How to Find the Determinant of the $3\times 3$ Matrix
Problem 138
Find the determinant of the matix
\[A=\begin{bmatrix}
100 & 101 & 102 \\
101 &102 &103 \\
102 & 103 & 104
\end{bmatrix}.\]
Find the determinant of the matix
\[A=\begin{bmatrix}
100 & 101 & 102 \\
101 &102 &103 \\
102 & 103 & 104
\end{bmatrix}.\]
Let $V$ be an $n$-dimensional vector space over a field $K$.
Suppose that $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ are linearly independent vectors in $V$.
Are the following vectors linearly independent?
\[\mathbf{v}_1+\mathbf{v}_2, \quad \mathbf{v}_2+\mathbf{v}_3, \quad \dots, \quad \mathbf{v}_{k-1}+\mathbf{v}_k, \quad \mathbf{v}_k+\mathbf{v}_1.\]
If it is linearly dependent, give a non-trivial linear combination of these vectors summing up to the zero vector.
Add to solve laterCalculate the determinants of the following $n\times n$ matrices.
\[A=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 &1 \\
1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & 1 & 1
\end{bmatrix}\]
The entries of $A$ is $1$ at the diagonal entries, entries below the diagonal, and $(1, n)$-entry.
The other entries are zero.
\[B=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 & -1 \\
-1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & -1 & 1
\end{bmatrix}.\]
The entries of $B$ is $1$ at the diagonal entries.
The entries below the diagonal and $(1,n)$-entry are $-1$.
The other entries are zero.
Let
\[ A=\begin{bmatrix}
2 & 0 & 10 \\
0 &7+x &-3 \\
0 & 4 & x
\end{bmatrix}.\]
Find all values of $x$ such that $A$ is invertible.
(Stanford University Linear Algebra Exam)
Add to solve laterFind all eigenvalues of the following $n \times n$ matrix.
\[
A=\begin{bmatrix}
0 & 0 & \cdots & 0 &1 \\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 &0\\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0&\cdots & 1& 0 \\
\end{bmatrix}
\]
Let $A$ and $B$ be $n\times n$ matrices.
Suppose that these matrices have a common eigenvector $\mathbf{x}$.
Show that $\det(AB-BA)=0$.
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Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that
(1) $$\det(A)=\prod_{i=1}^n \lambda_i$$
(2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$
Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.
Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
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Show that if $A$ and $B$ are similar matrices, then they have the same eigenvalues and their algebraic multiplicities are the same.
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