matrix Q: write A^2 in the form pA + qI where p and q are

[mathfun]

Hi. How do you do this matrix question?

If A = [. 3. 2 ]
. . . . .[ -2 -1 ]

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, but I don't know how to put it in pA + qI form. Any help will be appreciated. Thank you!

 

[galactus]

For \(\displaystyle A^{2}\), try

\(\displaystyle \L\\2\cdot\begin{bmatrix}3&2\\-2&-1\end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

For \(\displaystyle A^{-1}\):

Try:

\(\displaystyle \L\\-\begin{bmatrix}3&2\\-1&-1\end{bmatrix}+2\cdot\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

 

[pka]

\(\displaystyle \begin{array}{l}
A = \left[ {\begin{array}{cc}
3 & 2 \\
{ - 2} & { - 1} \\
\end{array}} \right]\quad \Rightarrow \quad A^2 = \left[ {\begin{array}{cc}
5 & 4 \\
{ - 4} & { - 3} \\
\end{array}} \right]\quad \& \quad A^{ - 1} = \left[ {\begin{array}{cc}
{ - 1} & { - 2} \\
2 & 3 \\
\end{array}} \right] \\
A^2 = \left( 2 \right)A + \left( { - 1} \right)I_2 \quad \& \quad A^{ - 1} = \left( { - 1} \right)A + \left( 2 \right)I_2 \\
\end{array}\)

 

[soroban]

Re: matrix

Hello, mathfun!

I don't understand the purpose of this exercise,
but here's part 1 . . . and no guessing is needed.

If \(\displaystyle \,A \:= \:\begin{bmatrix}3 & 2 \\ -2 & -1\end{bmatrix}\)

write \(\displaystyle A^2\) in the form \(\displaystyle pA\,+\,qI,\) where \(\displaystyle p\) and \(\displaystyle q\) are scalars.

\(\displaystyle A^2\;=\;\begin{bmatrix}3 & 2\\-2&-1\end{bmatrix}\begin{bmatrix}3 & 2\\-2 & -1\end{bmatrix}\;=\;\begin{bmatrix}5&4\\-4&-3\end{bmatrix}\)


We have: \(\displaystyle \,pA\,+\,qI\;=\;p\begin{bmatrix}3 & 2 \\ -2 & -1\end{bmatrix}\,+\,q\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\;=\;\begin{bmatrix}3p&2p\\-2p&-p\end{bmatrix}\,+\,\begin{bmatrix}q&0\\0&q\end{bmatrix} \;= \;\begin{bmatrix}3p\,+\,q & 2p \\ -2p & -p\,+\,q\end{bmatrix}\)


We want \(\displaystyle pA\,+\,qI\) to equal \(\displaystyle A^2\)

. . so we have: \(\displaystyle \:\begin{bmatrix}3p\,+\,q&2p\\-2p&-p\,+\,q\end{bmatrix} \;= \;\begin{bmatrix}5&4\\-4&-3\end{bmatrix}\)

and we have a system of equations: \(\displaystyle \:\begin{Bmatrix}3p\,+\,q\:=\:5 & 2p\:=\:4 \\ \;\;-2p\:=\:-4 & \;\;-p\,+\,q\:=\:-3\end{Bmatrix}\)

. . which has solutions: \(\displaystyle \,p\,=\,2\;q\,=\,-1\)


Therefore: \(\displaystyle \L\,pA\,+\,qI\;=\;2\begin{bmatrix}3 & 2\\-2 & -1\end{bmatrix}\,-\,\begin{bmatrix}1&0\\0&1\end{bmatrix} \;= \;\begin{bmatrix}5&4\\-4&-3\end{bmatrix}\;=\;A^2\)