1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 4 InterviewSolution

151.	3x-4y+2z=-1 2x+3y+5z=7 x+z=2A. 3B. 2C. 1D. `-2`
Answer» Correct Answer - A

Discussion

152.	Given that matrix `A[(x,3,2),(1,y,4),(2,2,z)]`. If `xyz=60` and `8x+4y+3z=20`, then A(adj A) is equal toA. `[(64,0,0),(0,64,0),(0,0,64)]`B. `[(88,0,0),(0,88,0),(0,0,88)]`C. `[(68,0,0),(0,68,0),(0,0,68)]`D. `[(34,0,0),(0,34,0),(0,0,34)]`
Answer» Correct Answer - C A adj. `A=\|A\| I` `\|A\|=xyz-8x-3(z-8)+2(2-2y)` `\|A\|=xyz-(8x+3z+4y)+28` `=60-20+28=68` `implies` A(adj. A)`=68 [(1,0,0),(0,1,0),(0,0,1)]=[(68,0,0),(0,68,0),(0,0,68)]`

Discussion

153.	If a,b,c, and c, are the roots of `x^(2)-4x+3=0,x^(2)-8x+15=0` and `x^(2)-6x+5=0,` `[{:(a^(2),+c^(2),a^(2)+b^(2)),(b^(2),+c^(2),a^(2)+c^(2)):}]+[{:(2ac,-2ab),(-2bc,-2ac):}]`
Answer» `therefore" " x^(2)-4x+3=0` `rArr" "(x-1)(x-3)=0 " " therefore x=1,3` `x^(2)-8x+15=0` `rArr " " (x-3)(x-5)=0 " "therefore x=3,5` and `x^(2)-6x+5=0` `rArr" " (x-5)(x-1)=0 " " therefore x=5,1` it is clear that a = 1,3 and c=5 Now, `[(a^(2)+c^(2),a^(2)+b^(2)),(b^(2)+c^(2),a^(2)+c^(2))]+[(2ac,-2ab),(-2bc,-2ac)]` `[(a^(2)+c^(2)+2ac,a^(2)+b^(2)-2ab),(b^(2)+c^(2)-2bc,a^(2)+c^(2)-2ac)]+[(2ac,-2ab),(-2bc,-2ac)]` `[((1+5)^(2),(1-3)^(2)),((3-5)^(2),(1-5)^(2))]=[(36,4),(4,16)]`

Discussion

154.	If `A=[-1 1 0-2]`, then prove that `A^2+3A+2I=Odot`Hence, find `Ba n dC`matrices of order 2 with integer elements, if `A=B^3+C^3dot`
Answer» `A=[(-1,1),(0,-2)]` `implies A^(2)=[(-1,1),(0,-2)][(-1,1),(0,-2)]=[(1,-3),(0,4)]` `implies A^(2)+3A+2I` `=[(1,-3),(0,4)]+3[(-1,1),(0,-2)]+2[(1,0),(0,1)]=[(0,0),(0,0)]` `implies A^(2)+3A+2I =O` (1) From (1), `A^(3)+3A^(2)+2A=O` `implies (A+I)^(3)-A=I^(3)` `implies A=(A+I)^(3)-I^(3)=(A+I)^(3)+(-I)^(3)` `implies B=A+I` and `C=-I` `:. B=[(-1,1),(0,-2)]+[(1,0),(0,1)]=[(0,1),(0,-1)]` and `C=[(-1,0),(0,-1)]`

Discussion

155.	If A is an invertible matrix of order 3 and \|A\|= 5, then find \|adj A\|.
Answer» \|adj A\| = \|A\|^{3 – 1} = 5² = 25 [∵\| adj A\| = \|A\|^{n – 1}, where n is order of A]

155.

If A is an invertible matrix of order 3 and |A|= 5, then find |adj A|.

Answer»

|adj A| = |A|^{3 – 1} = 5² = 25

[∵| adj A| = |A|^{n – 1}, where n is order of A]

Discussion

156.	Out of the following matrices, choose that matrix which is a scalar matrix :A. \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) B. \(\begin{bmatrix} 0 &0& 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] \end{bmatrix}\) C. \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] 0 & 0 \end{bmatrix}\) D. \(\begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}\)
Answer» (A). \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) ∵ Scalar Matrix is a matrix whose all off-diagonal elements are zero and all on-diagonal elements are equal. \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) Option (A) is the answer.

156.

Out of the following matrices, choose that matrix which is a scalar matrix :A. \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\) B. \(\begin{bmatrix} 0 &0& 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] \end{bmatrix}\) C. \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] 0 & 0 \end{bmatrix}\) D. \(\begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}\)

Answer»

(A). \(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\)

∵ Scalar Matrix is a matrix whose all off-diagonal elements are zero and all on-diagonal elements are equal.

\(\begin{bmatrix} 0 &0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}\)

Option (A) is the answer.

