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3

Hint: Think about inverting Matrix A and rewriting it as $X=A^{-1}B$. The inverse of a 2 by 2 matrix is simple to generate by hand. Your approach is the right one but you will have to introduce a 2 by 2 matrix for $X$ instead of a vector.

3

It's not possible to multiply $2\times 2$ matrix with $2\times 1$ matrix and give a $2\times 2$ matrix. But if in your question $B$ is a $2\times 1$ matrix, you can solve this system of equation by elimination: $$\begin{cases}2x+3y=a\\4x+5y=b\end{cases}$$

3

Let $$X=\left[\begin{matrix} a &b\\ c &d\end{matrix}\right].$$ Expanding the matrix product, you need to solve $$\begin{cases}2a+3c=3\\4a+5c=5\\2b+3d=4\\4b+5d=6\end{cases}$$ Hint: $$\begin{cases}2a+3c=3\\4a+5c=5\end{cases}$$ $$\begin{cases}2b+3d=4\\4b+5d=6\end{cases}$$ The real, efficient method is by solving the system (by Gaussian ...

2

The case $n=1$ is immediate. For the case $n\ge 2$, by user1551's comment, $${\rm adj}({\rm adj}{(A)})=(\det A)^{n-2}A.$$ Also, there is a fact that ${\rm adj}(kA)=k^{n-1}{\rm adj}(A)$ for any scalar $k\in\mathbb{C}$. Thus we have \begin{align} {\rm adj}((\det A)^{-\frac{n-2}{n-1}}{\rm adj}{(A)}) &=(\det A)^{-(n-2)}{\rm adj}({\rm adj}{(A)})\\ &=(\det ...

1

There is no solution over $\mathbb R$ if $n \ge 3$ is odd and $\det(A) < 0$. $\det(\text{adj}(B))= \det(\det(B) B^{-1}) = \det(B)^{n-1}$, which can't be negative in this case.

1

From Hadamard's determinant bound one can deduce that the absolute value of the determinant is at most $54$. This is done as follows. Let $M$ be an $n\times n$ matrix whose elements lie in the interval $[0,a]$, for some positive number $a$. Then $$\det M=\left(\frac{a}{2}\right)^n\det\left[\begin{array}{c|ccc}1 & 1 & 1 & 1\\ \hline 0 & ... 1 You should recall the formula$$ \chi_A(X) \overset{def}= \det(X\, \mathrm{id}_n - A) = \sum_{i=0}^n [(-1)^k \mathrm{tr}(\Lambda^k A)] X^{n-k}, $$which, if you have never seen it proved, I have written down here (Your question holds for a general commutative unital ring R, but if you only care about a vector space over a field or the real numbers, feel ... 1 Your answer should be given by scaling * rotation * translation. But your rotation matrix is wrong. From the figure, you need to rotate the object by -\pi/3, so your rotation matrix is$$\begin{pmatrix}0.5 && \sqrt 3 /2&&0\\-\sqrt 3 /2&& 1/2 &&0\\0&&0&&1\end{pmatrix}

1

In general, to find the derivative $Df(A)(H)$ you need to write $f(A+H)-f(A)=L(H)+O(\Vert H\Vert^\alpha)$, where $L(H)$ is linear on $H$, and $\alpha>1$. Let's do this with $g(A)=A^{-1}$: For all $H$ sufficiently small, $A-H$ in invertible, and ...

1

We consider the equations (1) $A^2+B^2=AB$, (2) $A^2+B^2=2AB$. A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$. Equ (1) has not the second property: indeed $A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}}{2}&-1\end{pmatrix}$ is a ...

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