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11

Determinants are not changed by adding multiples of a column to another. So $$ \begin{vmatrix}1&1&1\\ a&b&c\\ a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}1&0&0\\ a&b-a&c-a\\ a^3&b^3-a^3&c^3-a^3\end{vmatrix}=(b-a)(c^3-a^3)-(b^3-a^3)(c-a) $$ Edit: since in the comments to the other answer you were asking how to ...


9

Note that $$ \det(A^2 + xI) = \det(A + i\sqrt x I)\det(A - i\sqrt x I) $$ Since $A$ is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that $A$ has eigenvalues $\pm i \sqrt x$. Now, if $\lambda$ is an eigenvalue of $A$, then $\lambda^2 + \lambda + x$ is an eigenvalue of $A^2 + A + xI$. Thus, the matrix $A^2 + A + xI$ ...


7

You can compute the determinant of a generic $3\times 3$ matrix using a neat trick, if we have: $$\mathbf{A}=\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$ Then we have the sum of the diagonals (highlighted in green) minus the sum of the antidiagonals (highlighted in red) as follows: $$\det(\mathbf{A}) = ...


7

There's a slick way to discover the inverse by first solving the problem for (formal) power series. $$\begin{eqnarray} \rm (1-ab)^{-1} &=&\rm 1+ ab + a\color{#c00}{ba}b + a\color{#0a0}{baba}b +\,\cdots\\ &=&\rm 1+ a (1\, +\, \color{#c00}{ba}\ \ +\ \ \color{#0a0}{baba}\,\ +\,\cdots)b\\ &=&\rm 1+ a ...


5

If $AB-I$ is invertible with inverse $C$ then look at $-I+BCA$ as inverse of $BA-I$. \begin{align} (BA-I)(-I+BCA) &=-BA+BABCA+I-BCA\\ &= -BA+B(AB-I)CA+I\\ &=-BA+BA+I=I \end{align} and for the other side \begin{align} (-I+BCA)(BA-I)&=-BA+I+BCABA-BCA\\ &=-BA+BC(AB-I)A+I\\ &= -BA+BA+I=I \end{align} For the second part you can do ...


5

Here's a way to prove this statement: You want to prove that $BA-I$ is invertible if $AB-I$ is invertible. This is equivalent to proving that $AB-I$ is not invertible if $BA-I$ is not invertible. Can you relate non-invertibility to some condition on the eigenvalues of $AB$ and $BA$? Hint: $0$ is an eigenvalue of $AB-I$ if and only if $1$ is an eigenvalue ...


5

It is possible for $AB$ to be a square matrix but neither of $A$ nor $B$ is square, for example if $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 2$ matrix. In this case, it is possible that $\det(AB)$ will be zero, but the determinants of $A$ and $B$ are not defined. It may be that this is why your answer was marked as incorrect. However, if it was ...


5

Let $A$ and $B$ be two square matrices of equal size then$$\det(AB)=\det(A)\cdot\det(B)$$ so $$\det(A+B)^2=\det((A+B)\cdot(A+B))=\det(A+B)\cdot\det(A+B)=(\det(A+B))^2$$


5

Hints: $A^4 + 3A^3 + 2A^2 = A^2(A^2 + 3A + 2I) = A^2(A + I)(A + 2I)$ $\det XY = (\det X)(\det Y)$


4

By cofactor expansion along the fourth column, if you take the determinant $$\begin{vmatrix} 1&1&1&1\\ a&b&c&t\\ a^2&b^2&c^2&t^2\\ a^3&b^3&c^3&t^3 \end{vmatrix}$$ and find the coefficient of $t^2$, it will be the negative of the determinant you're after. This makes the problem a matter of understanding the ...


3

Call the matrix $M$. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix: \begin{align*} |M| &= -\begin{vmatrix} \sin x & \cos x & \sin 2x & \cos 2x \\ \cos x & ...


3

solve the recurrence relations $D_n = D_{n-1} - D_{n-2}$ with the initial condition $D_1 = 1 \mbox{ and} D_2 = 0.$ try $D_n = \lambda^n.$ the indicial equation is $\lambda^2 - \lambda + 1 = 0$ whose roots are $\lambda = {1 \pm i\sqrt 3 \over 2}.$ sso $D_n = k (\cos(n\pi/3 + \phi).$ requiring $D_2 = 0$ gives $\phi = -\pi/6$ and $D_! = 1$ shows $k = ...


3

You made a mistake in the first calculation. $\lambda I - A$ means $$\left[\begin{array}{cc}\lambda - a_{11} & -a_{12}\\-a_{21} & \lambda - a_{22}\end{array}\right];$$ the $\lambda$ doesn't appear in every term, only the diagonal.


2

A general linear algebra result says that for $S\succeq 0$, $R\succeq 0$ such that $R-S\succeq 0$, we may infer that $|S|\leq |R|$. Apply this with $$ S\equiv C^TC-C^TB(B^TB)^{-1}B^TC\succeq 0,\quad R\equiv C^TC\succeq0,\quad R-S\succeq0. $$


2

That's not quite right. You generally have the right idea though. Here's a hint: $$|\lambda I-PBP^{-1}| = |P(\lambda I - B)P^{-1}|.$$


2

Consider the zero matrix. It is upper triangular, the determinant is zero.


