Linear Algebra - Linear Algebra

46:

The eigen vectors of a real symmetric matrix corresponding to different eigen values are

A.	orthogonal
B.	singular
C.	non-singular
D.	none of these
	Answer Report Discuss
	Option: A Explanation : Let A be a real symmetric matrix, therefore A^T=A Let α₁and α₂ be different eigen values of matrix A, and X₁and X₂be the corresponding vectors, then AX₁= α₁X₁and AX₂ = α₂X₂ Taking transpose of the second equation (AX₂)^T= (α₂X₂) X_2TA^T= α₂.X_{2T 2}X_2T But A^T-A $\therefore X_{2}_{T}A= a^{2}.X_{2}_{T}$ Post multiply by X₁, we get X^T₂AX₁ = a₂X_2TX₁ But AX₁ = a₁X₁ $\therefore$ X^T₂a₁X₁ = a₂X_2TX₁ $\Rightarrow$ (a₁- a₂) X_2TX₁ = 0 Since a₁ $\neq$ a2, a₁- a₂ $\neq$ 0 $\therefore$ X_2TX₁ = 0 i.e. X₂and X₁are orthogonal. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

47:

An arbitary vector X is an eigen vector of the matrix

A.	(0,0)
B.	(1,1)
C.	(0,1)
D.	(1,2)
	Answer Report Discuss
	Option: B Explanation : Since matrix is triangular, the eigen values are a, a, b. (X₁,X₂,X₃) is an arbitary eigen vector, say corresponding to 1, then $\begin{bmatrix} 1&0&0\\0&a&0\\0&0&b \end{bmatrix}\begin{bmatrix} X_{1}\\X_{2}\\X_{3} \end{bmatrix}=1\begin{bmatrix} X_{1}\\X_{2}\\X_{3} \end{bmatrix}$ X₂ X₃being not zero, we have X₁= X₁; aX₂= X₂ which gives a = 1 and bX₃ = X₃ which gives b - 1 $\therefore$ (a, b) = (1, 1) Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

48:

For which value of k, the following system is consistent?

2x-5ky+6z=0

kx+2y-2z=0

2x+2y-kz=0

A.	1
B.	2
C.	3
D.	5
	Answer Report Discuss
	Option: B Explanation : $A=\begin{bmatrix} 2&-5k&6\\k&2&-2\\2&2&-k \end{bmatrix}=0$ Equations are consistent, if rank of A and that of k are equal. But in this case it is always true. Hence the equations will have a trivial solution if IAI $\neq$ 0 Therefore only non- trivial solution will exist if IAI = 0 $i.e. \begin{bmatrix} 2&-k&6\\k&2&-2\\2&2&k \end{bmatrix}=0$ $\Rightarrow$ $\Rightarrow$ -5k³+ 20k - 4k + 8 + 12k - 24 = 0 $\Rightarrow$ 5k³- 28k + 16 = 0 $\Rightarrow$ 5k³- 10k² + 10k² - 20k - 8k + 16 = 0 $\Rightarrow$ (5k²+ 10k - 8)(k - 2) = 0 $k=\frac{-10\pm \sqrt{100+160}}{2}$ or 2 $= -1\pm \sqrt{13/5}$ or 2 Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

49:

The value of λ for which the equations

2x + y + 2z = 0

x + y + 3z = 0

4x + y + λz = 0

have non-zero solution, is

A.	2
B.	4
C.	6
D.	8
	Answer Report Discuss
	Option: D Explanation : Equivalent matrix equation is $\begin{bmatrix} 2&1&2\\1&1&3\\4&3&\lambda \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=0$ In order that the given system of equations may have non-zero solution, the rank of A should be less than 3. This requires that $\begin{bmatrix} 2&1&2\\1&1&3\\4&3&\lambda \end{bmatrix}=0$ Interchanging R₁ and R₂ $\begin{bmatrix} 1&1&3\\2&1&2\\4&3&\lambda \end{bmatrix}=0$ By (R₂- 2R₁) and (R₃ - 4R₁), $\begin{bmatrix} 1&1&3\\1&-1&-4\\0&-1&\lambda \end{bmatrix}=0$ $\Rightarrow \begin{bmatrix} -1&-4\\-1&\lambda-12 \end{bmatrix}=0$ $\Rightarrow -1 ( \lambda - 12) - (-1) (4) = 0$ $\Rightarrow \lambda = 8$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

50:

The system of equations

a₁x + a₂y = 0

b₁x + b₂y = 0

where a₁, a₂ , b₁, b₂are real numbers,

has a non-trivial solutions if

A.	a₁b₁= a₂b₂
B.	a₁b₂= a₂b₁
C.	a₁a₂ = b₁b₂
D.	none of these
	Answer Report Discuss
	Option: B Explanation : $\frac{x}{y}=-\frac{a_{2}}{a_{1}}\\and \frac{x}{y}=-\frac{b_{2}}{b_{1}}$ Equations are conistent only if, $-\frac{a_{2}}{a_{1}}=-\frac{b_{2}}{b_{1}}$ $\Rightarrow$ a₁b₂= b₁a₂ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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