{"id":29398,"date":"2022-03-29T10:00:23","date_gmt":"2022-03-29T04:30:23","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=29398"},"modified":"2022-03-29T10:31:20","modified_gmt":"2022-03-29T05:01:20","slug":"ncert-solutions-for-class-12-maths-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-3-ex-3-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 3 Matrices Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-3-ex-3-2\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2<\/h2>\n

\"NCERT<\/p>\n

Ex 3.2 Class 12 Question 1.<\/strong>
\nLet
\nA = \\(\\left[\\begin{array}{ll} 2 & 4 \\\\ 3 & 2 \\end{array}\\right]\\)
\nB = \\(\\left[\\begin{array}{ll} 1 & 3 \\\\ -2 & 5 \\end{array}\\right]\\)
\nC = \\(\\left[\\begin{array}{ll} -2 & 5 \\\\ 3 & 4 \\end{array}\\right]\\)
\nF ind each of the following:
\ni. A + B
\nii. A – B
\niii. 3A – C
\niv. AB
\nv. BA
\nSolution:
\n\"Ex<\/p>\n

Class 12 Maths 3.2 Question 2.<\/strong>
\nCompute the following
\ni. \\(\\left[\\begin{array}{cc}
\na & b \\\\
\n-b & a
\n\\end{array}\\right]+\\left[\\begin{array}{ll}
\na & b \\\\
\nb & a
\n\\end{array}\\right]\\)
\nii. \\(\\left[\\begin{array}{cc}
\na^{2}+b^{2} & b^{2}+c^{2} \\\\
\na^{2}+c^{2} & a^{2}+b^{2}
\n\\end{array}\\right]+\\left[\\begin{array}{cc}
\n2 a b & 2 b c \\\\
\n-2 a c & -2 a b
\n\\end{array}\\right]\\)
\niii. \\(\\left[\\begin{array}{ccc}
\n-1 & 4 & -6 \\\\
\n8 & 5 & 16 \\\\
\n2 & 8 & 5
\n\\end{array}\\right]+\\left[\\begin{array}{ccc}
\n12 & 7 & 6 \\\\
\n8 & 0 & 5 \\\\
\n3 & 2 & 4
\n\\end{array}\\right]\\)
\niv. \\(\\left[\\begin{array}{cc}
\n\\cos ^{2} x & \\sin ^{2} x \\\\
\n\\sin ^{2} x & \\cos ^{2} x
\n\\end{array}\\right]+\\left[\\begin{array}{cc}
\n\\sin ^{2} x & \\cos ^{2} x \\\\
\n\\cos ^{2} x & \\sin ^{2} x
\n\\end{array}\\right]\\)
\nSolution:
\n\"Class<\/p>\n

\"NCERT<\/p>\n

Exercise 3.2 Class 12 Maths Question 3.<\/strong>
\nCompute the indicated products.
\n\"Exercise
\nSolution:
\n\"Ex<\/p>\n

iii. Number of columns of first matrix = Number of rows of second matrix.
\n\"Class<\/p>\n

\"Class<\/p>\n

Class 12 Maths Ncert Solutions Chapter 3 Matrices Exercise 3.2 Question 4.<\/strong>
\nIf A = \\(\\left[\\begin{array}{ccc}
\n1 & 2 & -3 \\\\
\n5 & 0 & 2 \\\\
\n1 & -1 & 1
\n\\end{array}\\right]\\), B = \\(=\\left[\\begin{array}{ccc}
\n3 & -1 & 2 \\\\
\n4 & 2 & 5 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\) and C = \\(\\left[\\begin{array}{ccc}
\n4 & 1 & 2 \\\\
\n0 & 3 & 2 \\\\
\n1 & -2 & 3
\n\\end{array}\\right]\\) then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A+ B) – C.
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Class 12 Ex 3.2 Question 5.<\/strong>
\nIf A = \\(\\left[\\begin{array}{ccc}
\n\\frac{2}{\\dot{3}} & 1 & \\frac{5}{3} \\\\
\n\\frac{1}{3} & \\frac{2}{3} & \\frac{4}{3} \\\\
\n\\frac{7}{3} & 2 & \\frac{2}{3}
\n\\end{array}\\right]\\) and B = \\(\\left[\\begin{array}{ccc}
\n\\frac{2}{5} & \\frac{3}{5} & 1 \\\\
\n\\frac{1}{5} & \\frac{2}{5} & \\frac{4}{5} \\\\
\n\\frac{7}{5} & \\frac{6}{5} & \\frac{2}{5}
\n\\end{array}\\right]\\), then compute 3A – 5B.
\nSolution:
\n\"NCERT<\/p>\n

