An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that

$AB=I$, and

$BA=I$,

where $I$ is the $n\times n$ identity matrix.

If such a matrix $B$ exists, then it is known to be unique and called the inverse matrix of $A$, denoted by $A^{-1}$.

In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition.
So if we know $AB=I$, then we can conclude that $B=A^{-1}$.

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix.

Since $AB=I$, we have
\begin{align*}
\det(A)\det(B)=\det(AB)=\det(I)=1.
\end{align*}
This implies that the determinants $\det(A)$ and $\det(B)$ are not zero.
Hence $A, B$ are invertible matrices: $A^{-1}, B^{-1}$ exist.

Now we compute
\begin{align*}
I&=BB^{-1}=BIB^{-1}\\
&=B(AB)B^{-1} &&\text{since $AB=I$}\\
&=BAI=BA.
\end{align*}
Hence we obtain $BA=I$.
Since $AB=I$ and $BA=I$, we conclude that $B=A^{-1}$.

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1 &x &x \\
x & x & x
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Let
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3 & -12 & 4 \\
-1 &0 &-2 \\
-1 & 5 & -1
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Then find all eigenvalues of $A^5$. If $A$ is invertible, then find all the eigenvalues of $A^{-1}$.
Proof.
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1 &0 &0 \\
2 & 1 & 1
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Prove the following statements.
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Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies
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Hint.
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Suppose $A$ is a positive definite symmetric $n\times n$ matrix.
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(b) Prove that $A^{-1}$ is symmetric.
(c) Prove that $A^{-1}$ is positive-definite.
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