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9

The matrix is the product \begin{eqnarray} \begin{pmatrix}\cos a_1& \sin a_1& 0\\ \cos a_2&\sin a_2& 0\\ \cos a_3&\sin a_3 &0\end{pmatrix}\begin{pmatrix}\cos b_1& \cos b_2&\cos b_3\\ \sin b_1&\sin b_2&\sin b_3\\ 0&0&0\end{pmatrix} \end{eqnarray} and thus the determinant is 0.


6

Hint: Replace $\det(A+B)$ on the right by $\det(A^T+B^T)$ (I trust you understand why that is allowed). Now use the product formula for determinants. You'll get a problem with signs, but notice that $A^TB-B^TA$ is skew symmetric!


5

Hint: Add all other rows to the first row. What row you'll obtain after that?


5

The matrix $A=\left[\begin{smallmatrix} 1 & 1\\ 0 & 1 \end{smallmatrix}\right]$, and its transpose $A^T$, have only one eigenvalue, namely $1$. However, the eigenvectors of $A$ are of the form $\left[\begin{smallmatrix} 0\\ a \end{smallmatrix}\right]$, whereas the eigenvectors of $A^T$ are of the form $\left[\begin{smallmatrix} a\\ 0 ...


4

$2A$ is the matrix obtained by multiply each entry of $A$ by $2$. Thus $|2A| = 2^3|A| = 8|A| = 8\times (-7) = -56$. Since $|2A|\times |(2A)^{-1}| = |I| = 1\Rightarrow |(2A)^{-1}| = \dfrac{1}{|2A|} = -\dfrac{1}{56}$


4

As we're looking for the determinant we may as well think about the transpose $$ T^T=\begin{bmatrix} 1& 1& 1& \cdots& 1& 0\\ a_{2}& a_{1}& a_{n}& \cdots& a_{4}& \varepsilon^{n-2}\\ a_{3}& a_{2}& a_{1}& \cdots& ...


4

Since $n$ is even, the characteristic polynomial of $T$ is given by $$ p(x) = \det(T - xI) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 $$ and $$ \lim_{x \to \infty}p(x) = \lim_{x \to -\infty}p(x) = \infty $$ We note that $p(0) = \det(T)$. If $p(0) = 0$, then $0$ is an eigenvalue. If $p(0) < 0$, then the intermediate value theorem tells us that $p$ ...


3

The simple mnemonic is this: When ever you are computing the Jacobian, elements of the same coordinate system must go on the same side of the equality sign What you did wrong: writing $x = y + u$ etc. When computing $\partial x / \partial u$ in computing the Jacobian, you are asked to compute "the partial derivative of $x$ relative to varying $u$ while ...


3

First we can subtract from each row the following row $$D= \begin{vmatrix} 1 & 2 & 3 & \ldots & n-2 & n-1 & n\\ 2 & 3 & 4 & \ldots & n-1 & n & n\\ 3 & 4 & 5 & \ldots & n & n & n \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ n & n & n ...


3

In general, determinant is not defined for "matrices" with infinitely many rows and columns. It's not surprising that you're running into paradoxes if you try to use identities that work for finite matrices on these "matrices". There are some rather special cases where you can define a determinant for an infinite matrix. For example, see this Math Overflow ...


2

When you multiply any row of a matrix by 2, the determinant gets multiplied by 2. Since A has 3 rows, $$|(2A)^{-1}|=\frac{1}{2^3*(-7)} = -\frac{1}{56}$$


2

The thing is that, unless the matrix is symmetric, $A$ and $A^T$ represent different systems of equations. Try with a simple example. When calculating the eigenvectors you solve the equations $(A-\lambda I)v =0$ and $(A^T-\lambda I)w=0$, which again are different systems.


2

Let me write $a=7x+42$ and $b=x-21$. Then by subtracting the first columns from all other columns we obtain $$ \det(A) = \det\pmatrix{ a & b & b & b & b \\ b & a & b & b & b \\ b & b & a & b & b \\ b & b & b & a & b \\ b & b & b & b & a } = \det\pmatrix{ a & b-a & b-a ...


2

HINT: Calculate the eigenvalues of $A=(x-21)O + (6x+63)I$, where $O$ is a matrix full of $1$s and $I$ is the unit matrix...


2

Since there isn't an answer with everyone's favourite counterexample yet, here's one: consider the matrix $$ A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$ Then $A$ has only one eigenvector, namely $(1,0)$, with eigenvalue $0$. Meanwhile, $$ A^T=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$ also has only one eigenvector, namely $(0,1)$. ...


2

The eigenvalue of the $n \times n$ matrix in which every entry is $1$ is $n$ (of multiplicity $1$) and $1$ (of multiplicity $n-1$). Proof here So, for such a matrix $A$, we have $\det(A - \lambda I_n) = 0$, and $\lambda = n$.


