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201.

Find the value of x and y, when (i) `[{:(x+y),(x-y):}]=[{:(8),(4):}]` (ii) `[{:(2x+5," "7),(" "0,3y-7):}]=[{:(x-3," "7),(" "0,-5):}]` (iii) `2[{:(x," "5),(7,y-3):}]+[{:(3,-4),(1," "2):}]=[{:(7,6),(15,14):}]`

Answer» (i) x=6, y=2
(ii) x=-8, `y=(2)/(3)`
(iii) x=2, y=4
Discussion
202.

Find the matrices of transformation `T_(1)T_(2) and T_(2)T_(1)` when `T_(1)` is rotation through an angle `60^(@) and T_(2)` is the reflection in the Y-asix Also, verify that `T_(1)T_(2)!=T_(2)T_(1).`

Answer» `T_(1)=[(cos60^(@),-sin60^(@)),(sin60^(@),cos60^(@))]=[((1)/(2),-(sqrt3)/(2)),((sqrt3)/(2),(1)/(2))]=(1)/(2)[((1)/(sqrt3),-(sqrt3)/(1))]`
and `T_(2)=[(-1,0),(0,1)]`
`therefore" " T_(1)T_(2)=(1)/(2)[(1,-sqrt3),(sqrt3,1)]xx[(-1,0),(0,1)]=(1)/(2)[(-1+0,0-sqrt3),(-sqrt3+0,0+1)]`
`=(1)/(2)[(1,-sqrt3),(-sqrt3,1)]=[(-(1)/(2),(-sqrt3)/(2)),((-sqrt3)/(2),(1)/(2))]`
`T_(1)T_(2)=[(-1,0),(0,1)]xx(1)/(2)[(1,-sqrt3),(sqrt3,1)]=(1)/(2)[(-1+0,sqrt3+0),(0+sqrt3,0+1)]`
`=(1)/(2)[(-1,sqrt3),(sqrt3,1)]=[(-(1)/(2),(sqrt3)/(2)),((sqrt3)/(2),(1)/(2))]`
it is clear from Eqs. (i) and (ii) , then
`T_(1)T_(2)!=T_(2)T_(1)`
Discussion
203.

The value of a for which system of equation , `a^3x+(a+1)^3y+(a+2)^3z=0, ax+(a+1)y+(a+2)z=0, x+y+z=0,` has a non-zero solution is:A. 2B. 1C. 0D. -1

Answer» Correct Answer - D
Discussion
204.

Let a, b, c be positive real numbers. The following system of equations in x,y and z `(x^(2))/(a^(2)) = (y^(2))/(b^(2)) - (z^(2))/(c^(2)) = 1, (x^(2))/(a^(2)) - (y^(2))/(b^(2)) + (z^(2))/(c^(2)) = 1, - (x^(2))/(a^(2)) + (y^(2))/(b^(2)) + (z^(2))/(c^(2)) = 1` hasA. no solutionB. unique solutionC. infinitely many solutionsD. finitely many solutions

Answer» Correct Answer - b
Discussion
205.

Find the values of `x , y , z`if the matrix `A=[0 2y z x y-z x-y z]`satisfy the equation `A^T A=I_3`.

Answer» It is given that `A=[(0,2y,z),(x,y,-z),(x,-y,z)]`
`:. A^(T)=[(0,x,x),(2y,y,-y),(z,-z,z)]`
Now, `A^(T)A=I`
`implies [(0,x,x),(2y,y,-y),(z,-z,z)][(0,2y,z),(x,y,-z),(x,-y,z)]=[(1,0,0),(0,1,0),(0,0,1)]`
`[(0+x^(2)+x^(2),0+xy-xy,0-xz+xz),(0+xy-xy,4y^(2)+y^(2)+y^(2),2yz-yz-yz),(0-xz+zx,2yz-yz-yz,z^(2)+z^(2)+z^(2))]`
or `[(2x^(2),0,0),(0,6y^(2),0),(0,0,3z^(2))]=[(1,0,0),(0,1,0),(0,0,1)]`
On comparing the corresponding elements, we have
`2x^(2)=1` or `x= pm 1/sqrt(2)`
`6y^(2)=1` or `y= pm 1/sqrt(6)`
`3z^(2)=1` or `z = pm 1/sqrt(3)`
Discussion
206.

