GO Classes DPP | GATE CS, DA | Linear Algebra | Diagonalization

Answer 1

Since $A$ has two distinct eigenvalues $1$ and $-2$, it is diagonalizable.

The corresponding eigenvectors are $\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}9\\1\end{pmatrix}$.

So take

$P=\begin{pmatrix}1&9\\2&1\end{pmatrix}$

and

$D=\begin{pmatrix}1&0\\0&-2\end{pmatrix}$.

Then

$A=PDP^{-1}$.

Now,

$\det(P)=1\cdot 1-9\cdot 2=1-18=-17$.

So,

$P^{-1}
=\frac{1}{-17}\begin{pmatrix}1&-9\\-2&1\end{pmatrix}
=\begin{pmatrix}
-\frac{1}{17} & \frac{9}{17}\\[4pt]
\frac{2}{17} & -\frac{1}{17}
\end{pmatrix}$.

Hence,

$A
=
\begin{pmatrix}1&9\\2&1\end{pmatrix}
\begin{pmatrix}1&0\\0&-2\end{pmatrix}
\begin{pmatrix}
-\frac{1}{17} & \frac{9}{17}\\[4pt]
\frac{2}{17} & -\frac{1}{17}
\end{pmatrix}$.

First compute

$DP^{-1}
=
\begin{pmatrix}1&0\\0&-2\end{pmatrix}
\begin{pmatrix}
-\frac{1}{17} & \frac{9}{17}\\[4pt]
\frac{2}{17} & -\frac{1}{17}
\end{pmatrix}
=
\begin{pmatrix}
-\frac{1}{17} & \frac{9}{17}\\[4pt]
-\frac{4}{17} & \frac{2}{17}
\end{pmatrix}$.

Now,

$A
=
\begin{pmatrix}1&9\\2&1\end{pmatrix}
\begin{pmatrix}
-\frac{1}{17} & \frac{9}{17}\\[4pt]
-\frac{4}{17} & \frac{2}{17}
\end{pmatrix}$

$=
\begin{pmatrix}
\frac{-1-36}{17} & \frac{9+18}{17}\\[4pt]
\frac{-2-4}{17} & \frac{18+2}{17}
\end{pmatrix}$

$=
\begin{pmatrix}
-\frac{37}{17} & \frac{27}{17}\\[4pt]
-\frac{6}{17} & \frac{20}{17}
\end{pmatrix}$.

Therefore, the sum of all elements of $A$ is

$-\frac{37}{17}+\frac{27}{17}-\frac{6}{17}+\frac{20}{17}
=\frac{4}{17}$.

Hence, the required answer is $\boxed{\frac{4}{17}}$.

Comments

As yo multiply DP^-1 first  what happen  when we multiply PD  first  as order of multiplication matter a lot in Linear Algebra ? @GO Classes

---

• Matrix multiplication is Associative.
   
• All three matrices $P, D,$ and $P^{-1}$ are of the same order.
   
• So, $(PD)P^{-1}$ and $P(DP^{-1})$ are both correct
   
• but you must keep the order $P, D, P^{-1}$ the same.

Answer 2

Answer: A