Discussion

157.	Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix.
Answer» Let a be a square matric. Then, `A=1/2 (A+A^(T))+1/2 (A-A^(T))` Here, `P=1/2 (A+A^(T))` is a symmetric matrix and `Q=1/2 (A-A^(T))` is a skew-symmetric matrix. Thus, matrix A can be expressed as sum of symmetric and skew-symmetric matrix.

Discussion

158.	Solve the system of equations `x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1` with the help of matrix inversion.
Answer» we have `x+2y+3z=1,2x+3y+2z=2 and 3x+3y+4z=1` the given system of equation in the matrix form are written as below. `[(1,2,3),(2,3,2),(3,3,4)]=[(,1),(,2),(,1)]` ` AX=B` `rArr X=A^(-1)B` where` " " A=[(1,2,3),(2,3,2),(3,3,4)],X=[(,x),(,y),(,z)]and B= [(,1),(,2),(,1)]` `\|A\|=1(12-6)-2(8-6)+3(6-9)` ` =6-4-9=-7!=0` ` therefore A^(-1)` exixts and has unique solution. Let C be the matrix of confactor of elements in `\|A\|.` Now, confactor along `R_(1)=6,-2,-3` confactor along `R_(2)=1,-5,3` and confacators along `R_(3)=-5,4,-1` `therefore " " C=[(6,-2,-3),(1,-5,3),(-5,4,-1)]` `therefore" " A=C^(T)` `rArr " " adjA=[(6,-2,-3),(1,-5,3),(-5,4,-1)]=[(6,1,-5),(-2,-5,4),(-3,3,-1)]` `rArr" " A^(-1)=(adjA)/(\|A\|)=-(1)/(7)[(6,1,-5),(-2,-5,4),(-3,3,-1)]` `=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))]` from Eq. (i),`X=A^(-1)B` `rArr " " [(,x),(,y),(,z)]=[(-(6)/(7),-(1)/(7),(5)/(7)),((2)/(7),(5)/(7),-(4)/(7)),((3)/(7),-(3)/(7),(1)/(7))][(,1),(,2),(,1)]=[(-(3)/(7)),((8)/(7)),(-(2)/(7))]` Hence `x=-(3)/(7),y=(8)/(7) and z=-(2)/(7)` is the required solution.

Discussion

159.	Determine whether the following equations will have non-trivial solutions, if so solve them: `x+3y-2z=0, 2x-y+4z=0,x-11y+14z=0`
Answer» We have, `x+3y-2z=0` ` 2x-y+4z=0` `x-11y+14z=0` The given system of equations in the matrix form are written as below. `[(1,3,-2),(2,-1,4),(1,-11,14)][(,x),(,y),(,z)]=[(,0),(,0),(,0)]` AX=O where `A=[(1,3,-2),(2,-1,4),(1,-11,14)],X=[(,x),(,y),(,z)]and O=[(,0),(,0),(,0)]` `therefore " " \|A\|=1(-14+44)-3(28-4)-2(-22+1)` `=30-72+42=0` and therefore the system has a non-trival solution, Now, we any write first two of the given equations `x+3y=2z and 2x-y=-4z` solving these equations in terms of z, we get Putting `x=-(10)/(7)z-(8)/(7)z` in third equation of the given system, we get , `LHS=-(10)/(7)z-(88)/(7)z=0=RHS` Now, if `z=7k,then x=-10k and y=8k.` Hence, `x=-10k,y=8k and z=7k` (where k is arbitary) are the requried solutions.

Discussion

160.	If `[a b c1-a]`is an idempotent matrix and `f(x)=x-^2=b c=1//4`, then the value of `1//f(a)`is ______.
Answer» Correct Answer - 0.25 `[(a,b),(c,1-a)]` is an idempotent matrix. `implies [(a,b),(c,1-a)]^(2)=[(a,b),(c,1-a)]` or `[(a,b),(c,1-a)][(a,b),(c,1-a)]=[(a,b),(c,1-a)]` or `[(a^(2)+bc,ab+b-ab),(ac+c-ac,bc+(1-a)^(2))]=[(a,b),(c,1-a)]` or `[(a^(2)+bc,b),(c,bc+(1-a)^(2))]=[(a,b),(c,1-a)]` or `a^(2)+bc=a` `a-a^(2)=bc=1//4` (given) `f(a)=1//4`

Discussion

161.	If B is an idempotent matrix, and `A=I-B`, thenA. `A^(2)=A`B. `A^(2)=I`C. `AB=O`D. `BA=O`
Answer» Correct Answer - A::C::D B is an idempotent matrix `:. B^(2)=B` Now, `A^(2)=(I-B)^(2)` `=(I-B) (I-B)` `=I-IB-IB+B^(2)` `=I-B-B+B^(2)` `=I-2B+B^(2)` `=I-2B+B` `=I-B` `=A` Therefore, A is idempotent. Again, `AB=(I-B)B=IB-B^(2)=B-B^(2)=B^(2)=B^(2)-B^(2)=O` Similarly, `BA=B(I-B)=BI-B^(2)=B-B=O`.

