# Matrix

• March 18th 2013, 01:14 AM
Tutu
Matrix
1.) If A is $\begin{bmatrix} 3 & 2 \\-2 & -1 \end{bmatrix}$, write A^2 in the form pA+qI where p and q are scalars. Hence write A^(-1) in the form rA+sI where r and s are scalars.

I know how to find A^2, I got $\begin{bmatrix} 5 & 4 \\-4 & -3 \end{bmatrix}$ but I do not know how to convert this matrix form into linear form pA+qI

2.) It is known that AB=A and BA=B where matrices A and B are not necessarily invertible.
Prove that A^2 = A.

When I first saw this, I thought B had to be I in AB=A and A in BA=B had to be I BUT they then added, NOTE: From AB=A, you cannot deduce that B=I. They asked me why, and I really dont know since I thought you could deduce that!
How do I then prove that A^2=A?
• March 18th 2013, 01:29 AM
Prove It
Re: Matrix
Quote:

Originally Posted by Tutu
1.) If A is $\begin{bmatrix} 3 & 2 \\-2 & -1 \end{bmatrix}$, write A^2 in the form pA+qI where p and q are scalars. Hence write A^(-1) in the form rA+sI where r and s are scalars.

I know how to find A^2, I got $\begin{bmatrix} 5 & 4 \\-4 & -3 \end{bmatrix}$ but I do not know how to convert this matrix form into linear form pA+qI

2.) It is known that AB=A and BA=B where matrices A and B are not necessarily invertible.
Prove that A^2 = A.

When I first saw this, I thought B had to be I in AB=A and A in BA=B had to be I BUT they then added, NOTE: From AB=A, you cannot deduce that B=I. They asked me why, and I really dont know since I thought you could deduce that!
How do I then prove that A^2=A?

\displaystyle \begin{align*} p\mathbf{A} + q\mathbf{I} &= p\left[ \begin{matrix} \phantom{-}5 & \phantom{-}4 \\ -4 & -3 \end{matrix} \right] + q \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \\ &= \left[ \begin{matrix} \phantom{-}5p & \phantom{-}4p \\ -4p & -3p \end{matrix} \right] + \left[ \begin{matrix} q & 0 \\ 0 & q \end{matrix} \right] \\ &= \left[ \begin{matrix} \phantom{-}5p + q & \phantom{-}4p \\ -4p & -3p + q \end{matrix} \right] \end{align*}

If this is equal to $\displaystyle \mathbf{A}^2$, then that means you can set each of the components equal and solve for p and q.
• March 18th 2013, 01:38 AM
Tutu
Re: Matrix
I see thank you so so much! ((:
Any ideas for the second question?

Thanks!
• March 18th 2013, 05:35 AM
Prove It
Re: Matrix
Quote:

Originally Posted by Tutu
1.) If A is $\begin{bmatrix} 3 & 2 \\-2 & -1 \end{bmatrix}$, write A^2 in the form pA+qI where p and q are scalars. Hence write A^(-1) in the form rA+sI where r and s are scalars.

I know how to find A^2, I got $\begin{bmatrix} 5 & 4 \\-4 & -3 \end{bmatrix}$ but I do not know how to convert this matrix form into linear form pA+qI

2.) It is known that AB=A and BA=B where matrices A and B are not necessarily invertible.
Prove that A^2 = A.

When I first saw this, I thought B had to be I in AB=A and A in BA=B had to be I BUT they then added, NOTE: From AB=A, you cannot deduce that B=I. They asked me why, and I really dont know since I thought you could deduce that!
How do I then prove that A^2=A?

$\displaystyle \mathbf{A}\mathbf{B} = \mathbf{A}$ and $\displaystyle \mathbf{B}\mathbf{A} = \mathbf{B}$. Then

\displaystyle \begin{align*} \mathbf{A}^2 &= \left( \mathbf{A}\mathbf{B} \right)^2 \\ &= \mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B} \\ &= \mathbf{A}\mathbf{B}\mathbf{B} \\ &= \mathbf{A}\mathbf{B} \\ &= \mathbf{A} \end{align*}