Find X and Y,if `2[{:(1,3),(0,x):}]+[{:(y,0),(1,2):}]=[{:(5,6),(1,8):}]`

Answer» Correct Answer - `x+y=6`
Discussion
207.

the matrix `[(0,1),(1,0)]` is the matrix reflection in the lineA. x=1B. x+y=1C. y=1D. x=y

Answer» Correct Answer - D
Discussion
208.

The number of solution of the set of equations `(2x^(2))/a^(2)-(y^(2))/(b^(2))-(z^(2))/(c^(2))=0,(x^(2))/(a^(2))+(2y^(2))/(b^(2))-(z^(2))/(c^(2))=0,(x^(2))/(a^(2))-(y^(2))/(b^(2))+(2z^(2))/(c^(2))=0` isA. 6B. 7C. 8D. 9

Answer» Correct Answer - D
Discussion
209.

If the matrix A is both symmetric and skew-symmetric, show that A is a zero matrix.

Answer»

Given that matrix A is both symmetric and skew symmetric, then,

We have A = A’ ……(i)

And A = -A’ ……(ii)

From (i) and (ii) we get,

A’ = -A’,

2A’ = 0

A’ = 0

Then, A = 0

Hence proved.
Discussion
210.

If `[{:(x-y," "2y),(2y+z,x+y):}]=[{:(1,4),(9,5):}]` then write the value of `(x+y+z)`.

Answer» Correct Answer - 10
Discussion
211.

If `A=[2-1 3 2]`and `B=[0 4-1 7]`, find `3A^2-2B+I`.

Answer» `[{:(4,-20),(38,-10):}]`
Discussion
212.

Compute AB and BA, whichever exists when (i) `A=[{:(2,-1),(3," "0),(-1," "4):}]" and "B=[{:(-2,3),(0,4):}]` (ii) `A=[{:(-1,1),(-2,2),(-3,3):}]" and "B=[{:(" "3,-2," "1),(" "0," "1," "2),(-3," "4,-5):}]` (iii) `A=[{:(0,1,-5),(2,4," "0):}]" and "B=[{:(1,3),(-1,0),(0,5):}]` (iv) `A=[1" "2" "3" "4]" and "B=[{:(1),(2),(3),(4):}]` (v) `A=[{:(2,1),(3,2),(-1,1):}]" and "B=[{:(1,0,1),(-1,2,1):}]`

Answer» (i) `AB=[{:(-4,2),(-6,9),(2,13):}]` and BA does not exist.
(ii) `BA=[{:(-2," "2),(-8," "8),(10,-10):}]` and AB does not exist.
(iii) `AB=[{:(-1,-25),(-2," "6):}]" and "BA=[{:(6,13,-5),(0,-1," "5),(10,20," "0):}]`
(iv) `AB=[30]" and "BA=[{:(1,2,3,4),(2,4,6,8),(3,6,9,12),(4,8,12,16):}]`
(v) `AB=[{:(1,2,3),(1,4,5),(-2,2,0):}]" and "BA=[{:(1,2),(3,4):}]`
Discussion
213.

Show that a matrix which is both symmetric and skew symmetric is a zero matrix.

Answer»

Let A = [aij] be a matrix which is both symmetric and skew symmetric.

Since A is a skew symmetric matrix, so A′ = –A.

Thus for all i and j, we have aij = – aji. .........(1)

Again, since A is a symmetric matrix, so A′ = A.

Thus, for all i and j, we have

aji = aij .............(2)

Therefore, from (1) and (2), we get

aij = –aij for all i and j

or 2aij = 0,

i.e., aij = 0 for all i and j. Hence A is a zero matrix.
Discussion
214.

Find a,p,q and b if `A={:[(a-3,6,b-q),(1,5,9)]={:[(0,2p-q,-4),((q-p)/(2),5,a+b)]:}`.

Answer» Correct Answer - a=3, b=6, p=8 and q=10
Discussion
215.