Discussion

162.	If Z is an idempotent matrix, then `(I+Z)^(n)`A. `I+2^(n)Z`B. `I+(2^(n)-1) Z`C. `I-(2^(n)-1)Z`D. none of these
Answer» Correct Answer - B Z is idempotent, then `Z^(2)=Zimplies Z^(3), Z^(4), ..., Z^(n)=Z` `:. (I+Z)^(n)=^(n)C_(0)I^(n)+ .^(n)C_(1)I^(n-1) Z+.^(n)C_(2)I^(n-2)Z^(2)+ +^(n)C_(n)Z^(n)` `=^(n)C_(0)I+ .^(n)C_(1)Z+ .^(n)C_(2)Z+ .^(n)C_(3)Z+ +^(n) C_(n)Z` `=I +(.^(n)C_(1)+.^(n)C_(2)+.^(n)C_(3)+ +.^(n)C_(n))Z` `=I+(2^(n)-1)Z`

Discussion

163.	If `A`is an idempotent matrix satisfying, `(I-0. 4 A)^(-1)=I-alphaA ,w h e r eI`is the unit matrix of the name order as that of `A ,`then th value of `\|9alpha\|`is equal to ________.
Answer» Correct Answer - 6 Given `A^(2)=A` Now `I=(I-0.4A) (I-alphaA)` `=I-IalphaA-0.4 AI+0.4 alpha A^(2)` `=I-Aalpha-0.4A+0.4 alpha A` `= I-A(0.4+alpha) + 0.4 alpha A` `implies 0.4 alpha=0.4+alpha` `implies alpha=-2//3` `implies \|9alpha\|=6`

Discussion

164.	the upper triangular matrix of the matrix `[{:(1,-1,2), (2," 1",3),(3, " 2",4):}]` isA. `[{:(1,-1," 2"), (0," 3",-1),(0," 0",(-1)/(3)):}]`B. `[{:(1,1,-2), (0, 3,-1),(0, 0,(-1)/(3)):}]`C. `[{:((-1)/(3)," 0",0), (" 3", -1,0),(-1, " 2",0):}]`D. `[{:(1," 1"," 2"), (0, -3,-1),(0, " 0",(-1)/(3)):}]`
Answer» Correct Answer - A

Discussion

165.	Let `A=[3x^2 1 6x],B=[abc],a n dC=[(x+2)^2 5x^2 2x5x^2 2x(x+2)^2 2x(x+2)^2 5x^2]`be three given matrices, where `a ,b ,ca n dx in Rdot`Given that `f(x)=a x^2+b x+c ,`then the value of `f(I)`is ______.
Answer» Correct Answer - 7 We have `AB=[(3ax^(2),3bx^(2),3cx^(2)),(a,b,c),(6ax,6bx,6cx)]` Now tr. `(AB)=` tr. (C) `implies 3ax^(2)+b+6cx=(x+2)^(2)+2x+5x^(2) AA x in R` (Identity) `implies 3ax^(2)+6cx+b=6x^(2)+6x+4` `implies a=2, c=1, b=4`

Discussion

166.	If A is a diagonal matrix of order `3xx3` is commutative with every square matrix of order `3xx3` under multiplication and trace (A)=12, then
Answer» Correct Answer - 64 A diagonal matrix is commutative with every square matrix if it is scalar matrix, so every diagonal element is 4. Therefore, `\|A\|=64`

Discussion

167.	Let `A=[a_("ij")]_(3xx3)` be a matrix such that `A A^(T)=4I` and `a_("ij")+2c_("ij")=0`, where `C_("ij")` is the cofactor of `a_("ij")` and `I` is the unit matrix of order 3. `\|(a_(11)+4,a_(12),a_(13)),(a_(21),a_(22)+4,a_(23)),(a_(31),a_(32),a_(33)+4)\|+5 lambda\|(a_(11)+1,a_(12),a_(13)),(a_(21),a_(22)+1,a_(23)),(a_(31),a_(32),a_(33)+1)\|=0` then the value of `lambda` is
Answer» Correct Answer - 0.4 Given that `A A^(T)=4I` `implies \|A\|^(2)=4` or `\|A\|= pm 2` So `A^(T)=4A^(-1)=4 ("adj A")/(\|A\|)` `implies [(a_(11),a_(21),a_(31)),(a_(12),a_(22),a_(32)),(a_(13),a_(23),a_(33))]=4/(\|A\|)[(c_(11),c_(21),c_(31)),(c_(12),c_(22),c_(32)),(c_(13),c_(23),c_(33))]` Now `a_("ij")=4/(\|A\|) c_("ij")` `implies -2c_("ij")=4/(\|A\|) c_("ij")" "("as "a_("ij")+2c_("ij")=0)` `implies \|A\|=-2` Now `\|A+4I\|=\|A+A A^(T)\|` `=\|A\|\|I+A^(T)\|` `=-2\|(I+A)^(T)\|` `=-2\|I+A\|` `implies \|A+4I\|+2\|A+I\|=0`, so on comparing, we get `5 lambda=2 implies lambda=2/5`

