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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
If `A[(-1,5),(-3,2)]`, then adj A=A. `[(2,3),(-5,-1)]`B. `[(1,-5),(3,-2)]`C. `[(2,-5),(3,-1)]`D. `[(-2,5),(-3,1)]` |
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Answer» Correct Answer - C |
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| 302. |
If A and B are two non-singular matrices of order 3 such that `A A^(T)=2I` and `A^(-1)=A^(T)-A`. Adj. `(2B^(-1))`, then det. (B) is equal toA. 4B. `4sqrt(2)`C. 16D. `16sqrt(2)` |
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Answer» Correct Answer - D `A A^(T)=2I` ...(1) `implies |A|^(2)=8` Now, `A^(-1)=A^(T)-A" adj."(2B^(-1))` ...(2) Multiplying with A, we get `I=A A^(T)-A^(2)"adj."(2B^(-1))` `:. I=2I-A^(2)"adj."(2B^(-1))` `implies A^(2)"adj." (2B^(-1))=I` `implies |A^(2)||"adj." (2B^(-1))|=1` `implies 8|2B^(-1)|^(2)=1` `implies 8. 64/(|B|^(2))=1` `implies |B|^(2)=64xx8` `:. |B|= pm (8xx2sqrt(2))= pm 16 sqrt(2)` |
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| 303. |
If `A=[{:(3,1,2),(1,2,-3):}]" and "B=[{:(-2," "0,4),(5,-3,2):}]`, find `(2A-B).` |
| Answer» `[{:(8,2," "0),(-3,7,-8):}]` | |
| 304. |
Construct a 2 × 3 matrix whose elements are aij = (i - 2j)2/2. |
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Answer» It is a (2 x 3) matrix. So, it has 2 rows and 3 columns. Given \(a_{ij}=\frac{(i-2j)^2}{2}\) So, a11 = \(\frac{1}{2},\) a12 = \(\frac{9}{2},\) a13 = \(\frac{25}{3},\) a21 = 0, a22 = 2, a23 = 8 So, the matrix = \(\begin{bmatrix}\frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\[0.3em]0 &2 & 8\\[0.3em]\end{bmatrix}\) Conclusion: Therefore, Matrix is \(\begin{bmatrix}\frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\[0.3em]0 &2 & 8\\[0.3em]\end{bmatrix}\) |
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| 305. |
Construct a 2 × 2 matrix whose elements are aij = (i + 2j)2/2. |
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Answer» It is a (2 x 2) matrix. So, it has 2 rows and 2 columns. Given \(a_{ij}=\frac{(i+2j)^2}{2}\) So, a11 = \(\frac{9}{2},\) a12 = \(\frac{25}{2},\) a21 = 8, a22 = 18 So, the matrix = \(\begin{bmatrix}\frac{9}{2} &\frac{25}{2} \\[0.3em]8 &18 \\[0.3em]\end{bmatrix}\) Conclusion: Therefore, Matrix is = \(\begin{bmatrix}\frac{9}{2} &\frac{25}{2} \\[0.3em]8 &18 \\[0.3em]\end{bmatrix}\) |
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| 306. |
If \(\begin{bmatrix}a + b&2\\[0.3em]7&ab\\[0.3em]-3&4\end{bmatrix} = \begin{bmatrix}6&2\\[0.3em]7&8\\[0.3em]-3&4\end{bmatrix}\) then find ‘a’ and ‘b’. |
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Answer» \(\begin{bmatrix}a + b&2\\[0.3em]7&ab\\[0.3em]-3&4\end{bmatrix} = \begin{bmatrix}6&2\\[0.3em]7&8\\[0.3em]-3&4\end{bmatrix}\) On comparing, a + b = 6 ……(i) ab = 8 From equation (i) and (ii), a(6 – a)= 8 ⇒ 6a – a2 – 8 = 0 ⇒ a2 – 2a – 4a + 8 = 0 ⇒ a2 – 2a – 4a + 8 = 0 (a – 2)(a – 4) = 0 So, a = 2, 4 From ab = 8 we get b = 4, 2 |
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| 307. |