Question 6.
\nSimplify cos \u03b8\\(\\left[\\begin{array}{ll}
\n\\cos \\theta & \\sin \\theta \\\\
\n-\\sin \\theta & \\cos \\theta
\n\\end{array}\\right]\\) + sin \u03b8\\(\\left[\\begin{array}{cc}
\n\\sin \\theta & -\\cos \\theta \\\\
\n\\cos \\theta & \\sin \\theta
\n\\end{array}\\right]\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 7.
\nFind X and Y, if
\ni. X + Y = \\(\\left[\\begin{array}{ll}
\n7 & 0 \\\\
\n2 & 5
\n\\end{array}\\right]\\) and X – Y = \\(\\left[\\begin{array}{ll}
\n3 & 0 \\\\
\n0 & 3
\n\\end{array}\\right]\\)<\/p>\n

ii. 2X + 3Y = \\(\\left[\\begin{array}{ll}
\n2 & 3 \\\\
\n4 & 0
\n\\end{array}\\right]\\) and 3X + 2Y = \\(\\left[\\begin{array}{ll}
\n2 & -2 \\\\
\n-1 & 5
\n\\end{array}\\right]\\)
\nSolution:
\n\"NCERT<\/p>\n

ii. 2X + 3Y = \\(\\left[\\begin{array}{ll}
\n2 & 3 \\\\
\n4 & 0
\n\\end{array}\\right]\\) …… (1)
\n3X + 2Y = \\(\\left[\\begin{array}{ll}
\n2 & -2 \\\\
\n-1 & 5
\n\\end{array}\\right]\\) …… (2)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nFind X, if Y = \\(\\left[\\begin{array}{ll}
\n3 & 2 \\\\
\n1 & 4
\n\\end{array}\\right]\\) and 2X + Y = \\(\\left[\\begin{array}{cc}
\n1 & 0 \\\\
\n-3 & 2
\n\\end{array}\\right]\\).
\nSolution:
\n\"NCERT<\/p>\n

Question 9.
\nFind x and y, if 2\\(\\left[\\begin{array}{ll}
\n1 & 3 \\\\
\n0 & x
\n\\end{array}\\right]\\) + \\(\\left[\\begin{array}{cc}
\ny & 0 \\\\
\n1 & 2
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n5 & 6 \\\\
\n1 & 8
\n\\end{array}\\right]\\)
\nSolution:
\n2\\(\\left[\\begin{array}{ll}
\n1 & 3 \\\\
\n0 & x
\n\\end{array}\\right]\\) + \\(\\left[\\begin{array}{cc}
\ny & 0 \\\\
\n1 & 2
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n5 & 6 \\\\
\n1 & 8
\n\\end{array}\\right]\\) \u21d2 \\(\\left[\\begin{array}{ll}
\n2 & 6 \\\\
\n0 & 2x
\n\\end{array}\\right]\\) + \\(\\left[\\begin{array}{cc}
\ny & 0 \\\\
\n1 & 2
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n5 & 6 \\\\
\n1 & 8
\n\\end{array}\\right]\\)
\n\u21d2 \\(\\left[\\begin{array}{cc}
\n2+y & 6 \\\\
\n1 & 2 x+2
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n5 & 6 \\\\
\n1 & 8
\n\\end{array}\\right]\\)
\nEquating the corresponding elements, we get
\n2 + y = 5 and 2x + 2 = 8
\n\u21d2 y = 3 and 2x = 6 \u21d2 x = 3
\n\u2234 x = 3, y = 3<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nSolve the equation for x, y, z and t, if 2\\(\\left[\\begin{array}{ll}
\nx & z \\\\
\ny & t
\n\\end{array}\\right]\\) + 3\\(\\left[\\begin{array}{ll}
\n1 & -1 \\\\
\n0 & 2
\n\\end{array}\\right]\\) = 3\\(\\left[\\begin{array}{ll}
\n3 & 5 \\\\
\n4 & 6
\n\\end{array}\\right]\\)
\nSolution:
\n2\\(\\left[\\begin{array}{ll}
\nx & z \\\\
\ny & t
\n\\end{array}\\right]\\) + 3\\(\\left[\\begin{array}{ll}
\n1 & -1 \\\\
\n0 & 2
\n\\end{array}\\right]\\) = 3\\(\\left[\\begin{array}{ll}
\n3 & 5 \\\\
\n4 & 6
\n\\end{array}\\right]\\)
\n\u21d2 \\(\\left[\\begin{array}{ll}
\n2x & 2z \\\\
\n2y & 2t
\n\\end{array}\\right]\\) + \\(\\left[\\begin{array}{cc}
\n3 & -3 \\\\
\n0 & 6
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n9 & 15 \\\\
\n12 & 18
\n\\end{array}\\right]\\)
\n\u21d2 \\(\\left[\\begin{array}{cc}
\n2x+3 & 2z-3 \\\\
\n2y & 2t+6
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n9 & 15 \\\\
\n12 & 18
\n\\end{array}\\right]\\)
\n\u21d2 2x + 3 = 9, 2z – 3 = 15
\n2y = 12, 2t + 6 = 18 \u21d2 x = 3, y = 6, t = 6, z = 9<\/p>\n

