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To verify that [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex] for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we follow these steps:
### Step 1: Define the matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Let
[tex]\[ A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \][/tex]
and
[tex]\[ B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}. \][/tex]
### Step 2: Calculate the inverse of [tex]\(A\)[/tex]
To find [tex]\(A^{-1}\)[/tex], we use the formula for the inverse of a 2x2 matrix:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \][/tex]
where [tex]\(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex].
For [tex]\(A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 2\)[/tex]
- [tex]\(\text{det}(A) = ad - bc = (1)(2) - (1)(0) = 2\)[/tex].
Thus,
[tex]\[ A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix}. \][/tex]
### Step 3: Calculate the inverse of [tex]\(B\)[/tex]
Similarly, for [tex]\(B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 3\)[/tex]
- [tex]\(\text{det}(B) = ad - bc = (1)(3) - (2)(0) = 3\)[/tex].
Thus,
[tex]\[ B^{-1} = \frac{1}{3} \begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix}. \][/tex]
### Step 4: Calculate the product of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Now, we compute [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + 1 \cdot 0) & (1 \cdot 2 + 1 \cdot 3) \\ (0 \cdot 1 + 2 \cdot 0) & (0 \cdot 2 + 2 \cdot 3) \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 0 & 6 \end{pmatrix}. \][/tex]
### Step 5: Calculate the inverse of [tex]\(AB\)[/tex]
For [tex]\(AB = \begin{pmatrix} 1 & 5 \\ 0 & 6 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 6\)[/tex]
- [tex]\(\text{det}(AB) = ad - bc = (1)(6) - (5)(0) = 6\)[/tex].
Thus,
[tex]\[ (AB)^{-1} = \frac{1}{6} \begin{pmatrix} 6 & -5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
### Step 6: Calculate [tex]\( B^{-1}A^{-1} \)[/tex]
Now, we compute [tex]\(B^{-1}A^{-1}\)[/tex]:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + -\frac{2}{3} \cdot 0) & (1 \cdot -\frac{1}{2} + -\frac{2}{3} \frac{1}{2}) \\ (0 \cdot 1 + \frac{1}{3} \cdot 0) & (0 \cdot -\frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2}) \end{pmatrix} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
### Step 7: Compare [tex]\((AB)^{-1}\)[/tex] and [tex]\(B^{-1}A^{-1}\)[/tex]
From our calculations, we see that:
[tex]\[ (AB)^{-1} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}, \][/tex]
and
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
Since [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex], we have verified that [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex].
### Step 1: Define the matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Let
[tex]\[ A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \][/tex]
and
[tex]\[ B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}. \][/tex]
### Step 2: Calculate the inverse of [tex]\(A\)[/tex]
To find [tex]\(A^{-1}\)[/tex], we use the formula for the inverse of a 2x2 matrix:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \][/tex]
where [tex]\(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex].
For [tex]\(A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 2\)[/tex]
- [tex]\(\text{det}(A) = ad - bc = (1)(2) - (1)(0) = 2\)[/tex].
Thus,
[tex]\[ A^{-1} = \frac{1}{2} \begin{pmatrix} 2 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix}. \][/tex]
### Step 3: Calculate the inverse of [tex]\(B\)[/tex]
Similarly, for [tex]\(B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 2\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 3\)[/tex]
- [tex]\(\text{det}(B) = ad - bc = (1)(3) - (2)(0) = 3\)[/tex].
Thus,
[tex]\[ B^{-1} = \frac{1}{3} \begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix}. \][/tex]
### Step 4: Calculate the product of [tex]\(A\)[/tex] and [tex]\(B\)[/tex]
Now, we compute [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + 1 \cdot 0) & (1 \cdot 2 + 1 \cdot 3) \\ (0 \cdot 1 + 2 \cdot 0) & (0 \cdot 2 + 2 \cdot 3) \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 0 & 6 \end{pmatrix}. \][/tex]
### Step 5: Calculate the inverse of [tex]\(AB\)[/tex]
For [tex]\(AB = \begin{pmatrix} 1 & 5 \\ 0 & 6 \end{pmatrix}\)[/tex]:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = 0\)[/tex]
- [tex]\(d = 6\)[/tex]
- [tex]\(\text{det}(AB) = ad - bc = (1)(6) - (5)(0) = 6\)[/tex].
Thus,
[tex]\[ (AB)^{-1} = \frac{1}{6} \begin{pmatrix} 6 & -5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
### Step 6: Calculate [tex]\( B^{-1}A^{-1} \)[/tex]
Now, we compute [tex]\(B^{-1}A^{-1}\)[/tex]:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix} \begin{pmatrix} 1 & -\frac{1}{2} \\ 0 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} (1 \cdot 1 + -\frac{2}{3} \cdot 0) & (1 \cdot -\frac{1}{2} + -\frac{2}{3} \frac{1}{2}) \\ (0 \cdot 1 + \frac{1}{3} \cdot 0) & (0 \cdot -\frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2}) \end{pmatrix} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
### Step 7: Compare [tex]\((AB)^{-1}\)[/tex] and [tex]\(B^{-1}A^{-1}\)[/tex]
From our calculations, we see that:
[tex]\[ (AB)^{-1} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}, \][/tex]
and
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 1 & -\frac{5}{6} \\ 0 & \frac{1}{6} \end{pmatrix}. \][/tex]
Since [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex], we have verified that [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex].
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