Let `a,lambda,mu in R,` Consider the system of linear equations `ax+2y=lambda 3x-2y=mu` Which of the flollowing statement (s) is (are) correct?A. (a) If a = -3, then the system has infinitely many solutions for all value of `lambda and mu`.B. If `a ne -3`, then the system has a unique solution fopor all values of `lambda and mu`.C. If `lambda+u=0`, then the system has infinitely many solutions for a = -3`.D. If `lambda +mu ne 0`, then the system has no solutions for a = -3. |
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Answer» Correct Answer - A If a = -3, then the two equations in the given system represent parallel lines for `-lambda ne mu i.e., lambda+mu ne 0` and so the system has no solution. So, option (d) is correct. If a = -3 and `lambda+mu=0 i.e., -lambda=mu`, then the two lines given by the above system are coincident and hence the system has infinitely many solutions. So, option (c) is correct for all values of `lambda and mu`. If `a ne -3`, then the two lines given in the system are not parallel i.e.,they are intersecting and hence the ststem has a unique solutions. So, option (b) is correct. Clearly, option(a) incorrect. |
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| 308. |
If matrix A = [1 2 3], write AA’. |
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Answer» Given A = [1 2 3] We will find A’ to calculate AA’, A' = \(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\) Now AA' = [1 2 3]\(\begin{bmatrix}1 \\[0.3em]2 \\[0.3em]3\end{bmatrix}\) \(\Rightarrow\) [1 + 4 + 9] \(\Rightarrow\) [14] |
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| 309. |
Find the values of a and b which `[{:(a,b),(-a,2b):}][{:(" "2),(-1):}]=[{:(5),(4):}].` |
| Answer» Correct Answer - a=1, b=-3 | |
| 310. |
If `A=[{:(" "3," "4),(-4,-3):}]`, find f(A), where `f(x)=x^(2)-5x+7.` |
| Answer» `f(A)=[{:(-15,-20),(" "20," "15):}]` | |
| 311. |
Give an example of twomatrices `A`and `B`such that`A!=O , B!=O , A B=O`and `B A!=O`(ii) `A!=O , B!=O , A B=B A=O`. |
| Answer» `A=[{:(1,0),(0,0):}],B=[{:(0,0),(1,0):}]` | |
| 312. |
Give an example ofthree matrices `A , B , C`such that `A B=A C`but `B!=C`. |
| Answer» `A=[{:(1,0),(0,0):}],B=[{:(0,0),(1,0):}]=C[{:(0,0),(1,0):}]` | |
| 313. |
If `A=[{:(1,0),(-1,7):}]" and "B=[{:(0,4),(-1,7):}]`, find `(3A^(2)-2B+I).` |
| Answer» `[{:(" "4,-8),(-22,134):}]` | |
| 314. |
If `A=[1 2 2 2 1-2a2b]`is a matrix satisfying the equation `AA^T=""9I`,where `I`is `3xx3`identity matrix, then the ordered pair (a,b) is equal to :(1) `(2,-1)`(2) `(-2,""1)`(3) (2, 1) (4) `(-2,-1)`A. (2, 1)B. (-2, -1)C. `(2, -1)`D. `(-2, 1)` |
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Answer» Correct Answer - B `therefore A A^(T) = 9 I` `[[1,2,2],[2,1,-2],[a,2,b]][[1,2,a],[2,1,2],[2,-2,b]]= 9 [[1,0,0],[0,1,0],[0,0,1]]` `rArr [[9,0,a+ 4+ 2b],[0,9,2a+2-2b],[a+4+2b,2a+2-2b,a^(2)+4+b^(2)]]= [[9,0,0],[0,9,0],[0,0,9]]` On comparing, we get `a + 2b + 4 = 0 " " (i)` ` 2a- 2b+2=0" "(ii)` From Eqs. (i) and (ii), we get `a = -2,` `b= -1` `therefore` Prdered pair is `(-2, -1)`. |
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| 315. |
Find the inverse of `[{:(1," "2,-3),(2," "3," "2),(3,-3,-4):}]` |
| Answer» `(1)/(67).[{:(-6,17,13),(14,5,-8),(-15,9,-1):}]` | |
| 316. |
Find the inverse of matrix `[{:(3,0,2),(1,5,9),(6,4,7):}]` |
| Answer» `(-1)/(55).[{:(-1,8,-10),(47,9,-25),(-26,-12," "15):}]` | |
| 317. |
Determine the matrix A, when `A=4[{:(1,2,3),(-1,-2,-3),(4,2,6):}]+2[{:(5,4,1),(3,2,4),(3,8,2):}]` |
