Question

Find the product of the following two matrices.
open square bracket,
begin matrix
row 1, column 1, 0
row 1, column 2, minus, 1
row 1, column 3, 0
row 2, column 1, minus, 5
row 2, column 2, minus, 2
row 2, column 3, minus, 5
row 3, column 1, 0
row 3, column 2, 0
row 3, column 3, 4
end matrix
, close square bracket, open square bracket,
begin matrix
row 1, column 1, 1
row 1, column 2, 0
row 1, column 3, 1
row 2, column 1, minus, 1
row 2, column 2, 0
row 2, column 3, minus, 1
row 3, column 1, 1
row 3, column 2, minus, 1
row 3, column 3, 0
end matrix
, close square bracket
​

0
−5
0
​

−1
−2
0
​

0
−5
4
​

​

​

1
−1
1
​

0
0
−1
​

1
−1
0
​

​

Answer 1

To find the product of the given matrices, we will perform matrix multiplication. Let's denote the first matrix (let's call it \(A\)) and the second matrix (let's call it \(B\)):

\[
A = \begin{bmatrix}
0 & -1 & 0 \\
-5 & -2 & -5 \\
0 & 0 & 4
\end{bmatrix}
\]

\[
B = \begin{bmatrix}
1 & 0 & 1 \\
-1 & 0 & -1 \\
1 & -1 & 0
\end{bmatrix}
\]

### Step 1: Dimension Check

Matrix \(A\) is of size \(3 \times 3\) and matrix \(B\) is also of size \(3 \times 3\). Hence, the resulting product \(C = A \times B\) will also be of size \(3 \times 3\).

### Step 2: Perform Multiplication

The entry in the resulting matrix \(C\) at position \(C[i][j]\) is computed as follows:

\[
C[i][j] = \sum_{k=1}^{3} A[i][k] \cdot B[k][j]
\]

Let's calculate each entry of the resulting matrix \(C\):

#### For \(C[1][1]\):

\[
C[1][1] = A[1][1] \cdot B[1][1] + A[1][2] \cdot B[2][1] + A[1][3] \cdot B[3][1]
\]
\[
= 0 \cdot 1 + (-1) \cdot (-1) + 0 \cdot 1 = 0 + 1 + 0 = 1
\]

#### For \(C[1][2]\):

\[
C[1][2] = A[1][1] \cdot B[1][2] + A[1][2] \cdot B[2][2] + A[1][3] \cdot B[3][2]
\]
\[
= 0 \cdot 0 + (-1) \cdot 0 + 0 \cdot (-1) = 0 + 0 + 0 = 0
\]

#### For \(C[1][3]\):

\[
C[1][3] = A[1][1] \cdot B[1][3] + A[1][2] \cdot B[2][3] + A[1][3] \cdot B[3][3]
\]
\[
= 0 \cdot 1 + (-1) \cdot (-1) + 0 \cdot 0 = 0 + 1 + 0 = 1
\]

#### For \(C[2][1]\):

\[
C[2][1] = A[2][1] \cdot B[1][1] + A[2][2] \cdot B[2][1] + A[2][3] \cdot B[3][1]
\]
\[
= (-5) \cdot 1 + (-2) \cdot (-1) + (-5) \cdot 1 = -5 + 2 - 5 = -8
\]

#### For \(C[2][2]\):

\[
C[2][2] = A[2][1] \cdot B[1][2] + A[2][2] \cdot B[2][2] + A[2][3] \cdot B[3][2]
\]
\[
= (-5) \cdot 0 + (-2) \cdot 0 + (-5) \cdot (-1) = 0 + 0 + 5 = 5
\]

#### For \(C[2][3]\):

\[
C[2][3] = A[2][1] \cdot B[1][3] + A[2][2] \cdot B[2][3] + A[2][3] \cdot B[3][3]
\]
\[
= (-5) \cdot 1 + (-2) \cdot (-1) + (-5) \cdot 0 = -5 + 2 + 0 = -3
\]

#### For \(C[3][1]\):

\[
C[3][1] = A[3][1] \cdot B[1][1] + A[3][2] \cdot B[2][1] + A[3][3] \cdot B[3][1]
\]
\[
= 0 \cdot 1 + 0 \cdot (-1) + 4 \cdot 1 = 0 + 0 + 4 = 4
\]

#### For \(C[3][2]\):

\[
C[3][2] = A[3][1] \cdot B[1][2] + A[3][2] \cdot B[2][2] + A[3][3] \cdot B[3][2]
\]
\[
= 0 \cdot 0 + 0 \cdot 0 + 4 \cdot (-1) = 0 + 0 - 4 = -4
\]

#### For \(C[3][3]\):

\[
C[3][3] = A[3][1] \cdot B[1][3] + A[3][2] \cdot B[2][3] + A[3][3] \cdot B[3][3]
\]
\[
= 0 \cdot 1 + 0 \cdot (-1) + 4 \cdot 0 = 0 + 0 + 0 = 0
\]

### Step 3: Construct the Result Matrix

Combining all of the results, we get the resulting matrix \(C\):

\[
C = \begin{bmatrix}
1 & 0 & 1 \\
-8 & 5 & -3 \\
4 & -4 & 0
\end{bmatrix}
\]

### Final Answer:

The product of the matrices \(A\) and \(B\) is:

\[
\begin{bmatrix}
1 & 0 & 1 \\
-8 & 5 & -3 \\
4 & -4 & 0
\end{bmatrix}
\]