Linear Algebra - Linear Algebra

36:

A.	-76
B.	-28
C.	+28
D.	+72
	Answer Report Discuss
	Option: B Explanation : $\begin{bmatrix}5&3&2\\1&2&6\\3&5&10\end{bmatrix}$ =5(20-30) - 3(10-18) + 2(5-6) = -50 + 24 - 2 = -28 Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

37:

The solution(s) to the equations
2x + 3y = 1
x- y = 4
4x - y = a
will exist if a is equal to

A.	-33
B.	0
C.	9
D.	12
	Answer Report Discuss
	Option: D Explanation : Equations are consistent when, $\begin{vmatrix} 2&3&-1\\1&-1&-4\\4&1&a \end{vmatrix}=0$ $\Rightarrow$ 5a - 7 - 52 = 0 $\Rightarrow a =\frac{59}{5}=11\frac{4}{5}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

38:

The system of simultaneous equations
x + 2y + z = 6
2x + y + 2z = 6
x + y + z = 5
has

A.	unique solution
B.	infinite number of solutions
C.	no solution
D.	exactly two solutions
	Answer Report Discuss
	Option: C Explanation : Given equations are; x + 2y + z = 6; 2x + y + 2z = 6; x + y + z = 5 Given system can be written as $\begin{bmatrix} 1&2&1\\2&1&2\\1&1&1 \end{bmatrix}.\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 6\\6\\5 \end{bmatrix}$ Applying row operation $R_{2}=R_{2}-2R_{1}, R_{3}=R_{3}-R_{2}$ , we get $\begin{bmatrix} 1&2&1\\0&-3&0\\0&-1&0 \end{bmatrix}.\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 6\\-6\\-1 \end{bmatrix}$ Applying $R_{3}= 3R_{3}-R_{2}$ , we get $\begin{bmatrix} 1&2&1\\0&-3&0\\0&0&0 \end{bmatrix}.\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 6\\-6\\3 \end{bmatrix}$ Since rank of co-efficient matrix is 2 and rank of argument matrix is 3, which is not equal. Hence system has no solution. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

39:

The inverse of A is

A.	$\begin{bmatrix} 1&0&-2\\0&\frac{1}{3}&0\\2&0&5 \end{bmatrix}$
B.	$\begin{bmatrix} 5&0&2\\0&-\frac{1}{3}&0\\\frac{1}{2}&0&1 \end{bmatrix}$
C.	$\begin{bmatrix} \frac{1}{5}&0&\frac{1}{2}\\0&\frac{1}{3}&0\\\frac{1}{2}&0&1 \end{bmatrix}$
D.	$\begin{bmatrix} \frac{1}{5}&0&-\frac{1}{2}\\0&\frac{1}{3}&0\\-\frac{1}{2}&0&1 \end{bmatrix}$
	Answer Report Discuss
	Option: A Explanation : $det A=\left \|A \right \|$ = 5[3 - 0] - 0[0 - 2] + 2[0 - 6] = 15 - 12 = 3 $Adj A = \begin{vmatrix} 3&0\\0&1 \end{vmatrix}-\begin{vmatrix} 0&2\\0&1 \end{vmatrix}\begin{vmatrix} 0&2\\3&0 \end{vmatrix}- \begin{vmatrix} 0&0\\2&1 \end{vmatrix}\begin{vmatrix} 5&2\\2&1 \end{vmatrix}\\-\begin{vmatrix} 5&2\\0&0 \end{vmatrix}\begin{vmatrix} 0&3\\2&0 \end{vmatrix}-\begin{vmatrix} 3&0\\2&0 \end{vmatrix}\begin{vmatrix} 5&0\\0&3 \end{vmatrix}\\Adj A = \begin{bmatrix} 3&0&-6\\0&1&0\\-6&0&15 \end{bmatrix}$ $\therefore A^{-1}=\frac{Adj A}{\left \| A \right \|}\\=\frac{1}{3}\begin{bmatrix} 3&0&-6\\0&1&0\\-6&0&15 \end{bmatrix}\\=\begin{bmatrix} 1&0&-2\\0&\frac{1}{3}&0\\-2&0&5 \end{bmatrix}$ Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

40:

Eigen values of the matrix

-1	4
4	-1

are

A.	3 and -5
B.	-3 and 5
C.	-3 and -5
D.	3 and 5
	Answer Report Discuss
	Option: A Explanation : Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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