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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 $\det(A+D) = 0$.

8

Consider for trivial counterexample $A=-D$

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 $0$· 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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