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If $B = A^T A^{-1}$, we have $\det(B) = 1$. Moreover, $1$ is an eigenvalue of $B$ because $B - I = (A^T - A) A^{-1}$, $A^T - A$ is antisymmetric, and any antisymmetric matrix in odd dimensions must be singular. If the eigenvalues of $B$ are $1, \lambda, 1/\lambda$, then $\text{tr}(B) = 1 + \lambda + 1/\lambda$ and $\det(I+B) = 2 (1 + \lambda) (1 + ... 8 Start from $$\begin{vmatrix}x & z & y \\ z & y & x \\ y & x &z\end{vmatrix}.$$ Multiply the colums by$y, x, \frac{xy}z$; multiply the rows by$\frac{z}{x},\frac z y, 1$. You have now multiplied the determinant with$xyz$. Edit: More symmetric: multiply the colums$yz, zx, xy$; multiply the rows by$\frac1x,\frac1y, \frac1z$. 6 If$c_i$is$i$th column of your second determinant, do$c_n= c_n-c_{n-1}$,$c_{n-1}=c_{n-1}-c_{n-2}$, ...,$c_2=c_2-c_1$to get: $$\left|\begin{array}{ccccccc} 1-x & x & 0 & 0 & \cdots & 0 & 0\\ 0 & 1-x & x & 0 & \cdots & 0 & 0\\ 0 & 0 & 1-x & x & \cdots & 0 & 0\\ \vdots & \vdots ... 5 Call the determinant D(m,k) (I'll consider n as fixed). If k>m the first row is zero, so D(m,k)=0 in this case. Now assume k\le m. Applying the identity$$ \binom{a}{b}=\frac{a}{b}\binom{a-1}{b-1} $$to each entry in the determinant, and taking out factors from rows and columns, we have$$ ... 5 Notice that$A^2=A$so$A^k=A,\; k\ge1$and then by the binomial formula we get $$(I+A)^{99}=\sum_{k=0}^{99}\binom{99}{k}A^kI^{99-k}=I+A\sum_{k=1}^{99}\binom{99}{k}=I+(2^{99}-1)A$$ 5 No. Notice that for a matrix$A\in\mathcal M_n(\Bbb F)$and$\lambda\in\Bbb F$we have $$\det(\lambda A)=\lambda^n\det(A)$$ 5 Hint $$\dfrac{1}{2}[(a+b)^2+(b+c)^2+(a+c)^2]=a^2+b^2+c^2+ab+ac+bc\le 0$$ so we have $$a=-b,b=-c,c=-a\Longrightarrow a=b=c=0$$ 4 This technique is called Cramer's rule. 4 hint: Use AM-GM inequality $$d_{1}+2d_{2}+4d_{3}+8d_{4}\ge 4\sqrt[4]{2^2\cdot 2^3\cdot 2 d_{1}d_{2}d_{3}d_{4}}=8\sqrt{2}\cdot\sqrt[4]{\det{(A)}}$$ so $$\det{(A)}\le 4$$ and since $$\tan{x}+\cot{x}\ge 2$$ so $$f(x)\le log_{2}{4}=2$$ 4 The formula$A \cdot (B \times C) = \textrm{Det}(A,B,C)$shows this the cross product can be thought of as the transpose of the linear map$\textrm{Det}(\cdot,B,C)$. Using the notation of riemannian geometry (hodge star, sharps, and flats) another way to say this is that$A \times B=\star(A^\flat \wedge B^\flat)^\sharp. This is the connection between the ... 3 Hint By adding all the rows to first you get $$\begin{vmatrix}x&1&0&0&0\\4&x&2&0&0\\0&3&x&3&0\\0&0&2&x&4\\0&0&0&1&x\end{vmatrix} = ... 3 If \det(A)=\pm 1 then A is invertible and$$A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)where \operatorname{adj}(A) is the adjugate matrix which is the transpose of the cofactor matrix so its entries are integers. The unicity of B comes from the unicity of the inverse of an invertible matrix. 3 Hints : The diagonal elements have to be 0 If the lower left triangle is given, the upper right triangle is determined by a_{ij}=-a_{ji} 3 This image was taken from Dune-Project.org Notice that each piece in fact has the same volume as the tetrahedron described in anorton's comment above (by connecting tips of the three vectors) by the same reason that the area of a triangle relies solely on base times height, a three dimensional object relies on the area of the base times the shortest ... 