2

$$\large V=|[2\vec b\times\vec c\quad3\vec c\times\vec a\quad4\vec a\times\vec b]|=\frac92\\\large 24[\vec a\quad\vec b\quad\vec c]^2=\frac92\implies |[\vec a\quad\vec b\quad\vec c]^2|=\frac3{16}$$ So: ...


2

Incomplete thoughts: Thinking of the elements as written in an $n \times m$ grid, send it to the element $\alpha = \bigotimes_{j=1}^m (v_{1j} \wedge \cdots \wedge v_{nj}) \otimes \bigotimes_{i=1}^n (w_{i1} \wedge \cdots \wedge w_{im})$. (so, putting the $v$'s together in columns and the $w$'s in rows.) To see that this is well-defined, it suffices to ...


2

You've factored the characteristic polynomial of $J_1$, so you know the eigenvalues: $0, 0, -\zeta, N \alpha - \beta - \rho$. $-\zeta < 0$, but $N \alpha - \beta - \rho$ could go either way. The criterion for the disease to "take off", i.e. for a small number of infectives to start an epidemic, is $N \alpha - \beta - \rho > 0$ (the zero eigenvalues ...


2

My guess is that your answer to 8 is incorrect. This has little to do with your grasp of the material and everything to do with the ambiguous wording of the question. I think that by "for all matrices $A$", they mean for arbitrary $n \times m$ matrices (that is, we are no-longer looking just at $n \times n$ matrices, only for the context of this question). ...


2

By the row operations $l_2\leftarrow l_2-xl_3$ and $l_4\leftarrow l_4-l_3$ and developing along the first column we get $$\Delta=\det\begin{bmatrix}x&1&2\\1-x^2 &1-x^2&0\\0&1-x&x-1\end{bmatrix}$$ Now $c_2\leftarrow c_2+c_3$ and we develop along the third column we get $$\Delta=(x-1)(x(1-x^2)-3(1-x^2)=-(x-1)^2(x-3)(x+1)$$


2

Multiplying rows and columns of $B$ by $-1$ as necessary, which at most changes the sign of $\det(B)$, we can assume $$B=\pmatrix{1&1&1&1\\-1&\pm1&\pm1&\pm1\\-1&\pm1&\pm1&\pm1\\-1&\pm1&\pm1&\pm1}$$ Adding the first row to each of the others gives ...


2

Running from the beginning, we have that: $$\chi_{\mathbf{A}}=\begin{vmatrix}3-\lambda & 1 & 0 \\ 1 & 3 - \lambda & 0 \\ 0 & 0 & 1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^{2}-6\lambda+8)$$ Thus we have that: $$\lambda=\{1,2,4\}$$ As you found. Now we can find the eigenvectors for each eigenvalue by solving $(\mathbf{A}-\lambda ...


2

Let $P(a,b,c)$ be the determinant in question. Then the general properties of determinants imply that: $P$ is a homogeneous polynomial of degree $4$. $P$ is alternating: that is, $P(a,b,c)=-P(b,a,c)=-P(a,c,b)=-P(c,b,a)$. In particular, $P$ vanishes whenever any two of $a,b,c$ are equal, and so it must have the linear homogeneous polynomials $(b-a)$, ...


2

Let $\lambda_{1,2}$ be the eigenvalues of $A$. Then $\lambda_{1,2}^2$ are the eigenvalues of $A^2$, and hence $-\lambda_{1,2}^2$ are the eigenvalues of $-A^2$ As $\det(xI-(-A^2))=0$, it follows that $x$ is an eigenvalue of $-A^2$. Therefore $x=-\lambda_j^2$ for $j=1$ or $j=2$. Without loss of generality $x=-\lambda_1^2$. This implies that $\lambda_1$ is ...


1

Let me try to do this without mentioning complex numbers or square roots. Given that $\det(A^2 + xI) = 0$, there must be some nonzero vector $v\in\ker(A^2+xI)$. It cannot be an eigenvector, as its eigenvalue would then have to be a real root of $X^2+x$ which does not exist (since $x>0$); therefore $v$ and $Av$ are linearly independent. Now $\ker(A^2+xI)$ ...


1

What happens if you type your matrix J2 in terms as $\{\{-\alpha N, 0, \zeta, 0\},\{\alpha N, -\beta-\rho, 0 , 0\},\cdots\}\}$


1

Remember cofactor expansion: $$\det \left[ {\begin{matrix} {\color{red}0} & {\color{red}0} & \ldots & {\color{red}0} & {\color{red}1}\\ 0 & 0 & \ldots & 1 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 1 & \ldots & 0 & 0\\ 1 & 0 & \ldots & 0 & 0\\ ...


1

Here's another solution. Your matrix $D_n$ is equal to Ones(n,n)+Diag($a_1,a_2,\ldots,a_n$). Well, I immediately wondered what theorems there are out there for taking the determinant of the sum of matrices. The first result I found was this: http://en.wikipedia.org/wiki/Matrix_determinant_lemma To apply this lemma, the Matrix-Determinant Lemma, you need ...


1

It seems that you can proceed farther as follows. Multiplying by $a_i^{-1}$ the $i$-th row and adding it to the last, you will kill “1” placed at $(n,i)$-cell. Then you can calculate the determinant as a product of diagonal cells of the obtained matrix. The cases when $a_i=0$ should be considered separately.



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