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Topic: Matrices, , Question bank with solutions, One mark question ( V S A), 1. Define matrix, 2. Define a diagonal matrix, 3. Define scalar matrix, 4. Define symmetric matrix, 5. Define skew-symmetric matrix, , 6.In a matrix, , 2, [ 35, , 5, −2, , 19, 2, , −17, 12 ], , √3, , 1, , −5, , 17, , 5, , find 1) order of the matrix, 2) Write the elements of 𝑎13 , 𝑎21 , 𝑎33 , 𝑎24 , 𝑎23, 7. If a matrix 8 elements what is the possible order it can have ?, 8. If a matrix 18 elements what is the possible order it can have?, 9. construct 2 × 2 matrix [𝑎𝑖𝑗 ] whose elements are given by, 1) 𝑎𝑖𝑗 = (𝑖 + 𝑗) 2, , 2) 𝑎𝑖𝑗 =, , (𝑖+𝑗) 2, 2, , 10. construct the 2 × 3 matrix whose elements are given by 𝑎𝑖𝑗 = |𝑖 − 𝑗|, 11. Construct the 3× 2 matrix whose elements are given by 𝑎𝑖𝑗 =, 12. Find x, y, z if [, , 4, 𝑥, , 13. Find x, y, z if [, , 3 𝑦, ]= [, 5 1, , 𝑥+𝑦, 5+𝑧, , 𝑧, ], 5, , 2, 6 2, ], ]= [, 𝑥𝑦 5 8, , 14. Find the matrix x such that 2A + B + X =0 where A = [, 1, 15. If A = [, 2, , 2 3, ], 3 1, , 𝑖, 𝑗, , 3 −1 3, B= [, ], −1 0 2, , Find 2A – B, , −1 2, 3 −2, ] and B = [, ], 3 4, 1 5

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16. Find X if Y =[, , 1 0, 3 2, ] and 2X+Y = [, ], 1 4, −3 2, , 17. Find X If X+Y = [, , 7 0, 3, ] and X-Y = [, 2 5, 0, , 18. Simplify cos 𝜃 [, 19. Find X If 2[, 1, 20. If A = [, 3, 1, 21. A = [, 0, , sin 𝜃, sin 𝜃, ] + sin 𝜃 [, cos 𝜃, cos 𝜃, , 𝑦, 1 3, ]+[, 0 𝑥, 1, , 0, 5, ]=[, 1, 2, , −cos 𝜃, ], sin 𝜃, , 6, ], 8, , 2, ] Find A + 𝐴1, 4, −2 3, ], 1 4, , sin 𝜃, 22. if A = [, −cos 𝜃, 23. if B = [, , cos 𝜃, −sin 𝜃, , 0, ], 3, , cos 𝜃, −sin 𝜃, , and B = [, , 0 2, 6 −3, , 5, ], 1, , Find 3A + 2B, , cos 𝜃, ] Verify A A1 = I, sin 𝜃, sin 𝜃, ] verify B B1= I, cos 𝜃, , 0, 24. If A = [1] B = [1, 2, 25. Compute 1) [, , 5 7] Find AB, , 1 −2 1, ][, 2 3 2, , 2 3, ], 3 1, , 2 −3, 3 −1 3, 2) [, ] [1 0 ], −1 0 2, 3 1, 26. Find X and Y [, , 2𝑥 + 𝑦, 6, , 3𝑦, 6 0, ] = [, ], 6 4, 4, , 27. What is the number of possible square matrix order 3 with each entries 0 or 1, 5−𝑥, 28. Find X and Y if [, 0, , 2𝑦 − 8, ] is a scalar matrix, 3, , 4, 𝑥+2, 29. Find X [, ] is a symmetric matrix, 2𝑥 − 3 𝑥 + 1