If `|{:(5, -3),(6, -a):}| =4`, then 5a-4 =_____.

Answer» Correct Answer - B
`|{:(a, b), (c, d):}|= ad-bc`
Discussion
216.

If `{:X=[(3,-4),(1,-1)]:}`, the value of `X^n` is equal toA. `[(3n,-4n),(n,-n)]`B. `[(2n+n,5-n),(n,-n)]`C. `[(3^(n),(-4)^(n)),(1^(n),(-1)^(n))]`D. None of these

Answer» Correct Answer - D
Discussion
217.

If `|{:(2, -3), (p-4, 2p-1):}| = -6`, then p =____.A. `(8)/(7)`B. `(7)/(8)`C. 5D. 0

Answer» Correct Answer - A
`|{:(a, b), (c, d):}|= ad-bc`
Discussion
218.

`"If A" = [{:(1, 2), (1, 3):}]` , then find `A^(-1) + A`. (a) I (b) 2I (c) 3I (d) 4I

Answer» ` A = [{:(1, 2), (1, 3):}], A^(-1) = [{:(3, -2), (-1, 1):}]`
`rArr A^(-1) + A = [{:(4, 0), (0, 4):}] = 4I.`
Discussion
219.

If `({:(1, 0),(0, 1):})({:(a, b+c),(b-c, d):}) = ({:(4, -5),(3, 2):})`, then (a-b) + (c-d)=____.A. -2B. 9C. 2D. -1

Answer» Correct Answer - D
Apply matrix multiplication concept and then equate the corresponding elements.
Discussion
220.

If the system of equations `ax+y=1,x+2y=3,2x+3y=5` are consistent, then a is given by

Answer» Correct Answer - A
Discussion
221.

Find X and Y, if(i) `X+Y=[[7, 0],[ 2, 5]]`and `X-Y=[[3, 0],[ 0 ,3]]`(ii) `2X+3Y=[[2, 3],[ 4, 0]]`and `3X+2Y=[[2,-2],[-1, 5]]`

Answer» (i) `X+Y = [[7,0],[2,5]]->(1)`
`X-Y = [[3,0],[0,3]]->(2)`
Adding (1) and (2),
`=> 2X = [[10,0],[2,8]]`
`=>X = [[5,0],[1,4]]`
`:. Y = [[7,0],[2,5]] - [[5,0],[1,4]]`
`=> y = [[2,0],[1,1]]`

(ii) `2X+3Y = [[2,3],[4,0]]->(1)`
`3X+2Y = [[2,-2],[-1,5]] ->(2)`
Multiplying (1) by `3` and (2) by `2` and then subtracting,
`6X+9Y-6X-4Y = [[6,9],[12,0]] - [[4,-4],[-2,10]]`
`=>5Y = [[2,13],[14,-10]] `
`=>Y = [[2/5,13/5],[14/5,-2]] `
Now, `3X+2Y = [[2,-2],[-1,5]]`
`=>X = 1/3([[2,-2],[-1,5]] - 2 [[2/5,13/5],[14/5,-2]])`
`=X = 1/3[[6/5,-36/5],[-33/5,12]]`
`=X = [[2/5,-12/5],[-11/5,4]].`
Discussion
222.

Find k, if the following matrices are singular. \(\begin{vmatrix}k-1&2&3\\3&1&2\\1&-2&4\end{vmatrix}\)[(k-1, 2, 3) (3, 1, 2) (1, -2, 4)]

Answer»

Let A = \(\begin{vmatrix}k-1&2&3\\3&1&2\\1&-2&4\end{vmatrix}\)

Since A is a singular matrix |A| = 0

∴\(\begin{vmatrix}k-1&2&3\\3&1&2\\1&-2&4\end{vmatrix}\)

∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0 

∴ 8k-8-20-21 =0 

∴ 8k = 49 

∴ k = 49/8
Discussion
223.