Discussion

168.	If A=`[{:(1," 2",-1),(3,-2," 5"):}]` , then `R_(1) harr R_(2)` and `C_(1) rarr C_(1) + 2C_(3)` givenA. `[{:(-13,2,-5),(-1,2,-1):}]`B. `[{:(-1,-2,-1),(" 13",-2," 5"):}]`C. `[{:(" 13",-2," 5"),( -1, " 2", -1):}]`D. `[{:(2,-13,-5),(2,-1,-1):}]`
Answer» Correct Answer - C

Discussion

169.	If `A=[(2,-3,3),(2,2,3),(3,-2,2)] ` then `C_(2)rarrC_(2)+2C_(1)` and then `R_(1)rarrR_(1)+R_(3)` givesA. `[(2,1,3),(2,6,3),(3,4,2)] `B. `[(2,-1,3),(2,4,3),(3,1,2)] `C. `[(5,5,5),(3,4,3),(2,6,3)] `D. `[(5,5,5),(2,6,3),(3,4,2)] `
Answer» Correct Answer - D

Discussion

170.	`[(1,3,-2),(-3,0,-5),(2,5,0)] =A [(1,0,0),(0,1,0),(0,0,1)]` then, `C_(2) rarr C_(2)-3C_(1) and C_(3) rarr C_(3) +2C_(1)` givesA. `[(1,0,0),(3,-9,11),(-2,1,4)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]`B. `[(1,0,0),(-3,-1,-4),(2,-9,11)] =A [(1,-3,2),(0,1,0),(0,0,1)]`C. `[(1,0,0),(-3,1,4),(-2,9,-11)] =A [(-1,3,2),(0,-1,0),(0,0,-1)]`D. `[(1,0,0),(-3,9,-11),(2,-1,4)] =A [(1,-3,2),(0,1,0),(0,0,1)]`
Answer» Correct Answer - D

Discussion

171.	If A is a matrix of order n such that `A^(T)A=I` and X is any matric such that `X=(A+I)^(-1) (A-I)`, then show that X is skew symmetric matrix.
Answer» We have `X=(A+I)^(-1) (A-I)` `:. X^(T)=((A+I)^(-1) (A-I))^(T)` `=(A^(T)-I)(A^(T)+I)^(-1)` (1) Now, `(A^(T)+I)(A^(T)-I)=(A^(T)-I) (A^(T)+I)` `:. (A^(T)+I)^(-1) (A^(T)+I) (A^(T)-I) (A^(T)+I)^(-1)` `=(A^(T)+I)^(-1) (A^(T)-I) (A^(T)+I) (A^(T)+I)^(-1)` `implies (A^(T)-I) (A^(T)+I)^(-1)=(A^(T)+I)^(-1) (A^(T)-I)` (2) From (1) and (2), we have `X^(T)=(A^(T)+I)^(-1) (A^(T)-I)` `=(A^(T)+A A^(T))^(-1) (A^(T)-A A^(T))` `=(A^(T) (I+A))^(-1) (A^(T) (I-A))` `=(I+A)^(-1) (A^(T))^(-1) (A^(T) (I-A))` `=(I+A)^(-1) ((A^(-1))^(T)A^(T)) (I-A)` `=(I+A)^(-1) (I-A)` `=-(A+I)^(-1) (A-I)` `=-X`

Discussion

172.	If matric A is skew-symmetric matric of odd order, then show that tr. A = det. A.
Answer» Since A is skew-symmetric matrix, tr. `A=0`. Also, order of A is odd, then det. `A=0`. Thus, tr. `A=` det. A

Discussion

173.	If `D=diag(d_1,d_2,d_3,…,d_n)" where "d ne 0" for all " I = 1,2,…,n," then " D^(-1)`is equal toA. DB. `diag(d_1^(-1)d_2^(-1),...,d_n^(-1))`C. InD. none of these
Answer» Correct Answer - B

Discussion

174.	The value of x, for which the matrix A = \(\begin{bmatrix}e^{x - 2} &e^{7 + x} \\[0.3em]e^{2 + x}&e^{2x + 3}\end{bmatrix}\) is singular is …(a) 9(b) 8 (c) 7 (d) 6
Answer» (b) 8 Given A is a singular matrix ⇒ \|A\| = 0 (i.e.) \(\begin{bmatrix}e^{x - 2} &e^{7 + x} \\[0.3em]e^{2 + x}&e^{2x + 3}\end{bmatrix}\)= 0 ⇒ e^{x - 2}.e^{2x + 3} – e^{2 + x}.e^{7 + x} = 0 ⇒ e^{3x + 1} – e^{9 + 2x} = 0 ⇒ e^{3x + 1} = e^{9 + 2x} ⇒ 3x + 1 = 9 + 2x 3x – 2x = 9 – 1 ⇒ x = 8

174.