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II. Two mark and Three marks questions (SA), 1.Radha , fauzia, simran are the student of 12th class Radha has 15 note book and 6 pens ,, Fauzia has 10 books 2 pens and Simran has 13 books and 5 pens express this in to matrix, forms., 2. Construct 3× 2 matrix whose elements are given by 𝑎𝑖𝑗 =, , 1, 2, , |𝑖 − 3𝑗|, , 𝑥+𝑦+𝑧, 9, 3. Find X,Y,Z from the equation [ 𝑥 + 𝑧 ] = [5], 𝑦+𝑧, 7, 4. Find a,b,c, d From the equation [, , 𝑎−𝑏, 2𝑎 − 𝑏, , 2𝑎 + 𝑐, −1 5, ]==[, ], 3𝑐 + 𝑑, 0 13, , 8 0, 2 −2, 5. If A = [4 −2] B = [ 4, 2 ] Find X such that 2A + 3X = 5B, 3 6, −5 1, 6. Find X and Y 2[, , 𝑥, 7, , 5, 7, 6, 3 −4, ]=[, ], ]+[, 𝑦−3, 1 2, 15 14, , 2, −1 10, 7. Find X and Y if x [ ] + 𝑦 [ ]= [ ], 3, 1, 5, 𝑥, 8. Given 3 [, 𝑧, 9. If 𝐴𝑋 = [, , 𝑦, 𝑥, ]=[, 𝑤, −1, , cos 𝑥, −sin 𝑥, , 4, 6, ]+ [, 2𝑤, 𝑧+𝑤, , 𝑥+𝑦, ] Fine the values of X,Y,Z and W, 3, , cos 𝑦, sin 𝑥, ] and 𝐴𝑌 = [, −sin 𝑦, cos 𝑥, , sin 𝑦, ] Show that 𝐴𝑋 𝐴𝑌 = 𝐴𝑋+𝑌, cos 𝑦, , 3, 10. If A = [, 4, , 1 0, −2, ] and I = [, ] Find K If A2 = KA – 2I, −2, 0 1, , 11. If A = [3, 4, , √3 2] and B =[2 −1 2] verify (A + B )1 = A1 +B1, 1 2 4, 2 0, , 12. For any matrix A with real number entries , A+ A1 is symmetric matrix and A – A1, Skew-symmetric matrix, 13. For any matrix A = [, , 1 5, ] verify that A+ A1 is symmetric matrix, 6 7, , 14. For any matrix A = [, , 1 5, ] verify that A – A1 Skew-symmetric matrix, 6 7, , 15. If A and B be the invertible matrices of same order then (AB)-1 = B-1A-1, 16. By using elementary operation Find the inverse of the matrix [, , 1 2, ], 2 −1

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17. By using elementary operation Find the inverse of the matrix [, , 1 3, ], 2 7, , 18. By using elementary operation Find the inverse of the matrix [, , 1 −2, ], 2 1, , 19. Find P-1 if it exists and P = [, , 10 −2, ], −5 1, , 3 1, 20. If A = [, ] Show that A2 -5A +7I = 0, −1 2, 2, 21. If A = [, 0, , 3, 1 5, ] and B = [, ] Show that (AB)1 = B1A1, −4, 2 0, , III. Five mark questions ( LA), 1, 1.If A = [2, 3, , 1 −1, 1 3, 1 2, 0, 3 ] B = [ 0 2] and C =[, 2 0, −1 2, −1 4, , 3, −2, , −4, ], 1, , Find A B , BC and show that (AB )C = A(BC), 0, 6 7, 0 1 1, 2, 2. If A = [−6 0 8] B = [1 0 2] C = [−2] calculate AC, BC and (A+B) C, 7 −8 0, 1 2 0, 3, Deduce that (A+B) C = AC + BC, 1 2 3, 3. If A = [3 −2 1] Show that A3 – 23A - 40 I = 0, 4 2 1, 1 2 −3, 3 −1 2, 4 1, 4. If A = [5 0, 2 ] B = [4 2 5] and C = [0 3, 2 0 3, 1 −2, 1 −1 1, verify A+ (B-C) = (A+B ) –C, 2, , 5. If A =, , 3, 1, 3, 7, , 1, 2, 3, , [3 2, , 5, , 2, , 3, , 3, 4, , 5, 1, , 5, 2, , 5, 7, , 5, 6, , 3, 2, , 3], , and B =, , [5, , 5, , 1, 4, 5, 2, , 5], , 2 0 1, 6. If A = [2 1 3] find A2 – 5 A + 6 I ?, 1 −1 0, , find 3A – 5B, , 2, 2], 3

