Certainly! Let's work through this step-by-step to verify that [tex]\( A A^{-1} = I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix.
1. Define Matrix [tex]\( A \)[/tex]:
The given matrix [tex]\( A \)[/tex] is:
[tex]\[
A = \begin{pmatrix}
3 & 2 & 1 \\
1 & 1 & 1 \\
5 & 1 & -1
\end{pmatrix}
\][/tex]
2. Find the Inverse of Matrix [tex]\( A \)[/tex]:
The inverse of matrix [tex]\( A \)[/tex], denoted as [tex]\( A^{-1} \)[/tex], is:
[tex]\[
A^{-1} = \begin{pmatrix}
-1.0000000000000007 & 1.5000000000000009 & 0.5000000000000002 \\
3.000000000000002 & -4.000000000000003 & -1.0000000000000007 \\
-2.0000000000000013 & 3.5000000000000018 & 0.5000000000000006
\end{pmatrix}
\][/tex]
3. Multiply [tex]\( A \)[/tex] by [tex]\( A^{-1} \)[/tex]:
To verify the identity, we need to compute the product of [tex]\( A \)[/tex] and its inverse [tex]\( A^{-1} \)[/tex]:
[tex]\[
AA^{-1} = \begin{pmatrix}
3 & 2 & 1 \\
1 & 1 & 1 \\
5 & 1 & -1
\end{pmatrix} \begin{pmatrix}
-1.0000000000000007 & 1.5000000000000009 & 0.5000000000000002 \\
3.000000000000002 & -4.000000000000003 & -1.0000000000000007 \\
-2.0000000000000013 & 3.5000000000000018 & 0.5000000000000006
\end{pmatrix}
\][/tex]
4. Result of the Multiplication:
The result of the multiplication is:
[tex]\[
AA^{-1} = \begin{pmatrix}
1.0000000000000013 & -8.881784197001252e-16 & -1.1102230246251565e-16 \\
4.440892098500626e-16 & 1.0 & 1.1102230246251565e-16 \\
0.0 & 0.0 & 0.9999999999999997
\end{pmatrix}
\][/tex]
5. Interpret the Result:
The resulting matrix is very close to the identity matrix:
[tex]\[
I = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\][/tex]
The computed matrix [tex]\( AA^{-1} \)[/tex] has values extremely close to the identity matrix, with very small numerical errors due to floating-point arithmetic.
6. Conclusion:
We can conclude that [tex]\( AA^{-1} \approx I \)[/tex], which verifies that the inverse of [tex]\( A \)[/tex], denoted [tex]\( A^{-1} \)[/tex], when multiplied by [tex]\( A \)[/tex], results approximately in the identity matrix [tex]\( I \)[/tex]. Hence, the computation confirms that [tex]\( A A^{-1} = I \)[/tex].
