Answer :
To verify that [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex] for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we will take the following steps:
1. Define the matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
2. Find the inverse of matrix [tex]\(A\)[/tex]:
Recall that for a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \][/tex]
The determinant of [tex]\(A\)[/tex] is:
[tex]\[ \text{det}(A) = 3 \cdot 5 - 2 \cdot 7 = 15 - 14 = 1 \][/tex]
Hence, the inverse of [tex]\(A\)[/tex] is:
[tex]\[ A^{-1} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \][/tex]
3. Find the inverse of matrix [tex]\(B\)[/tex]:
Similarly, for matrix [tex]\(B\)[/tex]:
[tex]\[ B = \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
The determinant of [tex]\(B\)[/tex] is:
[tex]\[ \text{det}(B) = (-2) \cdot 9 - 7 \cdot (-3) = -18 + 21 = 3 \][/tex]
Hence, the inverse of [tex]\(B\)[/tex] is:
[tex]\[ B^{-1} = \frac{1}{3} \begin{pmatrix} 9 & -7 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} 3 & -2.33333333 \\ 1 & -0.66666667 \end{pmatrix} \][/tex]
4. Calculate the product [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
Compute each element of the resulting matrix:
[tex]\[ AB = \begin{pmatrix} 3 \cdot (-2) + 2 \cdot (-3) & 3 \cdot 7 + 2 \cdot 9 \\ 7 \cdot (-2) + 5 \cdot (-3) & 7 \cdot 7 + 5 \cdot 9 \end{pmatrix} = \begin{pmatrix} -6 - 6 & 21 + 18 \\ -14 - 15 & 49 + 45 \end{pmatrix} = \begin{pmatrix} -12 & 39 \\ -29 & 94 \end{pmatrix} \][/tex]
5. Find the inverse of matrix [tex]\(AB\)[/tex]:
The determinant of [tex]\(AB\)[/tex] is:
[tex]\[ \text{det}(AB) = (-12) \cdot 94 - 39 \cdot (-29) = -1128 + 1131 = 3 \][/tex]
Hence, the inverse of [tex]\(AB\)[/tex] is:
[tex]\[ (AB)^{-1} = \frac{1}{3} \begin{pmatrix} 94 & -39 \\ 29 & -12 \end{pmatrix} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
6. Compute the product [tex]\(B^{-1}A^{-1}\)[/tex]:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 3 & -2.33333333 \\ 1 & -0.66666667 \end{pmatrix} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \][/tex]
Compute each element of the resulting matrix:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 3 \cdot 5 + (-2.33333333) \cdot (-7) & 3 \cdot (-2) + (-2.33333333) \cdot 3 \\ 1 \cdot 5 + (-0.66666667) \cdot (-7) & 1 \cdot (-2) + (-0.66666667) \cdot 3 \end{pmatrix} \][/tex]
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 15 + 16.33333331 & -6 - 7 \\ 5 + 4.66666669 & -2 - 2 \end{pmatrix} \][/tex]
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
7. Conclusion:
We observe that:
[tex]\[ (AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
This verifies that the relation [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex] holds true for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
1. Define the matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
2. Find the inverse of matrix [tex]\(A\)[/tex]:
Recall that for a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex], the inverse is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \][/tex]
The determinant of [tex]\(A\)[/tex] is:
[tex]\[ \text{det}(A) = 3 \cdot 5 - 2 \cdot 7 = 15 - 14 = 1 \][/tex]
Hence, the inverse of [tex]\(A\)[/tex] is:
[tex]\[ A^{-1} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \][/tex]
3. Find the inverse of matrix [tex]\(B\)[/tex]:
Similarly, for matrix [tex]\(B\)[/tex]:
[tex]\[ B = \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
The determinant of [tex]\(B\)[/tex] is:
[tex]\[ \text{det}(B) = (-2) \cdot 9 - 7 \cdot (-3) = -18 + 21 = 3 \][/tex]
Hence, the inverse of [tex]\(B\)[/tex] is:
[tex]\[ B^{-1} = \frac{1}{3} \begin{pmatrix} 9 & -7 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} 3 & -2.33333333 \\ 1 & -0.66666667 \end{pmatrix} \][/tex]
4. Calculate the product [tex]\(AB\)[/tex]:
[tex]\[ AB = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \begin{pmatrix} -2 & 7 \\ -3 & 9 \end{pmatrix} \][/tex]
Compute each element of the resulting matrix:
[tex]\[ AB = \begin{pmatrix} 3 \cdot (-2) + 2 \cdot (-3) & 3 \cdot 7 + 2 \cdot 9 \\ 7 \cdot (-2) + 5 \cdot (-3) & 7 \cdot 7 + 5 \cdot 9 \end{pmatrix} = \begin{pmatrix} -6 - 6 & 21 + 18 \\ -14 - 15 & 49 + 45 \end{pmatrix} = \begin{pmatrix} -12 & 39 \\ -29 & 94 \end{pmatrix} \][/tex]
5. Find the inverse of matrix [tex]\(AB\)[/tex]:
The determinant of [tex]\(AB\)[/tex] is:
[tex]\[ \text{det}(AB) = (-12) \cdot 94 - 39 \cdot (-29) = -1128 + 1131 = 3 \][/tex]
Hence, the inverse of [tex]\(AB\)[/tex] is:
[tex]\[ (AB)^{-1} = \frac{1}{3} \begin{pmatrix} 94 & -39 \\ 29 & -12 \end{pmatrix} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
6. Compute the product [tex]\(B^{-1}A^{-1}\)[/tex]:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 3 & -2.33333333 \\ 1 & -0.66666667 \end{pmatrix} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \][/tex]
Compute each element of the resulting matrix:
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 3 \cdot 5 + (-2.33333333) \cdot (-7) & 3 \cdot (-2) + (-2.33333333) \cdot 3 \\ 1 \cdot 5 + (-0.66666667) \cdot (-7) & 1 \cdot (-2) + (-0.66666667) \cdot 3 \end{pmatrix} \][/tex]
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 15 + 16.33333331 & -6 - 7 \\ 5 + 4.66666669 & -2 - 2 \end{pmatrix} \][/tex]
[tex]\[ B^{-1}A^{-1} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
7. Conclusion:
We observe that:
[tex]\[ (AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 31.33333333 & -13 \\ 9.66666667 & -4 \end{pmatrix} \][/tex]
This verifies that the relation [tex]\((AB)^{-1} = B^{-1}A^{-1}\)[/tex] holds true for the given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