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Answer» `A=[{:(4,8,12),(-4,-8,-12),(16,8,24):}]+[{:(10,8,2),(6,4,8),(6,16,4):}]` =`[{:(4+10,8+8,12+2),(-4+6,-8+4,-12+8),(16+6,8+16,24+4):}]=[{:(14,16,14),(2,-4,-4),(22,24,28):}]` |
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| 318. |
If A is a square matrix such that A2 = A, then write the value of (I + A)2 - 3A. |
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Answer» Given A2 = A (I+A)2-3A=? = I2 + A2 + 2IA - 3A { (a+b)2=a2+b2+2ab } IA=A (by property) Since A2=A and IA=A and I2=I = I+2A+A−3A = I |
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| 319. |
if `A=[{:(alpha,2),(2, alpha):}]` and `|A^(3)|=125` then the value of `alpha `isA. `pm=2`B. `pm=3`C. `pm=5`D. 0 |
| Answer» Correct Answer - B | |
| 320. |
If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.A. 4B. 5C. 6D. 7 |
| Answer» Correct Answer - B | |
| 321. |
if `A=[(1,2),(2,3)] and A^(2) -lambdaA-l_(2)=O,`then `lambda` is equal toA. `-4`B. `-2`C. 2D. 4 |
| Answer» Correct Answer - D | |
| 322. |
Let `A=[(0, alpha),(0,0)]` and `(A+I)^(50) A=[(a,b),(c,d)]`. Then the value of `a+b+c+d` isA. 1B. 2C. 4D. None of these |
| Answer» Correct Answer - B | |
| 323. |
If A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A). |
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Answer» We know that, (AB) adj (AB) = |AB|I = adj (AB)(AB) …(i) ⇒ (AB) (adj B . adj A) = A . B adj B . adj A = A(B adj B) adj A = A(|B|I) adj A [∵ B adj B = |B|I] = |B|(A . adj A) = |B||A|I [∵ A adj A = |A|I] = |A||B|I = |AB|I …(ii) From (i) and (ii), we get (AB) (adj AB) = AB (adj B . adj A) Pre-multiplying both sides by (AB)–1, we get (AB)–1 ((AB) adj AB) = (AB)–1 ((AB) adj B . adj A) ⇒ adj AB = adj B . adj A |
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| 324. |
The order of the matrix `[{:(1,2,3,4),(4,3,2,1),(2,4,3,1):}]` is _________ |
| Answer» Correct Answer - `3xx4` | |
| 325. |
The orders of the matrices `A=[{:(1,4),(-1,0),(5,2):}]` and `B=[{:(1,2,3),(-5,-2,1):}]` are ______ (equal/ not equal ) |
| Answer» Correct Answer - not equal | |
| 326. |
If A =\( \begin{pmatrix} 2& -4 \\[0.3em] 3 & 1 \\[0.3em] \end{pmatrix}\), then the adjoint of matrix A is :((2,-4)(3,1))(a) \( \begin{pmatrix} -1& 3 \\[0.3em] -4 & 1 \\[0.3em] \end{pmatrix}\) (b) \( \begin{pmatrix} 1& 4 \\[0.3em] -3 & 2 \\[0.3em] \end{pmatrix}\)(c) \( \begin{pmatrix} 1& 3 \\[0.3em] 4 & -2 \\[0.3em] \end{pmatrix}\)(d) \( \begin{pmatrix} -1& -3 \\[0.3em] -4 & 2 \\[0.3em] \end{pmatrix}\) |
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Answer» Option : (b) \( \begin{pmatrix} 1& 4 \\[0.3em] -3 & 2 \\[0.3em] \end{pmatrix}\) |
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| 327. |
The inverse of\( \begin{pmatrix} 0& 1 \\[0.3em] 1 & 0 \\[0.3em] \end{pmatrix}\)is :(A) \( \begin{bmatrix} 1& 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)(B) \( \begin{bmatrix} 0& 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\) (C) \( \begin{bmatrix} 1& 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}\) (D) None of these.((0,1)(1,0)) |
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Answer» Option : (B) \( \begin{bmatrix} 0& 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}\) |
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| 328. |
