Maths Q Help.. (1 Viewer)

Clara_Oswald · Mar 9, 2016

Okay hi guys,
I'm not sure if I'm posting this in the right place but I'm stuck on a problem and was wondering whether anybody could enlighten me on how to do this.
I would wait for my tut but that means i literally can't do the next 2 pages of questions and the tutor won't go through every single one that i personally have trouble with.
This is for MATH1031 by the way, and the topic is Inverse, Identity and Zero Matrices.

Also, I would ask people in my course however I don't know anyone/have anybody as a contact so :3

Clara_Oswald · Mar 9, 2016

Not sure if the photo link is working so http://m.imgur.com/NyN9H8O

I'm stuck on q17 (and 18) on how to find A^-1 and A^4

InteGrand · Mar 9, 2016

Clara_Oswald said:
Okay hi guys,
I'm not sure if I'm posting this in the right place but I'm stuck on a problem and was wondering whether anybody could enlighten me on how to do this.
I would wait for my tut but that means i literally can't do the next 2 pages of questions and the tutor won't go through every single one that i personally have trouble with.
This is for MATH1031 by the way, and the topic is Inverse, Identity and Zero Matrices.

Also, I would ask people in my course however I don't know anyone/have anybody as a contact so :3

$\noindent 17. We are given $A=\begin{bmatrix}3 &2\\ 7&5\end{bmatrix}$. Hence$

$\begin{align*}A^2 &= \begin{bmatrix}3 &2\\ 7&5\end{bmatrix}\begin{bmatrix}3 &2\\ 7&5\end{bmatrix} \\ &= \begin{bmatrix}3\cdot 3 + 2 \cdot 7 & 3 \cdot 2 + 2 \cdot 5 \\7\cdot 3 + 5\cdot 7 & 7\cdot 2 + 5 \cdot 5 \end{bmatrix} \\ &= \begin{bmatrix}23 & 16 \\56 & 39 \end{bmatrix}. \end{align*}$

$\noindent Also, $8A =\begin{bmatrix}24 & 16\\ 56 & 40\end{bmatrix}$. Thus$

$\begin{align*}A^2 -8A + I &= \begin{bmatrix}23 & 16 \\56 & 39 \end{bmatrix}-\begin{bmatrix}24 & 16\\ 56 & 40\end{bmatrix}+\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} \\ &= O \quad (\text{the zero matrix}), \text{ by simple computation}. \end{align*}$

$\noindent Now, assuming $A^{-1}$ exists (which it turns out indeed does), we can pre-multiply the above through by $A^{-1}$. Note that $A^{-1} A^2 = A^{-1} A A = IA = A$ (using associativity of matrix multiplication). Using distributivity rules then yields$

$\begin{align*}A^{-1} \left(A^2 -8A + I\right) &= A^{-1}O \\ \Rightarrow A^{-1} A^2 -8 A^{-1} A + IA^{-1} &= O \\ \Rightarrow A - 8I + A^{-1} &= O \\ \Rightarrow A^{-1} &= 8I - A \\ &=\begin{bmatrix}8 & 0\\ 0 & 8\end{bmatrix}-\begin{bmatrix}3 &2\\ 7&5\end{bmatrix} \\ &= \begin{bmatrix}5 &-2\\ -7&3\end{bmatrix}. \end{align*}$

$\noindent To find $A^4$, recall that we had $A^2 &= 8A - I$. Then squaring both sides,$

$\begin{align*}A^4 &= \left(8A -I\right)\left(8A -I\right) \\ &= 8^2A^2 - 8A - 8A + I \quad (\text{we used the facts that matrix multiplication with }I \text{ is commutative and }I^2 = I) \\ &= 8^2 \left(8A - I\right)-16A+I \\ &= 8^3 A-8^2 I - 16A +I \\ &= 496 A-63I \\ &= 496\begin{bmatrix}3 &2\\ 7&5\end{bmatrix} - \begin{bmatrix}63 & 0\\ 0 & 63\end{bmatrix} \\ &= \begin{bmatrix}1488 & 992 \\ 3472 & 2480\end{bmatrix} - \begin{bmatrix}63 & 0\\ 0 & 63\end{bmatrix} \\ &= \begin{bmatrix}1425 & 992 \\ 3472 & 2417\end{bmatrix}. \end{align*}$

$\noindent Question 18 is essentially just a more tedious version of 17, the way to do it is similar. Once you've shown the given identity, multiply both sides by $A$ to help find $A^4$. This results in an $A^2$ appearing on the R.H.S., which you would have computed already in computing $A^3$ to prove the identity initially. To help find $A^{-1}$, assuming it exists, multiply both sides of the identity by $A^{-1}$.$

Clara_Oswald · Mar 9, 2016

Omg. This is so good thank you so much.

Maths Q Help.. (1 Viewer)

Clara_Oswald

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InteGrand

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