Find the values of x, y and z from the following equations:(i) `[[4, 3],[x,5]]=[[y, z],[1, 5]]` (ii) `[[x+y,3],[ 5+z, x y]]=[[6, 2],[ 5 ,8]]` (iii) `[[x+y+z],[ x+z],[ y+z]]=[[9],[ 5],[ 7]]`

Answer» (i) From given matrices, we find that,
`y = 4`
`z = 3`
`x = 1`

(ii)From given matrices, we find that,
`x+y = 6 =>y = 6-x ->(1)`
`5+z = 5=> z =0`
`xy = 8 =>x(6-x) = 8`
`=>6x-x^2 = 8`
`=>x^2-6x+8 = 0`
`=>(x-4)(x-2) = 0`
`=> x = 4 or x = 2`
When `x = 4, y = 2`
When `x = 2, y = 4`
(iii) From given matrices, we find that,
`x+y+z = 9->(1)`
`x+z = 5->(2)`
`y+z = 7->(3)`
From (1) and (3),
`x+(y+z) = 9 => x+7 = 9 => x = 2`
From (1) and (2),
`(x+z)+y = 9 => 5+y = 9 => y = 4`
`4+z = 7=> z = 3`
Discussion
224.

Find k, if the following matrices are singular.\(\begin{vmatrix}4&3&1\\7&k&1\\10&9&1\end{vmatrix}\)[(4, 3, 1) (7, k, 1) (10, 9, 1)]

Answer»

Let A = \(\begin{vmatrix}4&3&1\\7&k&1\\10&9&1\end{vmatrix}\)

Since A is a singular matrix, |A|= 0

∴ \(\begin{vmatrix}4&3&1\\7&k&1\\10&9&1\end{vmatrix}\)

∴ 4(k -9) - 3(7 - 10) + 1(63 - 10k) = 0 

∴ 4k - 36 + 9 + 63 - 10k = 0 

∴ -6k + 36 = 0 

∴ 6k = 36 

∴ k = 6
Discussion
225.

Find k, if the following matrices are singular.\(\begin{vmatrix}7& 3\\-2&k\end{vmatrix}\)[(7, 3) (-2, k)]

Answer»

Let A = \(\begin{vmatrix}7& 3\\-2&k\end{vmatrix}\)

Since A is a singular matrix, |A|=0

∴\(\begin{vmatrix}7& 3\\-2&k\end{vmatrix}\)=0

∴ 7k + 6 = 0

∴ 7k = -6 

k = -6/7
Discussion
226.

Which of the following matrices are singular or non-singular?   \(\begin{vmatrix}7&5\\-4&7\end{vmatrix}\)[(7, 5) (-4, 7)]

Answer»

Let A = \(\begin{vmatrix}7&5\\-4&7\end{vmatrix}\)

∴ |A| = \(\begin{vmatrix}7&5\\-4&7\end{vmatrix}\)= 49 + 20 = 69 ≠ 0
Discussion
227.

Which of the following matrices are singular or non-singular?  \(\begin{vmatrix}3&5&7\\-2&1&4\\3&2&5\end{vmatrix}\)[(3, 5, 7) (-2, 1, 4) (3, 2, 5)]

Answer»

Let A = \(\begin{vmatrix}3&5&7\\-2&1&4\\3&2&5\end{vmatrix}\)

∴ |A| = \(\begin{vmatrix}3&5&7\\-2&1&4\\3&2&5\end{vmatrix}\)

= 3(5 -8) -5(-10 - 12) + 7(-4 - 3) 

= -9 + 110 - 49 = 52 ≠ 0 

∴ A is a non-singular matrix.
Discussion
228.

Find the image of the `(-2,-7)` under the transformations (x,y) to `(x-2y,-3x+y).`

Answer» Let `(x_(1),y_(1)` be the image of the point (x,y) under the given transormations, then
`{(x_(1)=x-2y=1.x+(-2).y),(y_(1)=-3x+y=(-3).x+1.y):}`
`rArr" " [(,x_(1)),(,y_(1))]=[(1,-2),(-3,1)][(,x),(,y)]`
`=[(1,-2),(-3,1)][(,-2),(,-7)][(-2+14),(6-7)]=[(,12),(,-1)]`
therefore, the required image is `(12,-1).`
Discussion
229.