If A is 2 × 3 matrix and B is a matrix such that ATB and BAT both are defined, then what is the order of B? |
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Answer» We are given that, Order of matrix A = 2 × 3 ATB and BAT are defined matrices. We need to find the order of matrix B. We know that the transpose of a matrix is a new matrix whose rows are the columns of the original. So, If the number of rows in matrix A = 2 And, number of columns in matrix A = 3 Then, The number of rows in matrix AT = number of columns in matrix A = 3 Number of columns in matrix AT = number of rows in matrix A = 2 So, Order of matrix AT can be written as, Order of matrix AT = 3 × 2 Thus, We have Number of rows of AT = 3 …(i) Number of columns of AT = 2 …(ii) If ATB is defined, that is, it exists, then Number of columns in AT = Number of rows in B ⇒ 2 = Number of rows in B [from (ii)] Or, Number of rows in B = 2 …(iii) If BAT is defined, that is, it exists, then Number of columns in B = Number of rows in AT Substituting value of number of rows in AT from (i), ⇒ Number of columns in B = 3 …(iv) From (iii) and (iv), Order of B = Number of rows × Number of columns ⇒ Order of B = 2 × 3 Thus, Order of B is 2 × 3 |
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| 329. |
If A is a 3 × 4 matrix and B is a matrix such that both ATB and BAT are defined, what is the order of the matrix B? |
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Answer» A is a matrix of order 3 × 4 So AT will be a matrix of order 4 × 3 ATB will be defined when B is a matrix of order 3 × n BAT will be defined when B is of order m × 4 From (1) and (2) we see that B should be a matrix of order 3 × 4 |
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| 330. |
if `A` and `B` are two matrices of order `3xx3 `so that `AB=A` and `BA=B` then `(A+B)^7=`A. `7(A+B)`B. `7. I_(3xx3)`C. `64 (A+B)`D. `128 I` |
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Answer» Correct Answer - C `AB=A, BA=B` `ABA=A^(2)` or `A(BA)=A^(2)` or `A=A^(2)` Similarly, `B^(2)=B` `(A+B)^(2)=A^(2)+B^(2)+AB+BA` `=A+B+A+B=2(A+B)` `(A+B)^(3)=(A+B)^(2) (A+B)=2(A+B)^(2)=2^(2) (A+B)` `implies (A+B)^(7)=2^(6)(A+B)=64 (A+B)` |
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| 331. |
The order of the matrix [c y x -x -y z] is ______A. `1xx4`B. `6xx1`C. `1xx6`D. `4xx1` |
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Answer» Correct Answer - C The given matrix has a row and 6 columns. `:.` The order of the matrix is `1xx6`. |
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| 332. |
If A = \( \begin{pmatrix}1& 2 \\[0.3em]3 & 4 \\[0.3em]\end{pmatrix}\), adj = \( \begin{pmatrix}4& a \\[0.3em]-3 & b \\[0.3em]\end{pmatrix}\) then the values of a and b are,(a) a = – 2, b = 1 (b) a = 2, b = 4 (c) a = 2, b = –1 (d) a = 1, b = –2 |
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Answer» Option : (a) a = – 2, b = 1 |
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| 333. |
If `A={:[(2,3),(-1,4)]:}andB={:[(-3,1),(4,-2)]:}`, then find A-B. |
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Answer» `A-B={:[(2,3),(-1,4)]:}-{:[(-3,1),(4,-2)]:}` `A-B={:[(2-(-3),3-1),(-1-4,4-(-2))]:}={:[(5,2),(-5,6)]:}` |
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| 334. |