The value of x, for which the matrix A = \(\begin{bmatrix}e^{x - 2} &e^{7 + x} \\[0.3em]e^{2 + x}&e^{2x + 3}\end{bmatrix}\) is singular is …(a) 9(b) 8 (c) 7 (d) 6

Answer»

(b) 8

Given A is a singular matrix ⇒ |A| = 0

(i.e.) \(\begin{bmatrix}e^{x - 2} &e^{7 + x} \\[0.3em]e^{2 + x}&e^{2x + 3}\end{bmatrix}\)= 0

⇒ e^{x - 2}.e^{2x + 3} – e^{2 + x}.e^{7 + x} = 0

⇒ e^{3x + 1} – e^{9 + 2x} = 0

⇒ e^{3x + 1} = e^{9 + 2x}

⇒ 3x + 1 = 9 + 2x

3x – 2x = 9 – 1 ⇒ x = 8

Discussion

175.	If ` [(7, -6), (8, -7)]^2008 = [(7, -6), (8, -7)]^2010` then `[(7, -6), (8, -7)]^2009` isA. `({:(7, -6), (8, -7):})^(2008) + ({:(7, -6), (8, -7):})^(2010)`B. `(1)/(2) [({:(7, -6), (8, -7):})^(2010) -({:(7, -6), (8, -7):})^(2008)]`C. `({:(1, 0), (0, 1):})`D. `({:(7, -6), (8, -7):})`
Answer» Correct Answer - D (i) Let `A =({:(7, -6), (8, -7):})` (ii) Calculate `A^(2)`

Discussion

176.	If `A = ({:(5, 5), (0, 0):})({:(0, 0), (5, 5):}) " and " A^(n) = ({:(5^(200), 5^(200)), (0, 0):})`, then find n.A. 100B. 50C. 25D. None of these
Answer» Correct Answer - A (i) Simplify the matrix A. (ii) Find `A^(2), A^(3),….., A^(n)`.

Discussion

177.	If `A = ({:(2, 3), (6, 9):})`, then find \|A\|.
Answer» Correct Answer - A Given `A = [{:(2, 3), (6, 9):}], \|A\| = \|{:(2, 3), (6, 9):}\| = (18-18)` `=0 ( therefore \|{:(a, b), (c, d):}\| = (ad-bc)).`

Discussion

178.	If `A = [{:(x^(2), y^(2)), ("log"_(1024)a, -9):}],a =16^(25) " and if " A = A^(-1), " then "[{:(x^(2), y), (1, x^(2) + y):}]^(-1)` = _____.A. `(1)/(65)[{:(7, 2),(-1, 9):}]`B. `(1)/(65)[{:(7, -2),(1, 9):}]`C. `(1)/(65)[{:(7, 2),(1, 9):}]`D. `(1)/(65)[{:(9, 2),(-1, 7):}]`
Answer» Correct Answer - A (i) Put `a = 16^(25) " in log"_(1024)` a and simplify. (ii) Then, find x and y values using the relation `A = A^(-1)`. (iii) Find the inverse of the given matrix.

Discussion

179.	The number of integral values of x for which the determinant of the matrix `[(5x+14,-2),(7x+8,x)]` is alwats less than 1 isA. 3B. 4C. 5D. 6
Answer» Correct Answer - B (i) Find the deteminant of matrix. (ii) Solve the inequation.

Discussion

180.	State whether the statements is true or false: AB = AC ⇒ B = C for any three matrices of same order.
Answer» Answer is False

Discussion

181.	In the matrix A = \(\begin{bmatrix} a&1&x \\[0.3em] 2&\sqrt3&x^2 - y \\0&5&-\frac{2}{5} \\[0.3em] \end{bmatrix}\), write(i) The order of the matrix A(ii) The number of elements(iii) Write elements a23, a31, a12
Answer» For the given matrix, (i) The order of the matrix A is 3 x 3. (ii) The number of elements of the matrix = 3 x 3 = 9 (iii) Elements: a₂₃ = x² – y, a₃₁ = 0, a₁₂ = 1

181.

In the matrix A = \(\begin{bmatrix} a&1&x \\[0.3em] 2&\sqrt3&x^2 - y \\0&5&-\frac{2}{5} \\[0.3em] \end{bmatrix}\), write(i) The order of the matrix A(ii) The number of elements(iii) Write elements a23, a31, a12

Answer»

For the given matrix,

(i) The order of the matrix A is 3 x 3.

