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

Answer 1

May be we can just use some Important points to answer this question, like sum of eigen values = Trace
and product is det
so only B works.

Answer 2

eigen value of M are 1, 4 then trace(M) = 1 + 4 = 5,  which eliminates option A, D. now determinant = 1*4 = 4 so option C is eliminated. so the answer is option B

Answer 3

Given eigenvalues $\lambda_1 = 1$, $\lambda_2 = 4$ with corresponding eigenvectors
\[
v_1 = \begin{pmatrix}1 \\ -1\end{pmatrix}, \quad
v_2 = \begin{pmatrix}2 \\ 1\end{pmatrix}
\]

Form the matrix $P$ using eigenvectors as columns:
\[
P = \begin{pmatrix}1 & 2 \\ -1 & 1\end{pmatrix}
\]

Form the diagonal matrix $D$:
\[
D = \begin{pmatrix}1 & 0 \\ 0 & 4\end{pmatrix}
\]

We use the formula:
\[
M = P D P^{-1}
\]

Step $1:$ Compute $P^{-1}$

\[
\det(P) = (1)(1) - (2)(-1) = 3
\]

\[
P^{-1} = \frac{1}{3}
\begin{pmatrix}
1 & -2 \\
1 & 1
\end{pmatrix}
\]

Step $2:$ Compute $PD$

\[
PD =
\begin{pmatrix}1 & 2 \\ -1 & 1\end{pmatrix}
\begin{pmatrix}1 & 0 \\ 0 & 4\end{pmatrix}
=
\begin{pmatrix}1 & 8 \\ -1 & 4\end{pmatrix}
\]

Step $3:$ Compute $M = PDP^{-1}$

\[
M =
\begin{pmatrix}1 & 8 \\ -1 & 4\end{pmatrix}
\cdot
\frac{1}{3}
\begin{pmatrix}1 & -2 \\ 1 & 1\end{pmatrix}
\]

\[
M = \frac{1}{3}
\begin{pmatrix}
1 + 8 & -2 + 8 \\
-1 + 4 & 2 + 4
\end{pmatrix}
=
\frac{1}{3}
\begin{pmatrix}
9 & 6 \\
3 & 6
\end{pmatrix}
\]

\[
M =
\begin{pmatrix}
3 & 2 \\
1 & 2
\end{pmatrix}
\]

Final Answer: $
\boxed{
M =
\begin{pmatrix}
3 & 2 \\
1 & 2
\end{pmatrix}
}$

 

 

$\underline{\textbf{Alternate Solution :}}$

$AX=\lambda X$

Now, let $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$

When $\lambda=1$:

$\begin{bmatrix} a & b\\ c & d \end{bmatrix}\begin{bmatrix} 1\\ -1 \end{bmatrix}
=1\begin{bmatrix} 1\\ -1 \end{bmatrix}$

$\therefore a-b=1$ and $c-d=-1$

When $\lambda=4$:

$\begin{bmatrix} a & b\\ c & d \end{bmatrix}\begin{bmatrix} 2\\ 1 \end{bmatrix}
=4\begin{bmatrix} 2\\ 1 \end{bmatrix}$

$\therefore 2a+b=8$ and $2c+d=4$

Solving simultaneously, we get:

$a-b=1$
$2a+b=8$

Adding, $3a=9 \Rightarrow a=3$

So, $b=2$

Also,

$c-d=-1$
$2c+d=4$

Adding, $3c=3 \Rightarrow c=1$

So, $d=2$

Hence,

$A=\begin{bmatrix} 3 & 2\\ 1 & 2 \end{bmatrix}$

Answer 4

Option B is correct 

Answer 5

Answer: B