If `A=[{:(1,0,-1),(2,1,3),(0,1, 1):}]` then verify that `A^(2)+A=(A+I)` , where I is `3xx3` unit matrix. |
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Answer» We have, `A=[{:(1,0,-1),(2,1,3),(0,1,1):}]` `therefore A^(2)=A.A` `=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(1,0,-1),(2,1,3),(0,1,1):}]=[{:(1,-1,-2),(4,4,4),(2,2,4):}]` `therefore A^(2)+A=[{:(1,-1,-2),(4,4,4),(2,2,4):}]+[{:(1,0,-1),(2, 1,3),(0,1,1):}]` `=[{:(2,-1,-3),(6,5,7) ,(2,3,5):}]` Now, `A+I=[{:(1,0,-1),(2,1,3),(0,1,1):}]+[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(2,0,-1),(2,2,3),(0,1,3):}]` and `A(A+I)=[{:(1,0,-1),(2,1,3),(0,1,1):}][{:(2,0,-1),( 2,2,3),(0,1,2):}]=[{:(2,-1,-3),(6,5,7),(2,3,5):}]` Thus, we see that `A^(2)+A=A(A+I)` |
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| 335. |
Let `A`and `B`be square matrices ofthe order `3xx3`. Is `(A B)^2=A^2B^2`? Give reasons. |
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Answer» Since A and B are square matrices of order `3xx3` `AB^(2)=AB AB` =AB AB =A A B B `[because AB =BA]` `=A^(2)B^(2)` So, `AB^(2)=A^(2)B^(2)` is true when AB=BA. |
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| 336. |
Find the inverse of `[0 1-1 4-3 4 3-3 4]`A. 2AB. `1/2A^(-1)`C. `1/2A`D. `A^2` |
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Answer» Correct Answer - A It is given that A is an involutory matrix. `:. A^2=I` `rArr A A =IrArr(1/2A)(2A)=IrArr (1/2A)^(-1)=(2A)` |
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| 337. |
In a `4xx4` matrix the sum of each row, column and both the main diagonals is `alpha`. Then the sum of the four corner elementsA. is also `alpha`B. may not be `alpha`C. is never equal to `alpha`D. none of these |
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Answer» Correct Answer - A Let `A=[a_(ij)]` be a `4xx4` matrix. It is given that `a_(i1)+a_(i2)+a_(i3)+a_(i4)=alpha" for" i=1,2,3,4` `a_(1j)+a_(2j)+a_(3j)+a_(4j)=alpha" for "j=1,2,3,4` `a_11+a_22+a_33+a_(44)=alpha` `anda_14+a_23+a_32+a_41=alpha` We have to find `a_11+a_14+a_41+a_44`. Clearly, `(a_11+a_12+a_13+a_14)+(a_14+a_42+a_43+a_44)` `+(a_11+a_22+a_33+a_44)+(a_14+a_23+a_32+a_41)` `-(a_12+a_22+a_32+a_42)-(a_13+a_23+a_33+a_43)` `=2(a_11+a_14+a_41+a_44)` `rArr4alpha-2alpha=2(a_11+a_14+a_41+a_44)` `rArr a_11+a_14+a41+a_44=alpha` |
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| 338. |
Let `A` and `B` be two non-singular square matrices such that `B ne I` and `AB^(2)=BA`. If `A^(3)-B^(-1)A^(3)B^(n)`, then value of `n` isA. `4`B. `5`C. `8`D. `7` |
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Answer» Correct Answer - C `(c )` `BA=AB^(2)` `impliesBA=AB^(2)` `impliesA=B^(-1)AB^(2)` `impliesA^(2)=(B^(-1)AB^(2))(B^(-1)AB^(2))` `=B^(-1)A(BA)B^(2)` `impliesB^(-1)A AB^(2)B^(2)` `=B^(-1)A^(2)B^(4)` `:.A^(3)=B^(-1)A^(3)B^(6)` |
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| 339. |
If `A={:[(2,3,-1),(5,6,1)]:}`, then find (a) -A , (b)3A , (c ) `(1)/(4)A`. |
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Answer» (a) `-A--{:[(2,3,-1),(5,6,1)]:}` `={:[(-1xx2,-1xx3,-1xx(-1)),(-1xx5,-1xx6,-1xx1)]:}={:[(-2,-3,1),(-5,-6,-1)]:}`. (b) `3A=3{:[(2,3,-1),(5,6,1)]:}` `={:[(3xx2,3xx3,3(-1)),(3xx5,3xx6,3xx1)]:}={:[(6,9,-3),(15,18,3)]:}` `(1)/(4)A=(1)/(4){:[(2,3,-1),(5,6,1)]:}` `={:[((1)/(4)xx2,(1)/(4)xx3,(1)/(4)xx(-1)),((1)/(4)xx5,(1)/(4)xx6,(1)/(4)xx1)]:}={:[((1)/(2),(3)/(4),(-1)/(4)),((5)/(4),(3)/(2),(1)/(4))]:}` |