Question 11.
\nIf x\\(\\left[\\begin{array}{l}
\n2 \\\\
\n3
\n\\end{array}\\right]\\) + y\\(\\left[\\begin{array}{l}
\n-1 \\\\
\n1
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{l}
\n10 \\\\
\n5
\n\\end{array}\\right]\\)
\nSolution:
\nx\\(\\left[\\begin{array}{l}
\n2 \\\\
\n3
\n\\end{array}\\right]\\) + y\\(\\left[\\begin{array}{l}
\n-1 \\\\
\n1
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{l}
\n10 \\\\
\n5
\n\\end{array}\\right]\\) \u21d2 \\(\\left[\\begin{array}{l}
\n2 x \\\\
\n3 x
\n\\end{array}\\right]+\\left[\\begin{array}{c}
\n-y \\\\
\ny
\n\\end{array}\\right]=\\left[\\begin{array}{c}
\n10 \\\\
\n5
\n\\end{array}\\right]\\) \u21d2 \\(\\left[\\begin{array}{l}
\n10 \\\\
\n5
\n\\end{array}\\right]\\)
\n\u21d2 2x – y = 10, 3x + y = 5
\nSolving, we get x = 3, y = – 4<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nGiven
\n\\(3\\begin{bmatrix} x & \\quad y \\\\ z & \\quad w \\end{bmatrix}=\\begin{bmatrix} x & \\quad 6 \\\\ -1 & \\quad 2w \\end{bmatrix}+\\begin{bmatrix} 4 & \\quad x+y \\\\ z+w & 3 \\end{bmatrix} \\)
\nfind the values of x,y,z and w.
\nSolution:
\n\\(3\\begin{bmatrix} x & \\quad y \\\\ z & \\quad w \\end{bmatrix}=\\begin{bmatrix} x & \\quad 6 \\\\ -1 & \\quad 2w \\end{bmatrix}+\\begin{bmatrix} 4 & \\quad x+y \\\\ z+w & 3 \\end{bmatrix} \\)
\n\u21d2 \\(\\begin{bmatrix} 3x & \\quad 3y \\\\ 3z & \\quad 3w \\end{bmatrix}=\\begin{bmatrix} x+4 & \\quad 6+x+y \\\\ -1+z+w & \\quad 2w+3 \\end{bmatrix}\\)
\n\u21d2 3x = x + 4 \u21d2 x = 2
\nand 3y = 6 + x + y \u21d2 y = 4
\nAlso, 3w = 2w + 3 \u21d2 w = 3
\nAgain, 3z = – 1 + z + w
\n\u21d2 2z = – 1 + 3
\n\u21d2 2z = 2
\n\u21d2 z = 1
\nHence x = 2 ,y = 4, z = 1, w = 3.<\/p>\n

Question 13.
\nIf F(x) = \\(\\left[\\begin{array}{ccc}
\n\\cos x & -\\sin x & 0 \\\\
\n\\sin x & \\cos x & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\) then show that F(x).F(y) = F(x+y)
\nSolution:
\n\"NCERT<\/p>\n

Question 14.
\nShow that
\n\"NCERT
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nFind A\u00b2 – 5A + 6I, if A = \\(\\left[\\begin{array}{ccc}
\n2 & 0 & 1 \\\\
\n2 & 1 & 3 \\\\
\n1 & -1 & 0
\n\\end{array}\\right]\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 16.
\nIf A = \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\) Prove that A\u00b3 – 6A\u00b2 + 7A + 2I = 0
\nSolution:
\n\"NCERT<\/p>\n