3 Hint: You know that \det(AB)=\det(A)\det(B) for all n\times n matrices A,B. So make a judicious choice for B... 3 Exploiting the commutativity of the determinant operation with multiplication is probably the easiest way. That said, here is another approach. The determinant of a square matrix is equal to the product of its eigenvalues. Now note that for an invertible matrix \mathbf A, \lambda\in\mathbb R is an eigenvalue of \mathbf A is and only if 1/\lambda is ... 3 The Laplace expansion is used to calculate the given determinant: we expand along the first row. 3 I don't think whoever wrote that section of the Wikipedia article meant that the first formula in your question follows from the second one. The author probably only meant to say that when A and B do not commute, we can still evaluate the determainant of the block matrix by \det(A+B)\det(A-B). At any rate, the two formulae hold because \begin{align} ... 3\#\operatorname{GL}(n,\mathbb{F}_2)=(2^n-1)(2^n-2)(2^n-4)\cdot\ldots\cdot(2^n-2^{n-1})$$hence:$$\nu_2\left(\#\operatorname{GL}(n,\mathbb{F}_2)\right) = 1+2+\ldots+(n-1)=\frac{n(n-1)}{2}$$and the number of 0-1 matrices with odd determinant (i.e. invertible) is a multiple of 2^{\binom{n}{2}}. 3 Hint 1. Try to find eigenvalues. Hint 2. If your matrix is anti-symmetric then what can you say about eigenvalues? 2 You may find the determinant by using elementary row operations. Subtract, simultaneously, the first row from the second row, the second row from the third row, etc., we can reduce A to$$ \left[\begin{array}{cccccc} x&a&a&\cdots&\cdots&\cdots&a\\ -x-a&x-a&0&0&\dots&\cdots&0\\ ... 2 The only result you need here is that\det(A)=\det(A^T)$and that$A^T B^T = (BA)^T$. 2 Let$X$be the$n \times n$-matrix obtained as follows: in the first row, the$i$-th entry equals$\mu(i)$where$\mu$denotes the Möbius function, and in the$j$-th row ($j>1$), the$i$-th entry is$1$if$i$divides$j$and$0$otherwise. For example, for$n=8we obtain $$X=\left( \begin{array}{cccccccc} 1 & -1 & -1 & 0 & -1 & 1 ... 2 HINT: Sarrus' scheme gives it immediately. 2 Once you got$$(p-a)(r(q-b)+b(r-c))+a(q-b)(r-c)=0$$Then divide this whole thing by (p-a)(q-b)(r-c) to get$$\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0.Now \begin{align*} \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} & =\frac{p}{p-a}+\frac{q}{q-b}-\frac{b}{q-b}-\frac{a}{p-a}\\ & ... 2 That is twice the signed area of the triangle \triangle PQR. See this for some more details. By "signed" I mean that the vertices must be taken in the counterclockwise direction to get the area. If taken clockwise, you get the negative of the area. 2 Consider the eigenvalues of x \cdot x^T + I x \cdot x^T v + v = \lambda v x \cdot x^T v = \lambda v - v v must be parallel to x or (\lambda = 1). wlog v = x ||x|| ^2 x= (\lambda - 1) x \lambda = ||x||^2 + 1 You can use the fact that the determinant is the product of eigenvalues (there should be n of them) 2 To expand on darij grinberg's comment, let X=A-I_n = \begin{bmatrix} x_1^2 &x_1x_2 & ... & x_1x_n \\ x_2x_1&x_2^2 &... & x_2x_n\\ ...& ... & ... &... \\ x_nx_1& x_nx_2 &... & x_n^2 \end{bmatrix}=(x_ix_j)_{1\leq i,j\leq n} $$Then all the lines of X are multiples of (x_1,x_2,\ldots,x_n) ; so ... 2 Here is a method without using (at least explicitely) the notion of eigenvalue. Call f(x_1,\dots,x_n) the wanted determinant. View the last column as$$\pmatrix{0\\\vdots\\ 0\\1 }+x_n\pmatrix{x_1\\\vdots\\ x_{n-1} \\x_n }.$$This gives by linearity with respect to the last column,$$f(x_1,\dots,x_n)=f(x_1,\dots,x_{n-1})+x_n\det\begin{bmatrix} 1+x_1^2 ... 2 HINT: Every complex matrix can be put in upper-triangular form. Try it for any diagonal matrixA\$ first.