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| 340. |
Let `A={:[(-3,1),(0,2)]:}B={:[(2,-1),(0,1)]:}`Then prove |
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Answer» `A={:[(-3(2)+1(0),-3(-1)+1(1)),((0)2+2(0),0(-1)+2(1))]:}={:[(-6,4),(0,2)]:}` `BA={:[(2,-1),(0,1)]:}{:[(-3,1),(0,2)]:}` `={:[(2(-3)+(-1)0,2(1)+(-1)2),(0(-3)+1(0),(0)1+1(2))]:}={:[(-6,0),(0,2)]:}`. |
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| 341. |
If `A=[{:(1,0,0),(1,0,1),(0,1,0):}]`, thenA. `A^(3)-A^(2)=A-I`B. `Det(A^(2010)-I)=0`C. `A^(50)=[{:(1,0,0),(25,1,0),(25,0,1):}]`D. `A^(50)=[{:(1,1,0),(25,1,0),(25,0,1):}]` |
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Answer» Correct Answer - A::B::C `(a,b,c)` `A^(2)=[{:(1,0,0),(1,0,1),(0,1,0):}][{:(1,0,0),(1,0,1),(0,1,0):}]=[{:(1,0,0),(1,1,0),(1,0,1):}]` `A^(3)=[{:(1,0,0),(1,1,0),(1,0,1):}][{:(1,0,0),(1,1,0),(1,0,1):}]=[{:(1,0,0),(2,0,1),(1,1,0):}]` `A^(3)-A^(2)=[{:(0,0,0),(1,-1,1),(0,1,-1):}]`and `A-I=[{:(0,0,0),(1,-1,1),(0,1,-1):}]` `impliesA^(3)-A^(2)=A-I` and det `(A-I)=0` `impliesDet|A^(n)-I)=Det((A-I)(1+A+A^(2)+...+A^(n-1)))` `=Det(A-I)Det(1+A+A^(2)+....+A^(n-1))=0` `A^(3)-A^(2)=A-I` ...........`(i)` `impliesA^(4)-A^(3)=A^(2)-A`..........`(ii)` `impliesA^(5)-A^(4)=A^(3)-A^(2)=A-I` (Using `(1)`) If `n` is even `A^(n)-A^(n-1)=A^(2)-A`...........`(iii)` If `n` is odd `A^(n)-A^(n-1)=A-I`..........`(iv)` Consider `n` is even `:.A^(n)-A^(n-1)=A^(2)-A`(Using `(iii)`) `A^(n-1)-A^(n-2)=A-I` (Using `(iv)`) On adding, we get `A^(n)-A^(n-2)=A^(2)-I` `impliesA^(n)-A^(n-2)+A^(2)-I` `=A^(n-4)+A^(2)-I)+A^(2)-I` `=(A^(n-6)+A^(2)-I)+2(^(2)-I)` `=(A^(2))+(n-2)/(2)(A^(2)-I)` `A^(n)=((n)/(2))A^(2)-((n-2)/(2))I` `:.A^(50)=25A^(2)-24I` `=25[{:(1,0,0),(1,1,0),(1,0,1):}]-24[{:(1,0,0),(0,1,0),(0,0,1):}]` `=[{:(1,0,0),(25,1,0),(25,0,1):}]` |
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| 342. |
Let `A={:[(2,-1),(1,7),(3,5)]:}_(3xx2)and{:[(-5,6,4),(9,11,8)]:}_(2xx3)`Find AB |
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Answer» `AB={:[(2(-5)(-1)9,2(6)+(-1)(11),2(4)+(-1)(8),),(1(-5)+7(9),(1)6+7(11),(1)4+7(8),),(3(-5)+5(9),(3)6+(5)11,3(4)+(5)8,)]:}_(3xx3)` `AB={:[(-10+(-9),12+(-11),8+(-8),),(-5+63,6+77,4+56,),(-15+45,18+55,12+40,)]:}_(3xx3)={:[(-19,1,0),(58,83,60),(30,73,52)]:}_(3xx3)` |
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| 343. |
Let `A` be a square matrix of order `3` so that sum of elements of each row is `1`. Then the sum elements of matrix `A^(2)` isA. `1`B. `3`C. `0`D. `6` |
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Answer» Correct Answer - B `(b)` Let `A=[{:(a,b,c),(p,q,r),(x,y,z):}]` Given `{:(a+b+c=1),(p+q+r=1),(x+y+z=1):}` `impliesA^(2)=[{:(a,b,c),(p,q,r),(x,y,z):}][{:(a,b,c),(p,q,r),(x,y,z):}]` `=[{:(a^(2)+bp+cx,,ab+bq+cy,,ac+br+cz),(pa+qp+rx,,qb+q^(2)+ry,,pc+qr+rz),(xa+yp+zx,,xb+yq+zy,,xc+yr+z^(2)):}]` Sum of elements of `R_(1)=a^(2)+bp+cx+ab+bq+cy+ac+br+cz` `=a(a+b+c)+b(p+q+r)+c(x+y+z)` `=a+b+c=0` Similarly sum of elements of `R_(2)=p(a+b+c)+q(p+q+r)+r(x+y+z)` `=p+q+r=1` `R_(3)=x(a+b+c)+y(p+q+r)+z(x+y+z)` `=x+y+z=1` `:.` sum of element of `A^(2)` is `3` |
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| 344. |