2

$6det(2(A^T)^2B^{-1}A^{-1})$ $=6det(2(A^T)^2)det(B^{-1})det(A^{-1})=$ $=6 \cdot 2^6 det(A)^2 det(B)^{-1} det(A)^{-1}=$ $= 6 \cdot 2^6 det(A) det(B)^{-1} = $ $= 6 \cdot 2^6 det(A) det(-\frac{2}{7}A)^{-1} = $ $= 6 \cdot 2^6 det(A) \cdot (-\frac{2}{7})^{-6} \cdot det(A)^{-1} =$ $= 6 \cdot 2^6 \cdot \frac{7^6}{2^6}=$ $= 6 \cdot 7^6=$ $=705894$ So your ...


2

If you add $3$ times the first row with $2$ times the second row, you end up with the third row. Thus the rows are linearly dependent, hence the matrix is singular for all values of $x$, so that the determinant is $0$ for all values of $x$ as well.


2

You know that $$\left((I-A)(I+A)^{-1} \right)^T(I-A)(I+A)^{-1} =I \Rightarrow \\ ((I+A)^{-1})^{T} (I-A)^T(I-A)(I+A)^{-1} =I \Rightarrow \\ (I-A)^T(I-A) =(I+A)^{T} (I+A)\Rightarrow \\ (I-A^T)(I-A) =(I+A^T)(I+A)\Rightarrow \\ I-A-A^T+AA^T =I+A+A^T+AA^T\Rightarrow \\ A+A^T=0$$


2

Assume that $A\in M_{nk,nk}$. Then let $(P_n)$ be the sequences of matrices defined as follows: $P_1=J_n,P_2={J_n}^2-I,P_q=J_nP_{q-1}-P_{q-2}$. Then $\det(A)=\det(P_k)$. EDIT. The eigenvalues of $J_n$ are the $(1+2\cos(\pi \dfrac{p}{n+1})),p=1\cdots n$; thus, if $Q$ is any polynomial, then we can calculate approximations of the eigenvalues of $Q(J_n)$, ...


2

Because this is the Laplace expansion for the determinant of a matrix with two identical rows.


2

If $A$ is your matrix, then $B=A+3I$ is the matrix all of whose entries are all $2$s. It is clear that the vector $(1,\dots,1)$ is an eigenvector of $B$ of eigenvalue $2n$. On the other hand, the rank of $B$ is obviously $1$, since the dimension of the vector space spanned by its rows is $1$: this means that $0$ is an eigenvalue of $B$ of multiplicity $n-1$. ...


2

Hint: the matrix $M = e e^T$ (where $e$ is a column vector consisting of $n$ $1$'s) satisfies $M^2 = n M$, so its eigenvalues are ...


2

Expanding all the terms isn't so bad in this example. ROWS-TRANSFORMATIONS $$A=\begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ n^2+4n+4&n^2+6n+9&n^2+8n+16 \end{pmatrix}\sim \begin{pmatrix} n^2&n^2+2n+1&n^2+4n+4\\ n^2+2n+1&n^2+4n+4&n^2+6n+9\\ 2n+3&2n+5&2n+7 \end{pmatrix}\sim\begin{pmatrix} ...


2

Note that $B\in\Bbb R^{8\times 5}$, so $B$ has $5$ columns and the rank of $B$ is at most $5$. ($C$ has $5$ rows so the rank of $C$ is also at most $5$.) Therefore the rank of $ABC$ is at most $5$. But $ABC$ is a $7\times 7$ matrix, so its rank is less than the number of its rows or columns. Therefore the determinant of $ABC$ is zero. Note that we need no ...


2

Yes, the problem is with interpretation. If we want to look at the partial derivative of the logdet with respect to a single element $(i,j)$ of $L$, we do this: $$\frac{\partial \log \det L}{\partial L_{ij}} = \left\langle \frac{d \log\det L}{d L}, E_{ij} \right\rangle = \langle L^{-T}, E_{ij} \rangle = \left(L^{-T}\right)_{ij} = \left(L^{-1}\right)_{ji}$$ ...


2

As you explained, performing a computation with the disk requires you to interpret a bunch of definitions and conventions for a rank-zero module. Probably not a very enlightening exercise. But Lickorish's definition holds for any Seifert surface we choose, so we can avoid using the disk altogether: Let $F$ be a torus with one boundary component, embedded in ...


2

Hint: Use Laplace expansion in the last column.


2

The "holes-digging" method might be interesting to prove this. On one hand, dig a hole at the lower-left corner of $A$, $$A := \begin{bmatrix}I & X \\ -Y^T & 1 \end{bmatrix} = \begin{bmatrix} I & 0 \\ Y^T & 1\end{bmatrix}\begin{bmatrix}I & X \\ 0 & 1 + Y^TX\end{bmatrix}$$ Take determinants on both sides to have $\det(A) = ...


2

Solution (Alternative Way): We were given the transformation $$u=x-y$$ $$v=xy$$ Notice that rewriting the equations in terms of $x$ and $y$ will make things harder (since you cannot solve for $x$ and $y$ explicitly.) However, we invoke the idea that $$J(u,v) J(x,y)=1$$ Thus instead of finding $J(u,v)$, we try to compute $J(x,y)$ since we already have ...



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