(ii) The number of elements of the matrix = 3 x 3 = 9

(iii) Elements: a₂₃ = x² – y, a₃₁ = 0, a₁₂ = 1

Discussion

182.	If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?
Answer» We know that in mathematics, a matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second. We are given with a matrix that has 28 elements. We know that, If a matrix has mn elements, then the order of the matrix can be given by m × n, where m and n are natural numbers. Therefore, for a matrix having 28 elements, that is, mn = 28, possible orders can be found out as follows: ∵ mn = 28 Take m and n to be any number, such that, when it is multiplied it gives 28. So, let m = 1 and n = 28. Then, m × n = 1 × 28 (=28) ⇒ 1 × 28 is a possible order of the matrix having 28 elements. Take m = 2 and n = 14. Then, m × n = 2 × 14 (=28) ⇒ 2 × 14 is a possible order of the matrix having 28 elements. Take m = 4 and n = 7. Then, m × n = 4 × 7 (=28) ⇒ 4 × 7 is a possible order of the matrix having 28 elements. Take m = 7 and n = 4. Then, m × n = 7 × 4 (=28) ⇒ 7 × 4 is a possible order of the matrix having 28 elements. Take m = 14 and n = 2. Then, m × n = 14 × 2 (=28) ⇒ 14 × 2 is a possible order of the matrix having 28 elements. Take m = 28 and n = 1. Then, m × n = 28 × 1 (=28) ⇒ 28 × 1 is a possible order of the matrix having 28 elements. Thus, the possible orders of the matrix having 28 elements are 1 × 28, 2 × 14, 4 × 7, 7 × 4, 14 × 2 and 28 × 1 If the matrix had 13 elements, then also we find the possible order in same way. Here, mn = 13. Take m and n to be any number, such that, when it is multiplied it gives 13. Take m = 1 and n = 13. Then, m × n = 1 × 13 (=13) ⇒ 1 × 13 is a possible order of the matrix having 13 elements. Take m = 13 and n = 1. Then, m × n = 13 × 1 (=13) ⇒ 13 × 1 is a possible order of the matrix having 13 elements. Thus, the possible orders of the matrix having 13 elements are 1 × 13 and 13 × 1

182.

If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

Answer»

We know that in mathematics, a matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second.

We are given with a matrix that has 28 elements.

We know that,

If a matrix has mn elements, then the order of the matrix can be given by m × n, where m and n are natural numbers.

Therefore, for a matrix having 28 elements, that is, mn = 28, possible orders can be found out as follows:

∵ mn = 28

Take m and n to be any number, such that, when it is multiplied it gives 28.

So, let m = 1 and n = 28.

Then, m × n = 1 × 28 (=28)

⇒ 1 × 28 is a possible order of the matrix having 28 elements.

Take m = 2 and n = 14.

Then, m × n = 2 × 14 (=28)

⇒ 2 × 14 is a possible order of the matrix having 28 elements.

Take m = 4 and n = 7.

Then, m × n = 4 × 7 (=28)

⇒ 4 × 7 is a possible order of the matrix having 28 elements.

Take m = 7 and n = 4.

Then, m × n = 7 × 4 (=28)

⇒ 7 × 4 is a possible order of the matrix having 28 elements.

Take m = 14 and n = 2.

Then, m × n = 14 × 2 (=28)

⇒ 14 × 2 is a possible order of the matrix having 28 elements.

Take m = 28 and n = 1.

Then, m × n = 28 × 1 (=28)

⇒ 28 × 1 is a possible order of the matrix having 28 elements.

Thus, the possible orders of the matrix having 28 elements are

1 × 28, 2 × 14, 4 × 7, 7 × 4, 14 × 2 and 28 × 1

If the matrix had 13 elements, then also we find the possible order in same way.

Here, mn = 13.

Take m and n to be any number, such that, when it is multiplied it gives 13.

Take m = 1 and n = 13.

Then, m × n = 1 × 13 (=13)

⇒ 1 × 13 is a possible order of the matrix having 13 elements.

Take m = 13 and n = 1.

Then, m × n = 13 × 1 (=13)

⇒ 13 × 1 is a possible order of the matrix having 13 elements.

Thus, the possible orders of the matrix having 13 elements are 1 × 13 and 13 × 1

Discussion

183.	If `P=[2" "2" "1],Q=[1" "2" "3],R={:[(1),(2),(3)]:},andS={:[(3),(2),(1)]:}`, then PR+QS=A. [19]B. [10]C. [15]D. [12]
Answer» Correct Answer - A

Discussion

184.	Given `A={:[(2,-3),(4,5)]:},B={:[(7,9),(8,1)]:}andC={:[(-2,3),(1,-5)]:}`, find (AB)C and A(BC). What do you notice ?
Answer» Correct Answer - `{:[(35,-105),(-95,-1)]:}`,

Discussion

185.	Match the following lists :
Answer» `a rarr q, b rarr p, c rarr s, d rarr r`. a. `\|A\|=2implies \|2A^(-1)\|=2^(3)//\|A\|=4` b. `\|"adj (adj (2A))"\|=\|2A\|^(4)=2^(12)\|A\|^(4)=2^(12)//2^(12)=1` c. `(A+B)^(2)=A^(2)+B^(2)` `implies AB+BA=O` `implies \|AB\|=\|-BA\|=-\|BA\|=-\|AB\|` `implies \|AB\|=0` `implies \|B\|=0` d. Product ABC is not defined.

