1. Weird Matrix Problem

The Matrix A is $\displaystyle \left[ \begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array} \right]$

(i) Find the matrix $\displaystyle B$ such that $\displaystyle B=A^2 -2I$ where $\displaystyle I$ is the identity matrix.

(ii) Find the inverse matrix $\displaystyle A^{-1}$

(iii) Use $\displaystyle A^{-1}$ to solve the simultaneous equations
[tex]

$\displaystyle 2x+2y=6$
$\displaystyle x+3y=8$

2. Shouldn't this go to the Linear Algebra subforum ?

3. Well, I am under the catergory "Pre University" and the description of Pre-Algebra and Algebra section is "Basic calculations, order of operations, basic matrices, solving for variables, functions, logs and exponents and more"

4. My mistake. Forgive me ...

5. Originally Posted by Punch
The Matrix is $\displaystyle \left[ \begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array} \right]$

(i) Find the matrix $\displaystyle B$ such that $\displaystyle B=A^2 -2I$ where $\displaystyle I$ is the identity matrix.

(ii) Find the inverse matrix $\displaystyle A^{-1}$

(iii) Use $\displaystyle A^{-1}$ to solve the simultaneous equations
$\displaystyle 2x+2y=6 x+3y=8$
Is this matrix you have posted matrix $\displaystyle A$?

If so...

$\displaystyle B = A^2 - 2I$

$\displaystyle = \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right]^2 - 2\left[\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}1 & 1\\ 1 & 3\end{matrix}\right] \left[\begin{matrix} 1 & 1\\ 1 & 3\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix} 0 & 4\\ 4 & 8\end{matrix}\right]$.

6. Originally Posted by Prove It
Is this matrix you have posted matrix $\displaystyle A$?

If so...

$\displaystyle B = A^2 - 2I$

$\displaystyle = \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right]^2 - 2\left[\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix}1 & 1\\ 1 & 3\end{matrix}\right] \left[\begin{matrix} 1 & 1\\ 1 & 3\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix} 0 & 4\\ 4 & 8\end{matrix}\right]$.
Thanks but I don't understand what the identity matrix is. May I ask how I can derive the "identity matrix"?

7. Originally Posted by Punch
The Matrix is $\displaystyle \left[ \begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array} \right]$

(i) Find the matrix $\displaystyle B$ such that $\displaystyle B=A^2 -2I$ where $\displaystyle I$ is the identity matrix.

(ii) Find the inverse matrix $\displaystyle A^{-1}$

(iii) Use $\displaystyle A^{-1}$ to solve the simultaneous equations
$\displaystyle 2x+2y=6 x+3y=8$
ii) $\displaystyle A = \left[\begin{matrix} 1 & 1\\ 1 & 3 \end{matrix}\right]$

$\displaystyle A^{-1} = \frac{1}{1\cdot 3 - 1 \cdot 1}\left[\begin{matrix}3 & -1 \\ -1 & 1\end{matrix}\right]$

$\displaystyle = \frac{1}{2}\left[\begin{matrix} 3 & -1\\ -1 & 1\end{matrix}\right]$

$\displaystyle = \left[\begin{matrix} \frac{3}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2}\end{matrix}\right]$.

8. Originally Posted by Punch
Thanks but I don't understand what the identity matrix is. May I ask how I can derive the "identity matrix"?
The Identity Matrix $\displaystyle I_n$ is a square $\displaystyle n \times n$ matrix with 1's down the main diagonal and 0's everywhere else.

It is usually abbreviated to $\displaystyle I$, and you have to remember what dimensions your matrices are. In your case, your matrices are $\displaystyle 2 \times 2$, so you will be using $\displaystyle I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]$.

It is the matrix equivalent of the number $\displaystyle 1$, because when you premultiply or postmultiply any matrix by $\displaystyle I$, you get that original matrix.

Also, if you multiply any nonsingular matrix by its inverse, you get the identity matrix. This gives us the ability to perform the matrix equivalent of division.

9. So basically an identity matrix is equals to one and I can use $\displaystyle I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]$ for all identity matrix?