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1 0 2, 7. If A = [0 2 1] prove that A3 – 6A2 + 7A + 2I = 0, 2 0 3, 2 −2 −4, 8. Express the matrix B = [−1 3, 4 ] Find the sum of symmetric and skew1 −2 −3, symmetric matrix, 6 −2 2, 9. Express the matrix B = [−2 3 −1] Find the sum of symmetric and skew2 −1 3, symmetric matrix, 2 0, 1 1, 1 2, 10. If A = [, ]B=[, ]C=[, ] calculate AB , BC, A(B+C), 2 1, 1 3, 2 3, Verify that AB + AC = A(B+C), cos 𝑥, 11. If F(x) = [ sin 𝑥, 0, , − sin 𝑥, cos 𝑥, 0, , 0, 0] show that F(x) F(y) = F(x+y), 1, , −2, 12. If A =[ 4 ] and B = [1 3 −6] verify (AB)1 = B1A1, 5, cos 𝜃, 13. If A = [, −sin 𝜃, , sin 𝜃, cos 𝑛𝜃, ] Prove that An = [, cos 𝜃, −sin 𝑛𝜃, , ********, , sin 𝑛𝜃, ], cos 𝑛𝜃

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Solutions, One mark questions (VSA), 1. The numbers arranged in rectangular array of rows and columns by the brockets is, called matrix, 2. A square matrix is said to be diagonal matrix if all non diagonals elements are zeros, 3. A diagonal matrix is said to be scalar marics if it’s diagonal elements are equal, 4. If a square matrix A = [𝑎𝑖𝑗 ]𝑚×𝑚 is said to be symmetric if and only if A1 = A, 5. If a square matrix A = [𝑎𝑖𝑗 ]𝑚×𝑚 is said to be skew-symmetric if and only if A1 = -A, 6. 1) order of the matrix is 3 × 4, , 2) 19, -2, -5 , 12,, , 5, 2, , 7. Possible orders are (1,8) (8,1) (2,4) (4,2) is 1X8 , 8X 1 , 2X 4 , 4X 2, 8. Possible orders are (1,18) (18,1) (3,6) (6,3) (2,9) (9,2) is 1 X18 , 18X1 , 3X6 , 6X3 ,, 2X9, 9X2 ,, 4 9, 9. 1) [𝑎𝑖𝑗 ] = [, ], 9 16, 10., 11., , 2, , 2) [𝑎𝑖𝑗 ] = [ 9, , 2, , 9, 2, , 8, , ], , 0 1 2, ], 1 0 2, 1, 1, 2, [ 2 1], 3, 3, , [, , 2, , 12. X= 1, 13. X = 2, , Y=4, Z= 3, Y=4, Z=0, 2 −4 3 −2, −1 −2, 14. X = -2A - B = [, ]-[, ]=[, ], −6 −8 1 5, −7 −13, −1 5 3, 15. 2A –B = [, ], 5 6 0, −1 −1, 16. By solving X = [, ], −2 −1, 2 0, 5 0, 17. By solving above matrix X = [, ] and Y = [, ], 1 1, 1 4, 1 0, 18. By multiplying we get the answer [, ]=I, 0 1, 19. 2+Y = 5 implies Y = 3 and 2x+2 = 8 implies x =3, 2 5, 20. A + A1 = [, ], 5 8

