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12

First note that, $$\begin{vmatrix} a & a^2 & a^3+1 \\ b & b^2 & b^3+1 \\ c & c^2 & c^3+1 \\ \end{vmatrix} = \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \\ \end{vmatrix} + \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \\ \end{vmatrix}=0$$ Then, ...


8

A non - trivial counterexample with $\det A = 0$ and $\det D \neq 0$: Let $$A =\begin{bmatrix} 0 & 0 \\ 1 & 2 \end{bmatrix}$$ and $$D = \begin{bmatrix} -1 & 0 \\ 0 & -2\end{bmatrix}.$$ Then, $$A+D =\begin{bmatrix} -1 & 0 \\1 & 0 \end{bmatrix},$$ with $\det(A+D) = 0$.


7

I guess it's easier to go for a reduction formula. I proceed along the generalization mentioned in comment: Call the determinant $M = M_n(x_1,x_2,\cdots, x_n)$ $$\displaystyle \begin{align}M &= \left|\begin{matrix}\dfrac{1}{x_1+x_1} & \cdots & \dfrac{1}{x_1+x_n}\\ \dfrac{1}{x_2+x_1} & \cdots & \dfrac{1}{x_2+x_n}\\ \cdot & \cdot ...


6

The hard part is to compute the determinants and factor them nicely: $ \det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} = (ab^2+bc^2+ca^2)- (b^2c+c^2a+a^2b) = (a-b)(b-c)(c-a) $ and $\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}$ $= (ab^2(1+c^3)+bc^2(1+a^3)+ca^2(1+b^3)) - ...


6

You are looking at a special case of the Matrix determinant Lemma. From the Wikipedia page, the proof for the case $A = I$ follows from the equality $$ \begin{bmatrix} I & 0 \\ v^T & 1 \end{bmatrix} \begin{bmatrix} I + uv^T & u \\ 0 &1 \end{bmatrix} \begin{bmatrix} I & 0 \\ -v^T & 1 \end{bmatrix} = \begin{bmatrix} I & u \\ ...


5

Let's rewrite the system as $$\mathbf r \times \mathbf P\mathbf{r} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\x&y&z\\z&x&y\end{vmatrix} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \equiv \mathbf f$$ where $\mathbf{r} = (x,y,z)^\top$ and $$ \mathbf P = \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 ...


5

Given $x^2-yz = 1, \quad y^2-xz = 2, \quad z^2-xy = 3$, we can sum all of these to get $$(x-y)^2+(y-z)^2+(z-x)^2 = 12 \tag{1}$$ OTOH, subtracting gives $(y^2-x^2)+z(y-x)=1 \implies (x+y+z)(y-x) = 1$ and similarly $(x+y+z)(z-y) = 1$, so we must have $y-x = z - y = a$, say. Using this in $(1)$, $$a^2+a^2+4a^2=12 \implies a = \pm \sqrt2$$ So we have $y = x ...


5

The main thing you need to use is: For any matrix $A$, $Ax = \lambda x$ is true if and only if $(A-I)x = (\lambda - 1)x$


5

We consider a general case. Let $$ |A|=\left|\begin{array}{}{\dfrac1{a_1+b_1}\cdots\dfrac1{a_1+b_n}\\ \vdots\hspace{20 mm} \vdots \\\dfrac1{a_n+b_1}\cdots\dfrac1{a_n+b_n}} \end{array}\right| $$ In your case, just set $a_i=i,b_j=j$. By multiplying $i^{th}$ row with $\prod\limits_{j=1}^{n}(a_i+b_j)$ each, we have $$ ...