Find the image of the point `(-sqrt2,sqrt2)` by the line mirror` y=x tan ((pi)/(8)).`

Answer» Let`(x_(1),y_(1))` be the image of `(-sqrt2,sqrt2)` about the line `y=xtan((pi)/(8)).`
On comparing `y=x tan ((pi)/(8)) by y = x tan theta`
` therefore" " theta=(pi)/(8)`
Now, `[(,x_(1)),(,y_(1))]=[(cos2theta,sin2theta),(sin2theta,-cos2theta)][(,-sqrt2),(,sqrt2)]`
`[((1)/sqrt2,(1)/sqrt2),((1)/sqrt2,(1)/sqrt2)][(,-sqrt2),(,sqrt2)]=[(,0),(,-2)]`
On comparing `x_(1)=0 and y_(1)=-2`
therefore, the requried image is `(0,-2)`
Discussion
230.

the image of the point `A(2,3)` by the line mirror y=x is the point B and the image of B by the line mirror y=0 is the point `(alpha,beta),` find `alpha and beta`

Answer» Let `B(x_(1),y_(1))` be the image of the point `A(2,3)` about the line y=x, then
`[(,x_(1)),(,y_(1))]=[(0,1),(1,0)][(,2),(,3)]=[(,3),(,2)]`
therefore , the image of A(2,3) by the line mirror y=x is B (3,2).
Given, image of B by the line mirror y=0 (X-axis) is `(alpha,beta),` then
`[(,alpha),(,beta)]=[(1,0),(0,-1)][(,3),(,2)]=[(,3),(,-2)]`
on comparing, we get `alpha=3 andbeta=-2`
Discussion
231.

If `A={:[(a,b),(2,-1)]:},B={:[(1,1),(4,-1)]and(A+B)^(2)=A^(2)+B^(2)`, then (b,a) = ______ .A. (1,-1)B. (-1,1)C. (1,1)D. (-1,0)

Answer» Correct Answer - B
`A+B{:[(a,b),(2,-1)]:}+{:[(1,1),(4,-1)]:}`
`={:[(a+1,b+1),(6,-2)]:}`.
`(A+B)^(2)={:[(a+1,b+1),(6,-2)]:}{:[(a+1,b+1),(6,-2)]:}`
`={:[((a+1)^(2)+6(b+1),(a+1)(b+1)-2(b+1)),(6(a+1)-12,6(b+1)+4)]:}`
`A^(2)={:[(a,b),(2,-1)]:}{:[(a,b),(2,-1)]:}`
`B^(2)={:[(1,1),(4,-1)]:}{:[(1,1),(4,-1)]:}`
`{:[(1+4,1-1),(4-4,4+1)]:}={:[(5,0),(0,5)]:}`.
`A^(2)+B^(2)={:[(5+a^(2)+2b,ab-b),(2a-2,6+2b)]:}`
Given `(A+B)^(2)=A^(2)+B^(2)`
`:.6(a+1)-12=2a-2`
`6a+6-12=2a-2`
`6a-6=2a-2`
4a=4
6(b+1)+4=6+2b
6b+6+4=6+2b
4b=-4
b=-1.
(b,a)=(-1,1).
Discussion
232.

If `A={:[(7,9),(3,8)]:}andB={:[(13),(18)]:}` then find the matrix X such that AX=B.A. `{:[(-2),(-3)]:}`B. `{:[(2),(3)]:}`C. `{:[(-2),(3)]:}`D. `{:[(2),(-3)]:}`

Answer» Correct Answer - C
`"Let X"={:[(x),(y)]:}`
`AX=BrArr{:[(7,9),(3,8)]:}{:[(x),(y)]:}={:[(13),(18)]:}`
`{:[(7x+9y),(3x+8y)]:}={:[(13),(18)]:}`
`7x+9y=13` (1)
`3x+8y=18` (2)
Solving Eqs. (1) and (2), we get
x=-2 and y=3
`:.x={:[(-2),(3)]:}`.
Discussion
233.