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3 −2 19, ], 12 −3 14, 1 0, sin 𝜃 cos 𝜃 sin 𝜃 −cos 𝜃, A A1 = [, ][, ]=[, ] = I after multiplying, 0 1, −cos 𝜃 sin 𝜃 cos 𝜃 sin 𝜃, 1 0, cos 𝜃 sin 𝜃 cos 𝜃 −sin 𝜃, BB 1 = [, ][, ]=[, ]= I after multiplying, 0 1, −sin 𝜃 cos 𝜃 sin 𝜃 cos 𝜃, 0 1 2, 1, (AB) = [0 5 10], 0 7 14, −3 −4 1, 14 −6, 1) [, ], 2) [, ] after multiplying, 8 13 3, 4, 5, 3 |𝐴| = K |𝐴| implies K = 3, Y = 0, X = 3 by solving, The square matrix of order 3X3 = 9 and 2 entries, , 21. 3 A +2 B = [, 22., 23., 24., 25., 26., 27., 28., , Then possible entries is 29 = 512, 5 − 𝑥 2𝑦 − 8 3 0, 29. [, ]= [, ] then X = 2 Y = 4, 0 3, 0, 3, 4, 𝑥+2, 4, 2𝑥 − 3, 30. [, ] = [, ] implies X = 5, 2𝑥 − 3 𝑥 + 1, 𝑥+2 𝑥+1, , Solutions : Two mark and Three marks questions (SA), 1., , books, Radha :, , 15, , Fauzia :, Simran:, , 10, 13, , 2. 𝑎𝑖𝑗 =, , 1, , 5, , 1, , 2, , 2, , [0, 3., , pens, 15 6, 15 10 13, 6 this can be expressed as [10 2] or [, ], 6, 2, 5, 13 5, 2, 5, , 2, , 3, , 2], , X+ Y + Z = 9, , X +Z = 5, , Y+Z=7, , 7+Z=9, , X+2=5, , Y+2=7, , X=3, , Y=5, , Z= 2, , 4. By solving equality a =1 , b= 2, c =3 and d = 4

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−2, 5. X =, , 4, [−, , −, , 10, , 3, 14, , 31, , 3, −7, , 3, , 3, , ], , 6. compare two matrices X = 2, Y = 9, 7. by solving we get X = 3 , Y = -4, 8. by solving and compare we get X = 2 , Y = 4,, cos 𝑥, 9. 𝐴𝑋 𝐴𝑌 = [, −sin 𝑥, , cos 𝑦, sin 𝑥, ] [, cos 𝑥 −sin 𝑦, , Z = 1,, , w=3, , cos(𝑥 + 𝑦) sin(x + 𝑦), sin 𝑦, ] =[, ] = 𝐴𝑋+𝑌, cos 𝑦, −sin(𝑥 + 𝑦) cos(𝑥 + 𝑦), , 10. A2 = KA – 2I, 1 −2, 3𝑘 − 2, −2𝑘, [, ] =[, ], 4 −4, 4𝑘, −2𝑘 − 2, Then 4K = 4, K =1, 5, 5, 5, 1, 1, 11. ( A+B) = [√3 − 1 4] and A + B = [√3 − 1, 4, 4, 4, 1, , 5, 4], 4, , Hence ( A+B)1 = A1 + B 1, 12. B = A + A1 , B1 = (A + A1 )1 = A1+A = B ∴ B = A +A1 is symmetric, C = A –A 1 , C 1 = (A -A1 )1 = A1-A = -(A- A1) = - C, 13. Z = A + A1 = [, , 1, 1 5, ] +[, 5, 6 7, , 6, 2 11, ]=[, ] = Z1, 7, 11 14, , ∴ C = A – A1 is skew- symmetric, ∴ Z = Z1 = A + A1 is symmetric, , 1 6 1, 0 −1 1, 1 5, 14. Z1 = (A - A1)1 = ([, ] − [, ]) = ([, ] ) = -Z, 5 7, 1 0, 6 7, skew- symmetric, 15., (AB) (AB)-1 = I, A-1(AB) (AB)-1 = A-1I, IA=A, B(AB)-1 = A-1, IA-1 = A-1, B-1B(AB)-1 = B-1A-1, AA-1 = I, (AB)-1 = B-1A-1, BB-1 I, , ∴ Z1 = - Z,, , A –A1