5

Define $C:=AB^{-1}$, which is possible since $B$ is invertible. Then $$CB=(AB^{-1})B=A(B^{-1}B)=AI=A$$ and $$\det(C)=\det(A)\det(B^{-1})=\frac{\det(A)}{\det(B)}=1$$


4

For part $(a)$ you are almost done. If $\mathrm{det}(A)=0$ and $$A=\begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{bmatrix},$$ then $(a_1,a_2,a_3)=c_1(a_4, a_5, a_6)+c_2(a_7,a_8,a_9)$ for some $c_1,c_2 \in \mathbb{R}$. Note that $c_1, c_2 \neq 0$ by assumption since $a_i \neq a_j$ whenever $i \neq j$. By ...


4

Hint: This is a rank one update of a diagonal matrix. In particular, we have $$ M = \pmatrix{ 1-n\\ &2-n\\ && \ddots\\ &&& 0 } + n \cdot xx^T $$ where $x = (1,\dots,1)^T$. We can find the determinant of this matrix using Sylvester's determinant theorem (more specifically, using the matrix determinant lemma). Full Solution: Let ...


4

There does exist a notion of exterior powers of modules over arbitrary (commutative) rings, and they can be defined almost exactly the same way as for vector spaces. Here is a reference that goes through the details. Using this definition, the proof that determinants are multiplicative over a field generalizes straightforwardly to arbitrary rings. ...


3

The resultant of $x^2-yz-1$ and $y^2-xz-2$ with respect to $z$ is $x^3-y^3-x+2y$. The resultant of $x^2 - yz - 1$ and $z^2 - xy - 3$ with respect to $z$ is $x^4-x y^3-2 x^2-3 y^2+1$. The resultant of $x^3-y^3-x+2y$ and $x^4-x y^3-2 x^2-3 y^2+1$ with respect to $x$ is $-y^4(18 y^2-1)$. So either $y = 0$ or $y = \pm 1/\sqrt{18}$. With $y=0$ we get $x^2 - ...


3

Let $z$ denote the vector consisting of only $1$s. We can write $A$ as $$ A = zz^T - I $$ In particular, $$ Ax = zz^Tx - Ix = \langle z,x \rangle z - x $$ Now, consider two cases: first, suppose that $x=z$. We then have $$ Ax = Az = \langle z,z \rangle z - z = (\langle z,z \rangle - 1)z $$ Next, suppose $x$ is perpendicular to $z$. We then have $$ Ax ...


3

All OK except $\det(2B)$. Remember that if you multiply a row/column of a matrix by $2$, the determinant is multiplied by $2$. If you multiply the entire matrix by $2$, then you are multiplying each row (or equivalently, each column) by $2$. Since there are $3$ rows/columns in the matrix, you are multiplying by $2$ three times, so $$\det(2B) = 2^3 \det(B) = ...


3

The determinant of a $3\times 3$ matrix is just the area of the parallelepiped spanned by it's column vectors. If you think about the problem geometrically, I think it's a bit easier to see why your answer should be correct.


3

Lemma. If positive numbers $a_1,a_2,a_3,a_4$ have at least two different values, then they can be put into a $2\times 2$ matrix so that the determinant is nonzero. Proof: Suppose that $0 < a_1\leq a_2\leq a_3\leq a_4$, where not all the $a_i$'s are the same. Then $a_1a_2 < a_3 a_4$, so the matrix $\begin{bmatrix}a_3 & a_1 \\ a_2 & ...


3

The answer is"No". Look at a diagonal matrix $D = \text{diag} (d_1, d_2, \ldots, d_n)$ with $d_i \ne 0$, $1 \le i \le n$; then in fact $\det D \ne 0$. Form $A$ from $D$ by replacing at least one $d_i$ with $-d_i$ and at least one $d_j$, $j \ne i$, by $0$ยท Then $\det A = 0$ and $D + A$ is a diagonal matrix with (at least) $(D + A)_{ii} = 0$, whence $\det ...


3

The block matrix (let us denote by $M$) can be expressed as the Kronecker product of matrices $A$ and $I$ (the fixed size identity matrix, of dimension $n$) as follows:- $$M=A\otimes I$$ where $A$ is the $3\times3$ matrix:- $$A=\left[\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ ...