Discussion

186.	If `A-B={:[(18,1,-9),(11,12,2)]:}andA+3B={:[(26,1,3),(-13,-16,10)]:}`, then find A.A. `{:[(20,1,6),(5,5,4)]:}`B. `{:[(20,1,6),(-5,-5,4)]:}`C. `{:[(20,1,-6),(5,5,4)]:}`D. `{:[(20,1,-6),(-5,-5,-4)]:}`
Answer» Correct Answer - C `A-B={:[(18,1,-9),(11,12,2)]:}` (1) `A+3B={:[(26,1,3),(-13,-16,10)]:}` (2) `3xx(1)rArr3A-3B={:[(54,3,-27),(33,36,6)]:}` (3) Add Eqs. (2) and (3), we get A=3B+3A-3B `={:[(26,1,3),(-13,-16,10)]:}+{:[(54,3,-27),(33,36,6)]:}` `4A={:[(80,4,-24),(20,20,16)]:}` `A{:[(20,1,-6),(5,5,4)]:}`.

Discussion

187.	The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512
Answer» Answer is (D) There are 9 elements in the matrix. Each elements can be placed in 2 ways is all possible matrices of order 3 x 3 is 2⁹ = 512

187.

The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512

Answer»

Answer is (D)

There are 9 elements in the matrix. Each elements can be placed in 2 ways is all possible matrices of order 3 x 3 is 2⁹ = 512

Discussion

188.	If `A= [(1,1,3),(5,2,6),(-2,-1,-3)]` then find `A^(14)+3A-2I`
Answer» `[(1,1,3),(5,2,6),(-2,-1,-3)][(1,1,3),(5,2,6),(-2,-1,3)]= [(1+5-6, 1+2-3, 3+6-9),(5+10-12, 5+4-6,15+12-18),(-2-5+6, -2-2+3, -6-6+9)]` `= [(0,0,0),(3,3,9),(-1,-1,-3)]` `A^4 = A^2 xx A^2 = [(0,0,0),(3,3,9),(-1,-1,-3)][(0,0,0),(3,3,9),(-1,-1,-3)]` `= [(0,0,0),(0,0,0),(0,0,0)]` so, `A^14 + 3A-2I = 3[(1,1,3),(5,2,6),(-2,-1,-3)]- 2[(1,0,0),(0,1,0),(0,0,1)]` `= [(3,3,9),(15,10,18),(-6,-3,-9)]-[(2,0,0),(0,2,0),(0,0,2)]` `= [(1,3,9),(15,8,18),(-6,-3,-11)]` Answer

Discussion

189.	If P = \(\begin{bmatrix} -3 & 1 \\[0.3em] 2 & 5\end{bmatrix},\) Q = \(\begin{bmatrix} 1 & 6 \\[0.3em] -4 & 0\end{bmatrix}\) and R = \(\begin{bmatrix} 4 & -1 \\[0.3em] 2 & 3\end{bmatrix}\) find the value of 4P – 2Q + 3R.
Answer» 4P - 2Q + 3R = 4\(\begin{bmatrix}-3 & 1 \\[0.3em] 2&5\end{bmatrix}\) - 2\(\begin{bmatrix}1 & 6 \\[0.3em] -4 & 0 \end{bmatrix}\) + 3\(\begin{bmatrix}4 & -1 \\[0.3em] 2&3\end{bmatrix}\) = \(\begin{bmatrix}-12 & 4 \\[0.3em] 8&20\end{bmatrix}\) - \(\begin{bmatrix}2 & 12 \\[0.3em] 8&0\end{bmatrix}\) + \(\begin{bmatrix}12 & -3 \\[0.3em] 6&9\end{bmatrix}\) = \(\begin{bmatrix}-12-2+12 &4-12-3 \\[0.3em]8-(-8)+6&20-0+9\end{bmatrix}\) = \(\begin{bmatrix}-2 & -11 \\[0.3em] 22&29\end{bmatrix}\)

189.

If P = \(\begin{bmatrix} -3 & 1 \\[0.3em] 2 & 5\end{bmatrix},\) Q = \(\begin{bmatrix} 1 & 6 \\[0.3em] -4 & 0\end{bmatrix}\) and R = \(\begin{bmatrix} 4 & -1 \\[0.3em] 2 & 3\end{bmatrix}\) find the value of 4P – 2Q + 3R.

Answer»

4P - 2Q + 3R

= 4\(\begin{bmatrix}-3 & 1 \\[0.3em] 2&5\end{bmatrix}\) - 2\(\begin{bmatrix}1 & 6 \\[0.3em] -4 & 0 \end{bmatrix}\) + 3\(\begin{bmatrix}4 & -1 \\[0.3em] 2&3\end{bmatrix}\)

= \(\begin{bmatrix}-12 & 4 \\[0.3em] 8&20\end{bmatrix}\) - \(\begin{bmatrix}2 & 12 \\[0.3em] 8&0\end{bmatrix}\) + \(\begin{bmatrix}12 & -3 \\[0.3em] 6&9\end{bmatrix}\)

= \(\begin{bmatrix}-12-2+12 &4-12-3 \\[0.3em]8-(-8)+6&20-0+9\end{bmatrix}\) = \(\begin{bmatrix}-2 & -11 \\[0.3em] 22&29\end{bmatrix}\)

Discussion

190.	If `A={:[(-1,2),(3,4)]:}andB=A^(T)," then "A^(T)+B^(T)=`A. `{:[(-3,4),(6,8)]:}`B. `{:[(-2,0),(0,4)]:}`C. `{:[(1,0),(0,1)]:}`D. `{:[(-2,5),(5,8)]:}`
Answer» Correct Answer - D Find `A^(T)andB^(T)` and add the two matrices.