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16., , 1 2, ], 2 −1, A = IA, 1 0, 1 2, [, ]=[, ]A, 2 −1, 0 1, 1 2, 1 0, [, ]=[, ]A, 0 −5, −2 1, 1 0, 1 2, [, ] = [ 2 − 1] A, 0 1, A=[, , 5, 1, , 𝑅2 = 𝑅2 − 2𝑅1, 1, , 𝑅2 = − 𝑅2, 5, , 𝑅1 = 𝑅1 − 2𝑅2, , 5, , 2, , 1 0, [, ] = [52 5 1] A, 0 1, −, 5, , 5, , 1, , 2, , ∴ A -1 = [52, , 5, , 5, , −, , 1], 5, , 17. By above process ∴A -1 = [, , 7 −3, ], −2 1, 1, , 5, 18. By above prose’s ∴A -1 = [−2, 5, , 2, 5, 1], 5, , 10 −2, ], −5 1, P = IP, 10 −2, 1 0, [, ] =[, ] P, −5 1, 0 1, By elementary operation, , 19. P = [, , 1, [, 0, , 20., 21., 22., 23., 24., , 25., , 1, , 0, −, 10, 5] = [ 1, ]p, 1, 0, 2, 1, , p-1 does not exists, 7 0, 0 0, 8 5, 15 5, A2 -5A +7I = [, ]-[, ]+[, ]= [, ]=0, 0 7, 0 0, −5 3 −5 10, By mathematical induction we get the solution, If A=A1 , B = B1 , (AB)1 = AB, (AB)1 = B1 A1 = BA, ∴ AB = BA, AB Is symmetric, 2, A = A A By product of two matrix get the solution, 8 −8, (AB)1 =[, ], 10 0, 8 −8, B1A1 = [, ], 10 0, ∴ (AB)1 = B1A1, By solving x = 2 , y = 4 , z = 3

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Solutions : Five mark questions (LA), 4, 4, 4 −7, (AB) C = [35 −2 −39, 22], 31 2 −27, 11, , 2, 1, 1.AB =[−1 18], 1, 5, 7 2 −3, BC = [4 0, −4, 7 −2 −11, , −1, 2], 8, , 4, 4, 4 −7, A(BC) = [35 −2 −39, 22], 31 2 −27, 11, , Hence (AB) C = A(BC), 2. (A +B) C =, , 10, [20], 28, , 9, 1, AC = [12] BC = [ 8 ], 30, −2, , 10, AC + BC = [20], 28, , Hence (A +B) C = AC + BC, 1, 3. A = [3, 4, , 2 3, −2 1], 2 1, , 19 4, 8, A = [ 1 12 8 ], 14 6 15, 2, , 63 46 69, A = [60 −6 23], 92 46 63, 3, , LHS = A3 – 23A – 40 I = 0 By simplification, 0, 4. A + (B – C) = [9, 2, , 0 −3, 0, −1 5 ] and (A+B) –C = [9, 1, 1, 2, , 0 −3, −1 5 ], 1, 1, , Hence A + (B – C) = (A+B) –C, 2 3, 5. 3A -5B = [1 2, 7 6, , 0 0 0, 5, 2 3 5, 4] - [ 1 2 4] = [ 0 0 0] = 0, 0 0 0, 2, 7 6 2, , 1, 6. A – 5A + 6I = [−1, −5, , −1 −3, −1 −10] by simplification, 4, 4, , 2, , 1 0, 7. If A = [0 2, 2 0, 2, 8. B = [−1, 1, , 2, 1] by calculating A2 , A3 take LHS = RHS, 3, , −2 −4, 3, 4 ] by theorem number 2, −2 −3, , 1, , 1, , 2, , 2, , B = (B +B 1) + (B -B 1) hence they are equal

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6, 9. B = [−2, 2, , −2 2, 3 −1] by theorem number 2, −1 3, , 1, , 1, , 2, , 2, , B = (B +B 1) + (B -B 1) hence they are equal, 4, 11. If AB = [, 5, , 6, ], 3, , cos 𝑥, 12. F(x).F(y) = [ sin 𝑥, 0, , 5 7, AC = [, ], 4 5, − sin 𝑥, cos 𝑥, 0, , 0 cos 𝑦, 0] [ sin 𝑦, 1, 0, , 9, A(B+C) = [, 9, − sin 𝑦, cos 𝑦, 0, , 0, 0], 1, , cos(𝑥 + 𝑦) − sin(𝑥 + 𝑦) 0, = [ sin(𝑥 + 𝑦) cos(𝑥 + 𝑦) 0] = F(x+y), 0, 0, 1, −2, 4, 5, 13. LHS = (AB) = [−6 12, 15 ] = B1A1 = RHS, 12 −24 −30, 1, , 14. By mathematical induction we get the solution, , 13, ] = AB + AC, 8