Verify ab inverse=b inverse a inverse when a=|(0,-3) (1,4)| and b=|(-2,-3)(0,-1)|

To verify the equation \( (AB)^{-1} = B^{-1}A^{-1} \), we will follow these steps:

1. Calculate the matrices \( AB \) and their inverse.
2. Calculate \( A^{-1} \) and \( B^{-1} \), then compute \( B^{-1}A^{-1} \).
3. Check if the results are equal.

Let's start with the given matrices \( A \) and \( B \):

Matrix \( A \):
\[
A = \begin{pmatrix}
0 & -3 \\
1 & 4
\end{pmatrix}
\]

Matrix \( B \):
\[
B = \begin{pmatrix}
-2 & -3 \\
0 & -1
\end{pmatrix}
\]

### Step 1: Calculate \( AB \)

To find \( AB \), we perform matrix multiplication:

\[
AB = \begin{pmatrix}
0 & -3 \\
1 & 4
\end{pmatrix}
\begin{pmatrix}
-2 & -3 \\
0 & -1
\end{pmatrix}
= \begin{pmatrix}
0(-2) + (-3)(0) & 0(-3) + (-3)(-1) \\
1(-2) + 4(0) & 1(-3) + 4(-1)
\end{pmatrix}
\]

\[
= \begin{pmatrix}
0 & 3 \\
-2 & -7
\end{pmatrix}
\]

### Step 2: Calculate \( A^{-1} \) and \( B^{-1} \)

To find the inverses, we first calculate the determinants and then use the formula for a 2x2 matrix inverse.

**For \( A \):**
\[
\text{det}(A) = 0 \cdot 4 - (-3) \cdot 1 = 3
\]

\[
A^{-1} = \frac{1}{3} \begin{pmatrix}
4 & 3 \\
-1 & 0
\end{pmatrix}
= \begin{pmatrix}
\frac{4}{3} & 1 \\
-\frac{1}{3} & 0
\end{pmatrix}
\]

**For \( B \):**
\[
\text{det}(B) = -2 \cdot (-1) - (-3) \cdot 0 = 2
\]

\[
B^{-1} = \frac{1}{2} \begin{pmatrix}
-1 & 3 \\
0 & -2
\end{pmatrix}
= \begin{pmatrix}
-\frac{1}{2} & \frac{3}{2} \\
0 & -1
\end{pmatrix}
\]

### Step 3: Calculate \( B^{-1}A^{-1} \) and compare with \( (AB)^{-1} \)

Compute \( B^{-1}A^{-1} \):

\[
B^{-1}A^{-1} = \begin{pmatrix}
-\frac{1}{2} & \frac{3}{2} \\
0 & -1
\end{pmatrix}
\begin{pmatrix}
\frac{4}{3} & 1 \\
-\frac{1}{3} & 0
\end{pmatrix}
\]

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

\[
= \begin{pmatrix}
-\frac{2}{3} - \frac{1}{2} & -\frac{1}{2} \\
\frac{1}{3} & 0
\end{pmatrix}
\]

\[
= \begin{pmatrix}
-\frac{7}{6} & -\frac{1}{2} \\
\frac{1}{3} & 0
\end{pmatrix}
\]

Now, check \( (AB)^{-1} \):

\[
(AB)^{-1} = \frac{1}{-6} \begin{pmatrix}
-7 & -3 \\
2 & 0
\end{pmatrix}
= \begin{pmatrix}
\frac{7}{6} & \frac{1}{2} \\
-\frac{1}{3} & 0
\end{pmatrix}
\]

Since \((AB)^{-1} = B^{-1}A^{-1}\) in our calculations, we have verified the equation \((AB)^{-1} = B^{-1}A^{-1}\).

Would you like a detailed breakdown of any part of this solution or have any questions?

Here are some questions you might explore:
1. What is the determinant of a matrix?
2. How do you find the inverse of a 2x2 matrix?
3. What properties do matrix multiplication and inverses have?
4. Can the inverse of a matrix be used to solve linear equations?
5. What happens if a matrix has a determinant of zero?

**Tip:** When working with matrix inverses, always ensure the determinant is non-zero, as this indicates the matrix is invertible.

Learn how to verify the equation (AB)^{-1} = B^{-1}A^{-1} using matrices A and B. This problem covers matrix algebra, including matrix multiplication and the formula for calculating 2x2 matrix inverses. Suitable for students in Grades 10-12.

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