If `A+B={:[(8,-9),(3,5)]:}andA-B={:[(6,5),(1,7)]:}`, the A= _______ .A. `{:[(7,-2),(2,6)]:}`B. `{:[(7,-2),(-2,6)]:}`C. `{:[(-7,2),(2,6)]:}`D. `{:[(-7,-2),(-2,-6)]:}`

Answer» Correct Answer - A
`A+B={:[(8,-9),(3,5)]:}` (1)
`A-B={:[(6,5),(1,7)]:}` (2)
Adding Eqs. (1) and (2) we get
`A+B+A-B={:[(8,-9),(3,5)]:}+{:[(6,5),(1,7)]:}`
`2A={:[(14,-4),(4,12)]:}`
`A={:[((14)/(2),(-4)/(2)),((4)/(2),(12)/(2))]:}={:[(7,-2),(2,6)]:}`.
Discussion
234.

If `x={: ((2,-1),(3,0)) :}andg(x)=x^(2)+x-2`, find g(x).

Answer» Correct Answer - `{:[(1,-3),(9,-5)]:}`
Discussion
235.

If`[5" "6]{:[(1,-3),(-1,x)]:}{:[(3),(4)]:}=[-15]`, then the value of x is __________ .A. 4B. 2C. 3D. 1

Answer» Correct Answer - B
`[5" "6]{:[(1,-3),(-1,x)]:}{:[(3),(4)]:}=[-15]`
`[5-6-15+6x]{:[(3),(4)]:}=[=15]`
`[-1-15+16x]{:[(3),(4)]:}=[15]`
`[-3+4(6x-15)]=[-15]`
`[-3+24x-60]=[-15]`
`[24x-63]=[-15]`
24x-63=-15
24x=63-15
24x=48
x=2.
Discussion
236.

If `A-B^(T)={:[(-2,2,1),(3,6,5)]:}andA^(T)+B={:[(-4,5),(8,-2),(9,1)]:}`, then matrix A = _______ .A. `{:[(3,-5,-5),(6,2,3)]:}`B. `{:[(2,6,5),(3,5,7)]:}`C. `{:[(9,5,7),(3,2,1)]:}`D. `{:[(-3,5,5),(4,2,3)]:}`

Answer» Correct Answer - D
`A-B^(T)={:[(-2,2,1),(3,6,5)]:}`
`(A-B^(T))^(T)={:[(-2,3),(2,6),(1,5)]:}`
`A^(T)-B={:[(-2,3),(2,6),(1,5)]:}` (1)
`andA^(T)+B={:[(-4,5),(8,-9),(9,1)]:}` (2)
Add Eqs. (1) and (2),
`A^(T)-B+A^(T)+B={:[(-2,3),(2,6),(1,5)]:}+{:[(-4,5),(8,-2),(9,1)]:}`
`2A^(T)={:[(-6,8),(10,4),(10,6)]:}`
`A^(T)={:[(-3,4),(5,2),(5,3)]:}`
`A={:[(-3,5,5),(4,2,3)]:}`
Discussion
237.

If `A={:[(1,-4),(3,2)]:}andB={:[(6,-1),(3,2)]:}`, then find the matrix C such that `C=AB+B^(2)`.A. `{:[(37,25),(16,9)]:}`B. `{:[(27,-17),(48,2)]:}`C. `{:[(4,5),(17,16)]:}`D. `{:[(3,-15),(6,18)]:}`

Answer» Correct Answer - B
`AB={:[(1,-4),(3,2)]:}={:[(6,-1),(3,2)]:}`
`={:[(-6,-9),(24,1)]:}`.
`B^(2)={:[(6,-1),(3,2)]:}={:[(6,-1),(3,2)]:}`
`={:[(36-3,-6-2),(18+6,-3+4)]:}`
`={:[(33,-8),(24,1)]:}`
`C=AB+B^(2)`
`={:[(-6+33,-9-8),(24+24,1+1)]:}`
`={:[(27,-17),(48,2)]:}`
Discussion
238.

If `5{:[(-3,1),(x,2)]:}+{:[(y,4),(3,2)]:}={:[(-15,9),(6,z)]:}`, then find x,y,z.

Answer» Correct Answer - `x=(3)/(5),y=0andz=12`
Discussion
239.

