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We add these equalities in this manner $$a+b+43+c+d+37=55+b+c+42+d+e$$ now we cancel we find $$\require{cancel}a+\cancel{b}+43+\cancel{c}+\cancel{d}+37=55+\cancel{b}+\cancel{c}+42+\cancel{d}+e$$ hence $$a+80=e+97\iff e=a-17$$ so each time you take a value of $a$ we find a value of $e$. Can you now answer your son?

6

The following people take part in a school trip: 55 boys and girls; 43 girls and fathers; 42 fathers and mothers and 37 mothers and teachers. How many teachers took part in the school trips? Assuming the classes are mutually exclusive (i.e. no teachers are also parents) If you assume there were zero teachers, you get 37 mothers, 5 fathers, 38 girls and ...

5

We have: $$b=55-a$$ $$c=43-b=43-(55-a)=a-12$$ $$d=42-c=42-(a-12)=54-a$$ $$e=37-d=37-(54-a)=a-17$$ For all these numbers to be non-negative we therefore need $$17\le a\le 54$$with the inequalities strict if we require all the numbers to be positive. It was just possible that the constraints that all the numbers are non-negative (or positive) integers would ...

4

Let $$A=\begin{bmatrix}I_{n-2} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & 0 & -1 \\ \mathbf{0} & 1 & 0\end{bmatrix},$$ then $$A^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & -1\end{bmatrix}.$$ Now make $B$ a matrix ...

3

Denoting the matrix by $A$, you will find that multiplying yields $$A^2=A^3$$ (assuming my program is working, that is). This implies $A^2=A^3=A^4=\cdots$. If $A$ were diagonalizable, we could write $A=S\Lambda S^{-1}$ so that $$S\Lambda^2 S^{-1}=S\Lambda^3 S^{-1}=\cdots$$ $$\Lambda^2=\Lambda^3=\cdots$$ which means all the eigenvalues must be $0$ or $1$. ...

2

Some methods: If the row or columns sums are all the same (say equal to $r$), then $(1,1,\ldots,1)$ is an eigenvector associated with $\lambda = r$. Also, Geršgorin's theorem tells us some good estimates, especially if the off diagonal entries are small. If your matrices have all positive entries, then the largest eigenvalue is positive (and real) and is ...

2

If $A$ is invertible, then $B=A^{-1}C$, so $B$ is unique. Otherwise, $B$ can certainly fail to be unique. For example, let $A=\left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right)$ and let $C=\left(\begin{array}{cc}1 & 1 \\ 0 & 0\end{array}\right)$. Then $B$ can be any matrix of the form $B=\left(\begin{array}{cc}* & * \\ 1 & ... 1 In principle, this could be almost anything and $$B=\begin{cases}219&\text{if }A=0\\ 224&\text{if }A=50\\ 231&\text{if }A=100\\ 246&\text{if }A=200\\ 255&\text{if }A=255\\ 42&\text{otherwise}\\ \end{cases}$$ would fit. Of course, one expects something "smoother", and there are still many options, for example a polynomial of degree ... 1 There are various methods of finding the matrix of$T$with respect to a basis$B$. One of them is to work out one column at a time: if the basis is$\{{\bf b}_1,{\bf b}_2\}$then you find the first column by calculating$T({\bf b}_1)$and finding its coordinate vector with respect to$B$, and similarly for the other column(s). Here you have ... 1 The key idea that this exercise might be trying to teach you is that eigenvalues are independent of any basis you choose. So let's choose a basis, write down the matrix of$T$in that basis and then compute the eigenvalues using the usual method. Why not choose the standard basis$b_1 = \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}, b_2 = ...

1

Let $C$ be a positive symmetric $n\times n$ matrix, and let $|\cdot|$ be the standard euclidean norm on $\Bbb R^n$. For all $X\in\Bbb R^n$, $$|CX|\leq\lambda_1(C)|X|$$ with equality iff $X$ is eigenvector of $C$ with eigenvalue $\lambda_1(C)$, so that, again, for $C$ positive symmetric, $$\lambda_1(C)=\max_{|X|\leq 1}|CX|=\|C\|$$ where $\|\cdot\|$ is the ...

1

$AB$ is similar to an SPD matrix, hence they have same eigenvalues: $$AB\sim B^{1/2}(AB)B^{-1/2}\quad\Rightarrow\quad\lambda_{1}(AB)=\lambda_1(B^{1/2}AB^{1/2}).$$ Using this and the variational characterisation of $\lambda_1$, we have $$\begin{split} \lambda_1(AB)&=\max_x\frac{x^TB^{1/2}AB^{1/2}x}{x^Tx} =\max_y\frac{y^TAy}{y^TB^{-1}y} ... 1 If you already know that you can compete a basis of a subspace to a basis for the whole space then you are practically done. Hint: Note that \alpha is a basis for V (this should give you W_{1}+W_{2}=V, why ?) and that the w_{i} are linearly independent of the u_{i} (this should show that W_{1}\cap W_{2}=\{0\}, why ?) Note: The way I see it, ... 1 Hints: \;A\; is diagonalizable and we can choose a basis \;B\; of \;\Bbb R^2\; of its eigenvectors so that we'll have$$A(=[A]_B)=\begin{pmatrix}\frac23&0\\0&\frac95\end{pmatrix}$$With respect to this same basis, we have for any \;\binom xy\in\Bbb R^2\; :$$Ax=\binom{\frac23x}{\frac95y}$$Now, just do some basic algebra with this. For ... 1 Taking the differential gives$$df = hh^Tdh + h(dh)^Th + dh (h^Th).$$df must be linear in dh, and usually for the derivative of vector-valued functions with respect to a vector it is possible through rearrangement to get that linear map in matrix form. For instance here, since (dh)^Th is a scalar, as is h^Th, the above can be rewritten$$df = hh^T ...

1

Your matrix is nothing but $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} \begin{bmatrix} x_1& x_2& x_3& x_4\end{bmatrix} + \begin{bmatrix} y_1\\ y_2\\ y_3\\ y_4\end{bmatrix} \begin{bmatrix} y_1& y_2& y_3& y_4\end{bmatrix}$$ Hence, the rank of the matrix is $2$, in general. If $y_k = ax_k$ for all $k \in \{1,2,3,4\}$, where $a$ ...

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