Question 17.
\nIf \\(\\left[\\begin{array}{ll}
\n3 & -2 \\\\
\n4 & -2
\n\\end{array}\\right]\\) and I = \\(\\left[\\begin{array}{ll}
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right]\\) find k so that A\u00b2 = kA – 2I
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 18.
\nIf A = \\(\\left[\\begin{array}{cc}
\n0 & -\\tan \\frac{\\alpha}{2} \\\\
\n\\tan \\frac{\\alpha}{2} & 0
\n\\end{array}\\right]\\) and I is the identity matrix of order 2, show that I + A = (I – A)\\(\\left[\\begin{array}{cc}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 19.
\nA trust has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bond if the trust fund obtains an annual total interest of
\n(a) Rs 1800
\n(b) Rs 2000
\nSolution:
\n(a) Let \u20b9 and \u20b9 y be the amount invested in two bonds such that x + y = 30000 … (1)
\nThen [x y]\\(\\left[\\begin{array}{l}
\n5 \\% \\\\
\n7 \\%
\n\\end{array}\\right]\\) = [1800]
\n\u21d2 [x \u00d7 5% + y x 7%] = [1800]
\n\u21d2 \\(\\frac{x \\times 5}{100}+\\frac{y \\times 7}{100}\\)
\n\u21d2 5x + 7y = 180000 … (2)
\n(1) x 5 = 5x + 5y = 150000 … (3)
\nSubtracting (3) from (2), we get
\n2y = 30000 \u21d2 y = 15000
\nFrom (1), we get x= 15000
\n\u2234 To get an interest of \u20b9 1800, the amount to be invested in each bonds are \u20b9 15000 and \u20b9 15000<\/p>\n

(b) Here x + y = 30000 … (4)
\n[x y]\\(\\left[\\begin{array}{l}
\n5 \\% \\\\
\n7 \\%
\n\\end{array}\\right]\\) = 2000
\n\u21d2 \\(\\frac{x \\times 5}{100}+\\frac{y \\times 7}{100}\\) = 2000
\n\u21d2 5x + 7y = 200000 … (5)
\n(4) x 5 \u21d2 5x + 5y = 150000 …. (6)
\nSubtracting (6) from (5), we get
\n2y = 50000 \u21d2 y = 25000
\nFrom (4) we get x = 5000
\nTo get an interest of \u20b9 2000, the amount to be invested in each bonds is \u20b9 5000 and \u20b9 25000.<\/p>\n

Question 20.
\nThe bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are \u20b9 80, \u20b9 60 and \u20b9 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
\nSolution:
\n1 dozen = 12
\n\u2234 10 dozen chemistry books = 10 x 12 = 120
\n8 dozen physics books = 8 x 12 = 96
\n10 dozen economics books = 10 x 12 = 120
\nTotal amount the bookshop will receive
\n= [120 96 120]\\(\\left[\\begin{array}{l} 80 \\\\ 60 \\\\ 40 \\end{array}\\right]\\)
\n= [120 x 80 + 96 x 60 + 120 x 40]
\n= [9600 + 5760 + 4800] = \u20b9 20160
\nAssume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k, respectively. Choose the correct answer in Exercises 21 and 22.<\/p>\n

\"NCERT<\/p>\n

Question 21.
\nThe restriction on n, k and p so that PY + WY will be defined are
\na. k = 3, p = n
\nb. k is arbitrary, p = 2
\nc. p is arbitrary, k = 3
\nd. k = 2, p = 3
\nSolution:
\nAnswer:
\nPY + WY = (P + W)Y
\n\u21d2 order of P = order of W
\n\u21d2 p k = n x 3
\n\u21d2 p = n and k = 3<\/p>\n

Question 22.
\nIf n = p, then the order of the matrix 7X – 5Z is
\na. p x 2
\nb. 2 x n
\nc. n x 3
\nd. p x n
\nSolution:
\nOrder of X = 2 x n
\nOrder of Z = 2 x p = 2 x n \u2235 (n = p)
\n\u2234 Order of 7X – 5Z is order of X or order of Z.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-3-ex-3-2\/ NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2 Ex 3.2 Class 12 Question 1. Let A = B = C = F ind each of the following: i. …<\/p>\n

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