2

Here $c,d$ are rows and not vectors. There is a formula that is valid for any $b$ (the last equality in kubek's proof is false). According to [Horn, Johnson "Matrix Analysis" (2013), 0.8.5.10 ], one has $\det(M)=b\det(A)+c. adj(A).d^T$, where $adj(A)$ is the classical adjoint of $A$.


2

why not just use the Laplace expansion of a determinant ? u can read how exactely it is done in here: https://en.wikipedia.org/wiki/Laplace_expansion in case you don't understand it, or you prefere a step by step tutorial on how to do it u can watch this. it's only 5 minutes: https://www.youtube.com/watch?v=6fmgMTJBWtY or if u are truly lazy, u can just ...


2

$$\det \begin{pmatrix} (b+c)^2 & a^2 &a^2\\ b^2 & (c+a)^2 & b^2\\ c^2&c^2&(a+b)^2 \end{pmatrix}$$ $$\det \begin{pmatrix} (b+c)^2-a^2 & 0 & a^2\\ 0 & (c+a)^2-b^2 & b^2\\ c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2 \end{pmatrix}$$ By $C_1-C_3$, $C_2-C_3$ $$\det \begin{pmatrix} (b+c+a)(b+c-a) & 0 & a^2\\ 0 & ...


2

Take the column $k$. If we denote $x^k=z$ then it looks like $$ \left[\matrix{z-1\\ z^2-1\\z^3-1\\ \vdots\\ z^n-1}\right]=(z-1)\left[\matrix{1\\ z+1\\z^2+z+1\\ \vdots\\ z^{n-1}+\ldots+z+1}\right]. $$ Now use the top row of ones to eliminate all other ones and $z$ to eliminate all other $z$ and so on $$ (z-1)\left[\matrix{1\\ z+1\\z^2+z+1\\ \vdots\\ ...


2

Call your matrix $A$. If $X=(a,b,c,\ldots,f)^T$ is a column vector then the polynomial $$ p(y) = ay+by^2+cy^3+\cdots+fy^n-(a+c+d+\cdots+f) $$ applied to $x, x^2, x^3, \ldots, x^n$ will produce the elements of $AX$. If $AX=0$, then accordingly $x, x^2, x^3, \ldots, x^n$ are all roots of $p$. We can also see directly that $1$ is a root of $p$. If the powers ...


2

Here is the difference between the two concepts: The Jacobian is an $m\times n$ matrix and it consists of first-order derivatives of all the variables of a given function $f$. The Jacobian matrix is an $m\times n$ matrix that gives the best linear approximation of $f$ near the point $x\in \mathbb{R}^n$. If we have a square matrix, then ...


2

$\det(-3A) = (-3)^5 \det(A)$. Because in general: if $A$ is an $n \times n$ matrix, we have $\det(xA) = x^n \det(A)$. Thus, $$\det(A) = \frac{-4}{3^5}$$ Your $\det(B)$ is correct. Finally, $\det(AB) = \det(A) \times \det(B)$.


2

Thought Samrat Mukhopadhyay"s answer is of course in principle correct, it provides no justification for the stated multiplicities of the eigenvalues. In what follows, I have tried to explain, amongst other things, just how these multiplicities arise. If $u = 0 \;\; \text{or} \;\; v = 0 \tag{1}$ we have $uv^T = 0 \tag{2}$ and $v^Tu = 0; \tag{3}$ then ...


2

This is the correct formulation for the $2\times 2$ case but you still need to prove that these formulas work. So, to finish you must show that \begin{array}{crcrcrcrcr} a_1\dfrac{D_1}{D} &+& b_1\dfrac{D_2}{D} &=& c_1 \\ a_2\dfrac{D_1}{D} &+& b_2\dfrac{D_2}{D} &=& c_2 \end{array} The first of these two equations is proved by ...



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