My Math Forum Master's Entrance Paper

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April 18th, 2017, 06:41 PM   #1
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Master's Entrance Paper

Hello all!
I have attached an image of some questions for which I just want to know whether they are correct or not. I will be telling my logic in solving them later. Let me know whether they are correct or not. Thank you so much. It will be helpful for me to prepare for my examination.
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 April 21st, 2017, 09:18 PM #2 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 236 Thanks: 2 Please someone help me with this!!!
 April 22nd, 2017, 10:47 AM #3 Senior Member   Joined: Aug 2012 Posts: 2,355 Thanks: 737 Perhaps try re-shooting with better lighting and making sure it's in focus. Current version's not readable. Thanks from topsquark and Joppy
April 22nd, 2017, 09:04 PM   #4
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Thank you Maschke. I have provided another image. Can you check once.
Many thanks í ½í¸Š
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April 22nd, 2017, 09:13 PM   #5
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Math Focus: Yet to find out.
Right side up.
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April 22nd, 2017, 09:14 PM   #6
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Quote:
 Originally Posted by Joppy Right side up.
Thank you ðŸ˜Š

 April 22nd, 2017, 10:04 PM #7 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,833 Thanks: 649 Math Focus: Yet to find out. And typed, if it helps (let me know if any typos). 2015 Paper 2) $\text{Let A be an$n \times n$nonzero matrix where A is not an identity matrix. If$A^2 = A$, then the eigenvalues of A are given by 0 and 1.}$ 3) $\text{Let A be a$7 \times 5$matrix over$\mathbb{R}$having at least 5 linearly independent rows: Then the dimension of the null space of A is at least 2.}$ 4) $\text{The dimension of the vector subspace$w$of$M_2(c)$given by }$ $w = \left\lbrace \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in c, \ a + b = c, \ b + c = d, \ c + a = d \right\rbrace$ $\text{is equal to 2}$. $\mathbf{a}: 4$ $\mathbf{b}: 3$ $\mathbf{c}: 2$ $\mathbf{d}: 1$ 5) $\text{If$|a - b| = |c - d|$then$a = b + c - d$.}$ $\mathbf{a}: \ a = b + c - d$ $\mathbf{b}: \ a = b - c + d$ $\mathbf{c}: \text{$a = b + c - d$and,$a = b - c + d$}$ $\mathbf{d}: \text{$a = b + c - d$or,$a = b - c + d$}$ 6) $\text{The set of all real numbers$x$for which there is some positive real number$y$such that$x < y$is equal to$\lbrace 0 \rbrace$}$ $\mathbf{a}: \mathbb{R}$ $\mathbf{b}: \text{The set of all negative real numbers}$ $\mathbf{c}: \lbrace 0 \rbrace$ $\mathbf{d}: \text{The empty set}$ 8.) $\text{If$\bar{\gamma} = x \bar{i} + y \bar{j} + z \bar{k}$,$\gamma = \sqrt{x^2 + y^2 + z^2}$and$n \in \mathbb{N}$then$\nabla \gamma^n$is equal to$n\gamma^{(n - 2)} \bar{\gamma}$}$ $\mathbf{a}: n\gamma^{(n - 1)} \bar{\gamma}$ $\mathbf{b}: (n - 1)\gamma^{(n - 2)} \bar{\gamma}$ $\mathbf{c}: n\gamma^{n - 2} \bar{\gamma}$ $\mathbf{d}: (n - 1)\gamma^{n} \bar{\gamma}$ P.S., Anyone know how to circle letters without the tikz package? Thanks from Maschke
April 22nd, 2017, 10:09 PM   #8
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Quote:
 Originally Posted by Joppy And typed, if it helps (let me know if any typos). 2015 Paper 2) $\text{Let A be an$n \times n$nonzero matrix where A is not an identity matrix. If$A^2 = A$, then the eigenvalues of A are given by 0 and 1.}$ 3) $\text{Let A be a$7 \times 5$matrix over$\mathbb{R}$having at least 5 linearly independent rows: Then the dimension of the null space of A is at least 2.}$ 4) $\text{The dimension of the vector subspace$w$of$M_2(c)$given by }$ $w = \left\lbrace \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in c, \ a + b = c, \ b + c = d, \ c + a = d \right\rbrace$ $\text{is equal to 2}$. $\mathbf{a}: 4$ $\mathbf{b}: 3$ $\mathbf{c}: 2$ $\mathbf{d}: 1$ 5) $\text{If$|a - b| = |c - d|$then$a = b + c - d$.}$ $\mathbf{a}: \ a = b + c - d$ $\mathbf{b}: \ a = b - c + d$ $\mathbf{c}: \text{$a = b + c - d$and,$a = b - c + d$}$ $\mathbf{d}: \text{$a = b + c - d$or,$a = b - c + d$}$ 6) $\text{The set of all real numbers$x$for which there is some positive real number$y$such that$x < y$is equal to$\lbrace 0 \rbrace$}$ $\mathbf{a}: \mathbb{R}$ $\mathbf{b}: \text{The set of all negative real numbers}$ $\mathbf{c}: \lbrace 0 \rbrace$ $\mathbf{d}: \text{The empty set}$ 8.) $\text{If$\bar{\gamma} = x \bar{i} + y \bar{j} + z \bar{k}$,$\gamma = \sqrt{x^2 + y^2 + z^2}$and$n \in \mathbb{N}$then$\nabla \gamma^n$is equal to$n\gamma^{(n - 2)} \bar{\gamma}$}$ $\mathbf{a}: n\gamma^{(n - 1)} \bar{\gamma}$ $\mathbf{b}: (n - 1)\gamma^{(n - 2)} \bar{\gamma}$ $\mathbf{c}: n\gamma^{n - 2} \bar{\gamma}$ $\mathbf{d}: (n - 1)\gamma^{n} \bar{\gamma}$ P.S., Anyone know how to circle letters without the tikz package?

