# Thread: Matrix help + binomial

1. ## Matrix help + binomial

Q1) Given that A = $\displaystyle \left(\begin{array}{cc}1&-1\\0&3\end{array}\right)$ and B = $\displaystyle \left(\begin{array}{cc}1&3\\3&4\end{array}\right)$, find the matrix $\displaystyle X$ such that $\displaystyle AX = B + 2I$, where $\displaystyle I$ is a 2 X 2 identity matrix

and

Q2) The coefficient of X in the expansion of $\displaystyle (2 + \frac{x}{3})^n$ is 3 times the coefficient of $\displaystyle X^2$. find the value of n.

i somehow worked my way through and got r = -2. is there something wrong?

Q3) The equation of a circle is X^2 + Y^2 - 10X - 4Y + 25 = 0
Find the set of values of K if the circle touches the line Y = K.
I subbed in the value of K into the equation but i have 2 unknowns. so how?

Hope someone could help me with this! thanks! (:

2. Hello, teddybear67!

1) Given that: .$\displaystyle A \:=\: \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix}\:\text{ and }\:B \:=\:\begin{pmatrix}1&3\\3&4\end{pmatrix}$,

find the matrix $\displaystyle X$ such that $\displaystyle AX \:=\: B + 2I$, where $\displaystyle I$ is the 2×2 identity matrix
$\displaystyle B + 2I \;=\;\begin{pmatrix}1&3\\3&4\end{pmatrix} + \begin{pmatrix}2&0\\0&2\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}$

Let $\displaystyle X \:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}$

Then: .$\displaystyle AX \;=\;B + 2I\quad\Rightarrow\quad \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}$

. . $\displaystyle \begin{pmatrix}a-c & b-d \\ 3c & 3d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix} \quad\Rightarrow\quad \begin{bmatrix}a-c\:=\:3 & b-d \:=\:3 \\ 3c \:=\:3 & 3d \:=\:6\end{bmatrix}$

And we have: .$\displaystyle X \;=\;\begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}4&5\\1&2\end{pmatrix}$

2) The coefficient of $\displaystyle x$ in the expansion of $\displaystyle \left(2 + \frac{x}{3}\right)^n$ is 3 times the coefficient of $\displaystyle x^2.$
Find the value of $\displaystyle n.$
The expansion begins: .$\displaystyle 2^n + n\!\cdot\!2^{n-1}\left(\frac{x}{3}\right) + \frac{n(n-1)}{2}\!\cdot\!2^{n-2}\left(\frac{x}{3}\right)^2 + \hdots$

The coefficient of $\displaystyle x$ is: .$\displaystyle \frac{n\!\cdot\!2^{n-2}}{3}$

The coefficient of $\displaystyle x^2$ is: .$\displaystyle \frac{n(n-1)\!\cdot\!2^{n-2}}{2\cdot3^2} \:=\:\frac{n(n-1)\!\cdot\!2^{n-3}}{9}$

We are told that: .$\displaystyle \frac{n\!\cdot\!2^{n-1}}{3} \;=\;3\bigg[\frac{n(n-1)\!\cdot\!2^{n-3}}{9}\bigg] \quad\Rightarrow\quad \frac{n\!\cdot\!2^{n-1}}{3} \;=\;\frac{n(n-1)\!\cdot\!2^{n-3}}{3}$

Assuming $\displaystyle n \neq 0$, multiply by $\displaystyle \frac{3}{n}\!:\quad 2^{n-1} \:=\:(n-1)\!\cdot\!2^{n-3}$

Divide by $\displaystyle 2^{n-3}\!:\quad 2^2 \:=\:n-1 \quad\Rightarrow\quad\boxed{ n \:=\:5}$

3. Originally Posted by Soroban
Hello, teddybear67!

$\displaystyle B + 2I \;=\;\begin{pmatrix}1&3\\3&4\end{pmatrix} + \begin{pmatrix}2&0\\0&2\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}$

Let $\displaystyle X \:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}$

Then: .$\displaystyle AX \;=\;B + 2I\quad\Rightarrow\quad \begin{pmatrix}1&\text{-}1\\0&3\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix}$

. . $\displaystyle \begin{pmatrix}a-c & b-d \\ 3c & 3d\end{pmatrix} \;=\;\begin{pmatrix}3&3\\3&6\end{pmatrix} \quad\Rightarrow\quad \begin{bmatrix}a-c\:=\:3 & b-d \:=\:3 \\ 3c \:=\:3 & 3d \:=\:6\end{bmatrix}$

And we have: .$\displaystyle X \;=\;\begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\;\begin{pmatrix}4&5\\1&2\end{pmatrix}$

The expansion begins: .$\displaystyle 2^n + n\!\cdot\!2^{n-1}\left(\frac{x}{3}\right) + \frac{n(n-1)}{2}\!\cdot\!2^{n-2}\left(\frac{x}{3}\right)^2 + \hdots$

The coefficient of $\displaystyle x$ is: .$\displaystyle \frac{n\!\cdot\!2^{n-2}}{3}$

The coefficient of $\displaystyle x^2$ is: .$\displaystyle \frac{n(n-1)\!\cdot\!2^{n-2}}{2\cdot3^2} \:=\:\frac{n(n-1)\!\cdot\!2^{n-3}}{9}$

We are told that: .$\displaystyle \frac{n\!\cdot\!2^{n-1}}{3} \;=\;3\bigg[\frac{n(n-1)\!\cdot\!2^{n-3}}{9}\bigg] \quad\Rightarrow\quad \frac{n\!\cdot\!2^{n-1}}{3} \;=\;\frac{n(n-1)\!\cdot\!2^{n-3}}{3}$

Assuming $\displaystyle n \neq 0$, multiply by $\displaystyle \frac{3}{n}\!:\quad 2^{n-1} \:=\n-1)\!\cdot\!2^{n-3}$

Divide by $\displaystyle 2^{n-3}\!:\quad 2^2 \:=\:n-1 \quad\Rightarrow\quad\boxed{ n \:=\:5}$

thanks! i think ive gotten it! (:
but just one question. Identity matrix = $\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$
then what is singular matrix?

4. You have a LaTex error. If you mean "identity matrix= $\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$", then you are right.

A "singular matrix" is a matrix that does not have a multiplicative inverse- which is the same as saying its determinant is 0.

The matrix, A, in your first problem does have an inverse (which is why you were able to find a unique solution to this problem) and you could have done this as $\displaystyle X= A^{-1}AX= A^{-1}(B+ 2I)$.

It is not difficult to calculate that $\displaystyle A^{-1}= \begin{pmatrix}1 & \frac{1}{3} \\ 0 & \frac{1}{3}\end{pmatrix}$ and, as Soroban said, $\displaystyle B+ 2I= \begin{pmatrix}3 & 3 \\ 3 & 6\end{pmatrix}$

So $\displaystyle X= \begin{pmatrix}1 & \frac{1}{3} \\ 0 & \frac{1}{3}\end{pmatrix} \begin{pmatrix}3 & 3 \\ 3 & 6\end{pmatrix}$

My method is a little simpler if you know how to find the inverse matrix, Soroban's method has the advantage that you don't have to know how to find the inverse.