GO Classes DPP | GATE CS, DA | Linear Algebra | Eigen Vectors

Answer 1

Answer: 0

Answer 2

$A=\begin{pmatrix}3&4\\4&-3\end{pmatrix}$.

We are given that the eigenvectors of $A$ are $\begin{pmatrix}a\\1\end{pmatrix}$ and $\begin{pmatrix}1\\b\end{pmatrix}$.

First find the eigenvalues of $A$. We have,

$\det(A-\lambda I)=
\begin{vmatrix}
3-\lambda & 4\\
4 & -3-\lambda
\end{vmatrix}$

$=(3-\lambda)(-3-\lambda)-16$

$=\lambda^2-25$.

So the eigenvalues are

$\lambda=\pm 5$.
 

Now for $\lambda=5$,

$A-5I=
\begin{pmatrix}
-2&4\\
4&-8
\end{pmatrix}$.

So the eigenvector satisfies

$-2x+4y=0$

which gives

$x=2y$.

Taking $y=1$, we get the eigenvector $\begin{pmatrix}2\\1\end{pmatrix}$.

Hence, $a=2$.

Now for $\lambda=-5$,

$A+5I=
\begin{pmatrix}
8&4\\
4&2
\end{pmatrix}$.

So the eigenvector satisfies

$2x+y=0$

which gives

$y=-2x$.

Taking $x=1$, we get the eigenvector $\begin{pmatrix}1\\-2\end{pmatrix}$.

Hence, $b=-2$.
 

Therefore, $a+b=2+(-2)=0$.

Hence, the required answer is $\boxed{0}$.

Answer 3

 If $X$ is an eigenvector of matrix $A$, then:

$$AX = \lambda X$$

where $\lambda$ is the corresponding eigenvalue. This means that multiplying the matrix by the vector only scales the vector without changing its direction.

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 Solve for $a$

Let the first eigenvector be $X_1 = \begin{bmatrix} a \\ 1 \end{bmatrix}$. Applying the definition $AX_1 = \lambda_1 X_1$:

$$\begin{bmatrix} 3 & 4 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} a \\ 1 \end{bmatrix} = \lambda_1 \begin{bmatrix} a \\ 1 \end{bmatrix}$$

This gives us a system of two equations:

1. $3a + 4 = \lambda_1 a$

2. $4a - 3 = \lambda_1$

Substitute equation (2) into equation (1):

$$3a + 4 = (4a - 3)a$$

$$3a + 4 = 4a^2 - 3a$$

$$4a^2 - 6a - 4 = 0$$

Divide by 2 to simplify: $2a^2 - 3a - 2 = 0$. Using the quadratic formula or factoring:

$$(2a + 1)(a - 2) = 0$$

So, $a = 2$ or $a = -1/2$.

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 Solve for $b$

Given matrix $A$ is symmetric ($A = A^T$), We know that  Eigenvectors of a symmetric matrix corresponding to distinct eigenvalues are orthogonal.

The dot product of $X_1$ and $X_2$ must be zero:

$$\begin{bmatrix} a \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ b \end{bmatrix} = 0$$

$$(a \times 1) + (1 \times b) = 0 \implies b = -a$$

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 Find $a + b$.

• If $a = 2$, then $b = -2$.

• If $a = -1/2$, then $b = 1/2$.

In either scenario, when we sum them:

$$a + b = a + (-a) = 0$$

 Answer:

The value of $a + b = 0$.

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