10. Also, I haven't learnt or seen any summing of matrix or subtraction of matrix..

How exactly do I solve $\displaystyle = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$

11. Originally Posted by Punch
The Matrix A is $\displaystyle \left[ \begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array} \right]$

(i) Find the matrix $\displaystyle B$ such that $\displaystyle B=A^2 -2I$ where $\displaystyle I$ is the identity matrix.

(ii) Find the inverse matrix $\displaystyle A^{-1}$

(iii) Use $\displaystyle A^{-1}$ to solve the simultaneous equations
[tex]

$\displaystyle 2x+2y=6$
$\displaystyle x+3y=8$
First, notice that you can rewrite the system of equations as

$\displaystyle x + y = 3$
$\displaystyle x + 3y = 8$

Write the system in matrix form.

$\displaystyle \left[\begin{matrix}1 & 1 \\ 1 & 3\end{matrix}\right] \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} 3 \\ 8\end{matrix}\right]$

I am going to write this equation as

$\displaystyle A\mathbf{x} = \mathbf{b}$.

Premultiply both sides of the equation by $\displaystyle A^{-1}$.

Then you have

$\displaystyle A^{-1}A\mathbf{x} = A^{-1}\mathbf{b}$.

Recall that if you multiply any matrix by its inverse, you get the identity matrix, so

$\displaystyle I\mathbf{x} = A^{-1}\mathbf{b}$

And if you multiply any matrix by the identity matrix, you get back to original matrix.

So $\displaystyle \mathbf{x} = A^{-1}\mathbf{b}$

$\displaystyle \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{3}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2} \end{matrix}\right] \left[\begin{matrix} 3 \\ 8 \end{matrix}\right]$

$\displaystyle \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{9}{2} - 4 \\ -\frac{3}{2} + 4\end{matrix} \right]$

$\displaystyle \left[\begin{matrix} x \\ y\end{matrix}\right] = \left[\begin{matrix} \frac{1}{2} \\ \frac{5}{2}\end{matrix}\right]$

Therefore $\displaystyle x = \frac{1}{2}$ and $\displaystyle y = \frac{5}{2}$.

12. Originally Posted by Punch
Also, I haven't learnt or seen any summing of matrix or subtraction of matrix..

How exactly do I solve $\displaystyle = \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right]$
How can you be given questions on these things without having been taught them? Matrix addition, Matrix multiplication and the Identity Matrix are the most important things to know when dealing with matrices.

Matrix addition is performed component-wise. So you add (or subtract) corresponding components.

$\displaystyle \left[\begin{matrix} 2 & 4\\ 4 & 10\end{matrix}\right] - \left[\begin{matrix} 2 & 0\\ 0 & 2\end{matrix}\right] = \left[\begin{matrix} 2 - 2 & 4 - 0\\ 4 - 0 & 10 - 2\end{matrix}\right]$

13. Originally Posted by Punch
So basically an identity matrix is equals to one and I can use $\displaystyle I_2 = \left[\begin{matrix} 1 & 0 \\ 0 & 1\end{matrix}\right]$ for all identity matrix?
No, an identity matrix is not equal to 1. It ACTS LIKE a 1 when you are dealing with matrices.

And you can't ALWAYS use $\displaystyle I_2$ because the matrices you are working with might not be $\displaystyle 2 \times 2$. It depends on the dimensions of the matrices you are working with. In this case you use $\displaystyle I_2$ because the matrices you are working with are $\displaystyle 2 \times 2$.

But what if you were dealing with $\displaystyle 3 \times 3$ matrices?

Which $\displaystyle I$ would you use then?

14. Well, I found this question in an assesment book and am not sure if this type of question is in the syllabus.

15. Originally Posted by Prove It
No, an identity matrix is not equal to 1. It ACTS LIKE a 1 when you are dealing with matrices.

And you can't ALWAYS use $\displaystyle I_2$ because the matrices you are working with might not be $\displaystyle 2 \times 2$. It depends on the dimensions of the matrices you are working with. In this case you use $\displaystyle I_2$ because the matrices you are working with are $\displaystyle 2 \times 2$.

But what if you were dealing with $\displaystyle 3 \times 3$ matrices?

Which $\displaystyle I$ would you use then?
hmm, 3$\displaystyle I_2$?

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