Thank you so much for your work!

 April 25th, 2017, 05:25 AM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 (2) Let A be a non-zero matrix that is not the identity matrix. Then if $\displaystyle A^2= A$, the eigenvalues are 0 and 1. Let $\displaystyle \lambda$ be an eigenvalue of A with corresponding eigenvector v. Then $\displaystyle A^2v= A(Av)=A(\lambda v)= \lambda Av= \lambda (\lambda v)= \lambda^2 v$. That is, the eigenvalues of $\displaystyle A^2$ are the squares of the eigenvalues of A with the same eigenvectors. This is true for all matrices. But since $\displaystyle A^2= A$, we must have $\displaystyle A^2v= Av$ so that $\displaystyle \lambda^2v= \lambda v$, $\displaystyle \lambda^2v- \lambda v= (\lambda^2- \lambda)v= 0$. For v non-zero, we have $\displaystyle \lambda^2- \lambda= 0$ which has only 0 and 1 as solutions. (I am surprised that they specifically exempt the 0 and identity matrices since they obviously also satisfy this theorem.) 4, 5, 6, and 8 are strange- they give an answer then list possible answers! Was the correct answer added? (4) asks you to show that the dimension of the subspace of 2 by 2 matrices, $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ with the conditions that a+ b= c, b+ c= d, and c+ a= d is 2 but then gives a list of possible answers, one of which (c) is 2! In any case, to prove that, would use the given equations to solve for as many of a, b, c, and d, in terms of the others as possible. For example, from a + b = c and b + c = d, I can subtract the second from the first to get a - c = c - d so that a = 2c - d. Then a + b = 2c- d + b = c so b = d - c. The third equation, c + a = d can be written c+ a = c+ 2c - d = 3c - d = d so that 2d = 3c and d = (3/2)c. From that b = d - c = (3/2)c - c = (1/2)c and a = 2c - d = 2c- (3/2)c = (1/2)c. a, b, and d can all be written in terms of c so the matrix can be written as $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}(1/2)c & (1/2)c \\ c & (3/2)c\end{pmatrix}= c\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ 1 & \frac{3}{2}\end{pmatrix}$ That is, every matrix in this set is a number, c, times that matrix so the subspace has dimension 1, not 2. (More simply, the set of 2 by 2 matrices has dimension 4 and putting 3 independent conditions on it reduces the dimension to 4 - 3 = 1. The calculations I did were really to show that the conditions were independent.) Last edited by skipjack; April 25th, 2017 at 11:24 PM.
April 26th, 2017, 12:15 AM   #10
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Quote:
 Originally Posted by Country Boy (2) Let A be a non-zero matrix that is not the identity matrix. Then if $\displaystyle A^2= A$, the eigenvalues are 0 and 1. Let $\displaystyle \lambda$ be an eigenvalue of A with corresponding eigenvector v. Then $\displaystyle A^2v= A(Av)=A(\lambda v)= \lambda Av= \lambda (\lambda v)= \lambda^2 v$. That is, the eigenvalues of $\displaystyle A^2$ are the squares of the eigenvalues of A with the same eigenvectors. This is true for all matrices. But since $\displaystyle A^2= A$, we must have $\displaystyle A^2v= Av$ so that $\displaystyle \lambda^2v= \lambda v$, $\displaystyle \lambda^2v- \lambda v= (\lambda^2- \lambda)v= 0$. For v non-zero, we have $\displaystyle \lambda^2- \lambda= 0$ which has only 0 and 1 as solutions. (I am surprised that they specifically exempt the 0 and identity matrices since they obviously also satisfy this theorem.) 4, 5, 6, and 8 are strange- they give an answer then list possible answers! Was the correct answer added? (4) asks you to show that the dimension of the subspace of 2 by 2 matrices, $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ with the conditions that a+ b= c, b+ c= d, and c+ a= d is 2 but then gives a list of possible answers, one of which (c) is 2! In any case, to prove that, would use the given equations to solve for as many of a, b, c, and d, in terms of the others as possible. For example, from a + b = c and b + c = d, I can subtract the second from the first to get a - c = c - d so that a = 2c - d. Then a + b = 2c- d + b = c so b = d - c. The third equation, c + a = d can be written c+ a = c+ 2c - d = 3c - d = d so that 2d = 3c and d = (3/2)c. From that b = d - c = (3/2)c - c = (1/2)c and a = 2c - d = 2c- (3/2)c = (1/2)c. a, b, and d can all be written in terms of c so the matrix can be written as $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}(1/2)c & (1/2)c \\ c & (3/2)c\end{pmatrix}= c\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ 1 & \frac{3}{2}\end{pmatrix}$ That is, every matrix in this set is a number, c, times that matrix so the subspace has dimension 1, not 2. (More simply, the set of 2 by 2 matrices has dimension 4 and putting 3 independent conditions on it reduces the dimension to 4 - 3 = 1. The calculations I did were really to show that the conditions were independent.)
Thanks for the analysis

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