If `(A+B)^(2)=A^(2)+B^(2)` and `|A| ne 0` , then `|B|=` (where `A` and `B` are matrices of odd order)A. `2`B. `-2`C. `1`D. `0` |
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Answer» Correct Answer - D `(d)` `(A+B)^(2)=A^(2)+B^(2)` `impliesAB+BA=0` `impliesAB=-BA` `implies|AB|=|-BA|` `implies|A||B|=-|B||A|` (`A` and `B` are odd ordered matices) `implies|B|=-|B|(|A|=2)` `implies|B|=0` |
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| 345. |
If A and B are matrices of same order, then (AB’ – BA’) is aA. skew symmetric matrixB. null matrixC. symmetric matrixD. unit matrix |
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Answer» Let C = (AB’ – BA’) C’ = (AB’ – BA’)’ ⇒ C’ = (AB’)’ – (BA’)’ ⇒ C’ = (B’)’A’ – (A’)’B’ ⇒ C’ = BA’ – AB’ ⇒ C’ = -C ∴ C is a skew-symmetric matrix. Clearly Option (A) matches with our deduction. ∴ Option (A) is the correct. |
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| 346. |
The element in the third column of matrix `[{:(1,5,10),(3,6,11),(4,7,12):}]` are _____A. 1,3,4B. 10,11,12C. 5,6,7D. 4,7,12 |
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Answer» Correct Answer - B The numbers in the third column are numbers on the third vertical line. |
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| 347. |
Anitha, Nikita and Ankitha have purchased some books, pencils and pens, This can be respresented in the following matrix. `{:("Anitha"),("Nikitha"),("Ankitha"):}[{:("Books","Pens","Pencils"),(5,7,8),(4,3,2),(7,6,0):}]` The total number of items purchased by Ankitha isA. 9B. 7C. 13D. 10 |
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Answer» Correct Answer - C (i) The sum of element of third row. (ii) The sum of elements in the third row represent the number of articles purchased by Ankitha. |
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| 348. |
Let `A` be a `2xx3` matrix, whereas `B` be a `3xx2` amtrix. If `det.(AB)=4`, then the value of `det.(BA)` isA. `-4`B. `2`C. `-2`D. `0` |
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Answer» Correct Answer - D `(d)` Let `A=[{:(a_(1),a_(2),a_(3)),(a_(4),a_(5),a_(6)):}]`, `B=[{:(b_(1),b_(2)),(b_(3),b_(4)),(b_(5),b_(6)):}]` So, det `(BA)=|{:(b_(1)a_(1)+b_(2)a_(4),,b_(1)a_(2)+b_(2)a_(5),,b_(1)a_(3)+b_(2)a_(6)),(b_(3)a_(1)+b_(4)a_(4),,b_(3)a_(2)+b_(4)a_(5),,b_(3)a_(3)+b_(4)a_(6)),(b_(3)a_(1)+b_(6)a_(4),,b_(5)a_(2)+b_(6)a_(5),,b_(5)a_(3)+b_(6)a_(6)):}|` `=|{:(a_(1),,a_(2),,a_(3)),(a_(4),,a_(5),,a_(6)),(0,,0,,0):}|xx|{:(b_(1),,b_(2),,0),(b_(3),,b_(4),,0),(b_(5),,b_(6),,0):}|=0` (column by row) |
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| 349. |
The element in the second row of the matrix `[{:(a,b,c),(1,2,3),(5,6,d):}]` are ______A. 1,2,3B. a,1,5C. b,2,6D. 5,6,d |
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Answer» Correct Answer - A The number in the second row are numbers in second horizontal line. |
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| 350. |
The order of the matrix `[{:(a),(b),(c ),(d):}]` is _______A. `5xx1`B. `2xx3`C. `1xx4`D. `4xx1` |
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Answer» Correct Answer - D Order of a matrix = Number of rows `xx` Number of columns. |
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