Discussion

191.	If A, B and C are three matrices of order `3xx4,3xx2andaxx2` respectively, then find the order of matrix `A^(T)BC^(T)`.
Answer» Correct Answer - `4xx1`

Discussion

192.	If `X={:[(1,2),(3,2)]:}andX^(2)-3X=`A. IB. 3IC. 4ID. `-4I`
Answer» Correct Answer - C (i) Evaluate `X^(2),3X`. (ii) `X^(2)-3XrArrX(X-3)`.

Discussion

193.	If X and Y are two matrices such that `X+Y={:[(4,-3),(5,2)]:}andX-Y={:[(6,-5),(3,2)]:}`, then find the matrices X and Y.
Answer» Correct Answer - `X={:[(5,-4),(4,2)]:},Y={:[(-1,1),(1,0)]:}`

Discussion

194.	If `A-B^(T){:[(2,3,4,-1),(0,3,-1,5)]:}andA^(T)+B={:[(5,3),(1,0),(2,5),(3,-1)]:}`, then find the matrices A and B.
Answer» Correct Answer - `A={:[(7//2,2,3,1),(3//2,3//2,2,2)]:}` `B={:[(3//2,3//2),(-1,-3//2),(-1,3),(2,-3)]:}`

Discussion

195.	If `A={:[(-1,3),(4,5)]:}`, then find (A-I)(A-2I).
Answer» Correct Answer - `(A-I)(A-2I)={:[(18,3),(4,24)]:}`

Discussion

196.	If `A={:[(7,0),(0,7)]:}," then find "A^(n)(ninN)` ________ .A. `{:[(7,0),(0,7)]:}`B. `{:[(7^(n),0),(0,7^(n))]:}`C. `{:[(7,7^(n)),(0,7^(n))]:}`D. `{:[(0,7^(n)),(7^(n),0)]:}`
Answer» Correct Answer - B Given `A={:[(7,0),(0,7)]:}` `A-7={:[(1,0),(0,1)]:}` `A=7I` `A^(n)=7^(n)I^(n)` `A^(n)=7^(n)I(becauseI^(n)=I)` `=7^(n){:[(1,0),(0,1)]:}` `={:[(7^(n),0),(0,7^(n))]:}`.

Discussion

197.	If `A={:[(1,2),(3,4)]:}andB={:[(5,6),(7,8)]:}`, then find 3A+7B.
Answer» Correct Answer - `{:[(38,48),(58,68)]:}`

Discussion

198.	`A={:[(-5,-3,4),(3,2,-4)]:}andB={:[(-4,5,-2),(3,1,5)]:}`, then find X such that 3A-2B+X=0.A. `{:[(7,19,16),(3,4,22)]:}`B. `{:[(-7,-19,16),(3,4,22)]:}`C. `{:[(7,19,-16),(-3,-4,22)]:}`D. `{:[(7,19,16),(-3,-4,22)]:}`
Answer» Correct Answer - C (i) X=2B-3A (ii) `3A-2B+X=0rArrX=2B-3A`. (iii) Multiply each element of B with 2 to get 2B. (iv) Multiply each element of A with 3 to get 3B.

Discussion

199.	If `{:A=[(3,1),(-1,2)]:}," then "A^(-2)=`A. `{:[(8,-5),(-5,3)]:}`B. `{:[(8,-5),(5,3)]:}`C. `{:[(8,-5),(-5,-3)]:}`D. `{:[(8,5),(-5,3)]:}`
Answer» Correct Answer - d

Discussion

200.	From the matrix equation AB = AC we can conclude B = C provided thatA. A is singularB. A is non-singular.C. A is symmetric.D. A is square
Answer» Correct Answer - B

Discussion

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The value of x, for which the matrix A = \(\begin{bmatrix}e^{x - 2} &e^{7 + x} \\[0.3em]e^{2 + x}&e^{2x + 3}\end{bmatrix}\) is singular is …(a) 9(b) 8 (c) 7 (d) 6

In the matrix A = \(\begin{bmatrix} a&1&x \\[0.3em] 2&\sqrt3&x^2 - y \\0&5&-\frac{2}{5} \\[0.3em] \end{bmatrix}\), write(i) The order of the matrix A(ii) The number of elements(iii) Write elements a23, a31, a12

If P = \(\begin{bmatrix} -3 & 1 \\[0.3em] 2 & 5\end{bmatrix},\) Q = \(\begin{bmatrix} 1 & 6 \\[0.3em] -4 & 0\end{bmatrix}\) and R = \(\begin{bmatrix} 4 & -1 \\[0.3em] 2 & 3\end{bmatrix}\) find the value of 4P – 2Q + 3R.