If `A={:[(2,-3),(5,7)]:}andB={:[(20,31),(44,53)]:}`, then find the matrix X such that A+X=B.

Answer» Correct Answer - `{:[(18,34),(39,46)]:}`
Discussion
240.

If `A+B={:[(7,6),(-3,2)]:}andA-B={:[(1,2),(3,6)]:}`, then find A.A. `{:[(4,4),(0,4)]:}`B. `{:[(8,8),(0,8)]:}`C. `{:[(-4,-4),(0,-4)]:}`D. `{:[(-8,-8),(0,-8)]:}`

Answer» Correct Answer - A
(i) Add the given matrix equations.
(ii) Add(A-B) to (A+B) to find matrix A.
Discussion
241.

If `2A-3B={:((-27,4,5),(7,6,-15)):}and5A-2B={:((-40,-1,18),(12,15,-21)):}`, then B=A. `{:((55,-22,-11),(-11,0,33)):}`B. `{:((5,-2,-1),(-1,0,3)):}`C. `{:((-55,22,11),(11,0,-33)):}`D. `{:((-5,-2,1),(1,0,3)):}`

Answer» Correct Answer - B
(i) Solve as linear equations.
(ii) Multiply 2A-3B by 5.
Multiply 5A-2B by 2 and then add to get matrix B.
Discussion
242.

If a matrix has 15 elements, then find the possible order of the matrix.

Answer» Correct Answer - `1xx15,15xx1,3xx5,5xx3`
Discussion
243.

Find the possible square roots of the two-rowed unit matrix I.

Answer» Let `A=[(a,b),(c,d)]` be a square root of the matrix `I`.
Then `A^(2)=I`, where `I=[(1,0),(0,1)]`
`implies [(a,b),(c,d)][(a,b),(c,d)]=[(1,0),(0,1)]`
`implies [(a^(2)+bc,ab+bd),(ac+cd,cb+d^(2))]=[(1,0),(0,1)]`
`implies a^(2)+bc=1` (1)
`ab+bd=0` (2)
`ac+cd=0` (3)
`cb+d^(2)=1` (4)
Case I : `a+d=0`
The above four equations hold simultaneously if `d=-a` and `a^(2)+bc=1`.
Hence, one possible square roots of `I` is `A=[(alpha,beta),(gamma,-alpha)]`, where `alpha, beta, gamma` are the three numbers related by the condition `alpha^(2)+beta gamma=1`.
Case II : `a+d+ ne 0`
The above four equation hold simultaneously
if `b=0, c=0, a=1, d=1` or if `b=0, c=0, a=-1, d=-1`
Hence, `[(1,0),(0,1)], [(-1,0),(0,-1)]` i.e., `pm I` are other possible equare roots of I.
Discussion
244.

`A=[1t a n x-t a n x1]a n df(x)`is defined as `f(x)=d e tdot(A^T A^(-1))`en the value of `(f(f(f(ff(x))))_`is `(ngeq2)`_________.

Answer» Correct Answer - 2
`because A = [[1,tan x],[-tan x,1]]`
`therefore det A + [[1, tan x],[-tan x, 1]]= (1+ tan^(2)x) = ""^(2)x`
`rArr det A^(T) = det A = sec^(2) x `
Now, `f(x) = det (A^(T) A^(-1)) = (det A^(T)) (detA^(-1))`
`= ( det A^(T) ) (det A)^(-1) = (det(A^(T)))/(detA)= 1 `
`therefore underset("n times")(underbrace(lambda = f(f(f(f...f(x))))))=1 [because f(x) = 1]`
Hence, `2^(lambda)= 2^(1) = 2`
Discussion
245.

If `B=[(2,3),(-4,3)]` , then co-factor `a_(21)=`A. 2B. `-3`C. `-4`D. 3

Answer» Correct Answer - B
Discussion
246.

