This problem covers essential linear algebra concepts, including matrix multiplication and identity matrices, perfect for undergraduate students.

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