Linear Algebra - Linear Algebra

41:

,then eigen values of the matrix, I+A+A2 ,where I denotes the identity matrix are

A.	3,7,11
B.	3,7,12
C.	3,7,13
D.	3,9,16
	Answer Report Discuss
	Option: C Explanation : $A=\begin{bmatrix} 1&3&5\\0&2&-1\\0&0&3 \end{bmatrix}\\\Rightarrow A^{2}=\begin{bmatrix} 1&8&17\\0&4&-5\\0&0&9 \end{bmatrix}\\\therefore A^{2}+A+I=\begin{bmatrix} 3&11&22\\0&7&-6\\0&0&13 \end{bmatrix}$ A + I is a triangular matrix. Since eigen values of a triangular matrix are is diagonal elements, therefore eigen values of $A^{2}+A+2$ are 3, 7, 13. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

42:

Rank of the matrix

A.	0
B.	1
C.	3
D.	4
	Answer Report Discuss
	Option: C Explanation : Given matrix A possesses a minor of order 3, $viz \begin{vmatrix} 1&3&1\\2&4&0\\3&1&5 \end{vmatrix}\\=\begin{vmatrix} 0&0&1\\2&4&0\\-2&-14&5 \end{vmatrix}$ Replacing $C_{1}-C_{3} and C_{2}-3C_{3}$ $=\begin{vmatrix} 2&4\\-2&-14 \end{vmatrix}$ expanding with respect to R₁ = 2(-14) - (4)(-2) = -28 + 8 ≠ 0 $\therefore p(A)\geq 3$ .............................................(i) Also A does not possess any minor of order 4, i.e. 3 + 1 $\therefore p(A )\leq3$ ..................................................(ii) From Equations (i) and (ii), we get p(A) = 3 i.e. rank of A is 3. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

43:

Eigen values of the matrix

A.	1,1,1
B.	1,4,2
C.	1,4,4
D.	1,2,4
	Answer Report Discuss
	Option: C Explanation : Characteristic equation is $\left [ A-\lambda I \right ]=0$ $\Rightarrow \begin{bmatrix} 3-\lambda &-1&-1\\-1&3-\lambda &-1\\ -1&-1&3-\lambda \end{bmatrix}=0$ $\Rightarrow (3-\lambda)\left \{ (3-\lambda )^{2}-1\right \}+\left \{ (-3-\lambda ) -1\right \}-\left \{ 1+(3-\lambda ) \right \}=0 \\\Rightarrow (3-\lambda )\left \{ (3-\lambda ) -1\right \}\left \{ (-3-\lambda )+1 \right \}-2\left \{ (3-\lambda )+1 \right \}=0\\\Rightarrow \left \{ (3-\lambda )+1 \right \}\left \{ (3-\lambda )(3-\lambda -1)-2 \right \}=0 \\\Rightarrow (4-\lambda )(6-3\lambda -2\lambda +\lambda ^{2}-2)=0\\\Rightarrow (4-\lambda )(4-4\lambda -\lambda +\lambda ^{2}-2)=0\\\Rightarrow (4-\lambda )(\lambda -1)(\lambda -4)=0$ $\therefore$ λ = 1, 4, 4 are the eigen values. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

44:

Eigen values of the matrix

are

A.	1,1,1
B.	1,4,2
C.	1,4,4
D.	1,2,4
	Answer Report Discuss
	Option: C Explanation : Characteristic equation is $\left [ A-\lambda I \right ]=0$ $\Rightarrow \begin{bmatrix} 3-\lambda &-1&-1\\-1&3-\lambda &-1\\ -1&-1&3-\lambda \end{bmatrix}=0$ $\Rightarrow (3-\lambda)\left \{ (3-\lambda )^{2}-1\right \}+\left \{ (-3-\lambda ) -1\right \}-\left \{ 1+(3-\lambda ) \right \}=0 \\\Rightarrow (3-\lambda )\left \{ (3-\lambda ) -1\right \}\left \{ (-3-\lambda )+1 \right \}-2\left \{ (3-\lambda )+1 \right \}=0\\\Rightarrow \left \{ (3-\lambda )+1 \right \}\left \{ (3-\lambda )(3-\lambda -1)-2 \right \}=0 \\\Rightarrow (4-\lambda )(6-3\lambda -2\lambda +\lambda ^{2}-2)=0\\\Rightarrow (4-\lambda )(4-4\lambda -\lambda +\lambda ^{2}-2)=0\\\Rightarrow (4-\lambda )(\lambda -1)(\lambda -4)=0$ $\therefore$ λ = 1, 4, 4 are the eigen values. λ - Wiktionary Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

45:

then

A.	A' = -A
B.	A' = A
C.	$A'= -\frac{1}{A}$
D.	none of this
	Answer Report Discuss
	Option: A Explanation : $A'= \begin{bmatrix} 0&-2&-3\\2&0&-5\\3&5&0 \end{bmatrix}= -\begin{bmatrix} 0&2&3\\-2&0&5\\-3&-5&0 \end{bmatrix}=-A$ This is a skew-symmetric matrix. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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