If `A=[{:(" 1",1," 1"),(" 2", 1," -3"),(-1,2," 3"):}]` , then co-factor of `3^(rd)` row areA. `4, -5, 1`B. `-4, 5, -1`C. `-4, 5, 1`D. `4, 5 , -1`

Answer» Correct Answer - B
Discussion
247.

find the cofactor of` a_(23)` in `[(3,1,4),(0,2,-1),(1,-3,5)]`

Answer» Let `A=[(3,1,4),(0,2,-1),(1,-3,5)]`
`therefore" cofactor of" " "a_(23) =-D`
where `D=|(3,1),(1,-3)|`
[after crossing the 2nd row and 3rd column]
`=-9-1=-10`
Hence, cofactor of `a_(23)=-(-10)=10`
Discussion
248.

If `A=[{:(2,-1," 3"),(4," 2"," 5"),(0," 4",-1):}]` , then cofactor `A_(32)` isA. `-2`B. `-8`C. 4D. 2

Answer» Correct Answer - D
Discussion
249.

If the matrix `A = [[lambda_(1)^(2), lambda_(1)lambda_(2), lambda_(1) lambda_(3)],[lambda_(2)lambda_(1),lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]]` is idempotent, the value of `lambda_(1)^(2) + lambda_(2)^(2) + lambda _(3)^(2)` is

Answer» Correct Answer - 1
`because A^(2) = Acdot A = [[lambda_(1)^(2), lambda_(1) lambda_(2), lambda_(1) lambda_(3) ],[lambda_(2)lambda_(1), lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]] [[lambda_(1)^(2), lambda_(1) lambda_(2), lambda_(1) lambda_(3) ],[lambda_(2)lambda_(1), lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]]`
` = [[lambda_(1)^(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)), lambda_(1) lambda_(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)), lambda_(1) lambda_(3)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)) ],[lambda_(2)lambda_(1)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)), lambda_(2)^(2) (lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)),lambda_(2)lambda_(3)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2))],[lambda_(3)lambda_(1)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)),lambda_(3)lambda_(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)),lambda_(3)^(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2))]]`
`= (lambda_(1)^(2) + lambda_(2)^(2) + lambda_(3)^(2) ) A`
Given, A is idempotent
`rArr A^(2) = A`
`= lambda_(1)^(2) + lambda_(2)^(2) + lambda_(3)^(2) = 1`
Discussion
250.

Let A be a `3xx3` matrix given by ` A = [a_(ij)].` If for every column vector `X, X^(T)AX=O and a_(23)=-1008, ` the sum of the digits of `a_(32)` is

Answer» Correct Answer - 9
Let `X= [[x_(1)],[x_(2)],[x_(3)]]` and given `X^(T) AX = O`
`rArr [x_(1) x_(2) x_(3)] [[a_(11), a_(12), a_(13)],[a_(21),a_(22),a_(23)],[a_(31), a_(32), a_(33) ]][[x_(1)],[x_(2)],[x_(3)]]= O`
`rArr [x_(1) x_(2) x_(3)] [[a_(11)x_(1)+ a_(12)x_(2)+ a_(13)x_(3)],[a_(21)x_(1)+a_(22)x_(2)+a_(23)x_(3)],[a_(31)x_(1)+ a_(32)x_(2)+ a_(33)x_(3) ]]= O`
` a_(11)x_(1)^(2)+ a_(12)x_(1)x_(2)+ a_(13)x_(1)x_(3)+a_(21)x_(1)x_(2)+a_(22)x_(2)^(2)+a_(23)x_(2)x_(3)`
`+a_(31)x_(1) x_(3) + a_(32)x_(2)x_(3)+a_(33) x_(3)^(2)= 0`
`rArr a_(11) x_(1)^(2) + a_(22) x_(2) ^(2) + a_(33) x_(3) ^(2) + (a_(12)+ a_(21))x_(1) x_(2) + (a_(23) + a_(32)) x_(2) x_(3) + (a_(31)+a_(13)) x_(3) x_(1) = 0`
it is true for every `x_(1), x_(2), x_(3)` then
`a_(11) = a_(22) =a_(33) = 0 and a_(12) = -a_(21), a_(23)=-a_(32),a_13=-a_(31)`
Now , as `a_(23) = - 1008 rArr a_(32) = 1008` `therefore ` Sum of digits `= 1+ 0 + 0 + 8 =9`
Discussion