5. Let $$ A=\left[\begin{array}{ll}1 & 1 \ 0 & 2\end{array} ight] $$ and $$ B=\left[\begin{array}{ll}1 & 2 \ 0 & 3\end{array} ight] $$, verify that $$ (A B)^{-1}=B^{-1} A^{-1} $$

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5. Let $$ A=\left[\begin{array}{ll}1 & 1 \ 0 & 2\end{array}ight] $$ and $$ B=\left[\begin{array}{ll}1 & 2 \ 0 & 3\end{array}ight] $$, verify that $$ (A B)^{-1}=B^{-1} A^{-1} $$

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To verify that $$ (AB)^{-1} = B^{-1} A^{-1} $$, we will first compute the product $$ AB $$, then find its inverse $$ (AB)^{-1} $$, and finally compute $$ B^{-1} $$ and $$ A^{-1} $$ and their product $$ B^{-1} A^{-1} $$.

### Step 1: Compute $$ AB $$

Given:
$$
A = \begin{bmatrix} 1 & 1 \ 0 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}
$$

Now, we compute the product $$ AB $$:
$$
AB = \begin{bmatrix} 1 & 1 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix}
$$

Calculating the entries:
- First row, first column: $$ 1 \cdot 1 + 1 \cdot 0 = 1 $$
- First row, second column: $$ 1 \cdot 2 + 1 \cdot 3 = 2 + 3 = 5 $$
- Second row, first column: $$ 0 \cdot 1 + 2 \cdot 0 = 0 $$
- Second row, second column: $$ 0 \cdot 2 + 2 \cdot 3 = 0 + 6 = 6 $$

Thus,
$$
AB = \begin{bmatrix} 1 & 5 \ 0 & 6 \end{bmatrix}
$$

### Step 2: Compute $$ (AB)^{-1} $$

To find the inverse of $$ AB $$, we use the formula for the inverse of a $$ 2 \times 2 $$ matrix:
$$
\text{If } M = \begin{bmatrix} a & b \ c & d \end{bmatrix}, \text{ then } M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}
$$

For our matrix $$ AB = \begin{bmatrix} 1 & 5 \ 0 & 6 \end{bmatrix} $$:
- $$ a = 1, b = 5, c = 0, d = 6 $$
- The determinant $$ ad - bc = 1 \cdot 6 - 5 \cdot 0 = 6 $$

Thus,
$$
(AB)^{-1} = \frac{1}{6} \begin{bmatrix} 6 & -5 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{5}{6} \ 0 & \frac{1}{6} \end{bmatrix}
$$

### Step 3: Compute $$ A^{-1} $$ and $$ B^{-1} $$

**For $$ A $$**:
$$
A^{-1} = \frac{1}{1 \cdot 2 - 1 \cdot 0} \begin{bmatrix} 2 & -1 \ 0 & 1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & -1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} \ 0 & \frac{1}{2} \end{bmatrix}
$$

**For $$ B $$**:
$$
B^{-1} = \frac{1}{1 \cdot 3 - 2 \cdot 0} \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 & -2 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -\frac{2}{3} \ 0 & \frac{1}{3} \end{bmatrix}
$$

### Step 4: Compute $$ B^{-1} A^{-1} $$

Now we compute $$ B^{-1} A^{-1} $$:
$$
B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{2}{3} \ 0 & \frac{1}{3} \end{bmatrix} \begin{bmatrix} 1 & -\frac{1}{2} \ 0 & \frac{1}{2} \end{bmatrix}
$$

Calculating the entries:
- First row, first column: $$ 1 \cdot 1 + (-\frac{2}{3}) \cdot 0 = 1 $$
- First row, second column: $$ 1 \cdot (-\frac{1}{2}) + (-\frac{2}{3}) \cdot \frac{1}{2} = -\frac{1}{2} - \frac{1}{3} = -\frac{3}{6} - \frac{2}{6} = -\frac{5}{6} $$
- Second row, first column: $$ 0 \cdot 1 + \frac{1}{3} \cdot 0 = 0 $$
- Second row, second column: $$ 0 \cdot (-\frac{1}{2}) + \frac{1}{3} \cdot \frac{1}{2} = 0 + \frac{1}{6} = \frac{1}{6} $$

Thus,
$$
B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \ 0 & \frac{1}{6} \end{bmatrix}
$$

### Conclusion

We have:
$$
(AB)^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \ 0 & \frac{1}{6} \end{bmatrix} \quad \text{and} \quad B^{-1} A^{-1} = \begin{bmatrix} 1 & -\frac{5}{6} \ 0 & \frac{1}{6} \end{bmatrix}
$$

Thus, we conclude that:
$$
(AB)^{-1} = B^{-1} A^{-1}
$$
This verifies the statement.
The statement $$ (AB)^{-1} = B^{-1} A^{-1} $$ is verified by calculating the inverses and showing that they are equal.

5. Let \( A=\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right] \) and \( B=\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right] \), verify that \( (A B)^{-1}=B^{-1} A^{-1} \)

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