{"id":29762,"date":"2024-01-18T17:12:18","date_gmt":"2024-01-18T11:42:18","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=29762"},"modified":"2024-01-19T17:58:15","modified_gmt":"2024-01-19T12:28:15","slug":"ncert-solutions-for-class-12-maths-chapter-3-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-3-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 3 Matrices Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

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

\"NCERT<\/p>\n

Question 1.
\nLet A = \\(\\left[\\begin{array}{ll} 0 & 1 \\\\ 0 & 0 \\end{array}\\right]\\) show that (aI + bA)\u207f = a\u207fI + nan-1<\/sup> bA, where I is the identity matrix of order 2 and n \u2208 N .
\nSolution:
\nLet P(n) : (aI + bA)\u207f = a\u207fI + nan-1<\/sup> bA
\nP(1) : (aI + bA)1 <\/sup>= a1<\/sup>I + 1a\u00b0bA
\n\u21d2 P(1) : aI + bA = aI + bA
\n\u2234 P(1) is true
\nLet us assume that P(k) is true.
\ni.e., P(k) : (aI + bA)k<\/sup> = ak<\/sup>I + kak-1<\/sup>bA
\nP(k+ 1) : (aI + bA)k+1<\/sup>
\n= (aI + bA)k<\/sup>.(aI + bA)
\n= (ak<\/sup>I + kak-1<\/sup>A) (aI + bA)
\n= ak<\/sup>.a.I.l + ak<\/sup>.b. I.A + kak-1<\/sup>.b.a. AI + kak-1<\/sup>b.bA.A
\n= ak+1<\/sup>I + ak<\/sup>bA + kak<\/sup>bA + kak-1<\/sup>b\u00b2A\u00b2
\n= ak+1<\/sup>I + (k + 1)ak<\/sup>bA + kak-1<\/sup>b\u00b2A\u00b2 … (1)
\nA\u00b2 = \\(\\left[\\begin{array}{ll}
\n0 & 1 \\\\
\n0 & 0
\n\\end{array}\\right]\\left[\\begin{array}{ll}
\n0 & 1 \\\\
\n0 & 0
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ll}
\n0 & 0 \\\\
\n0 & 0
\n\\end{array}\\right]\\) = 0
\n\u2234 (1) \u21d2 P(k + 1) = ak+1<\/sup> I + (k + 1)ak<\/sup>bA + kakk-1<\/sup>b\u00b2 x 0
\n= ak+1<\/sup> I + (k + 1)a(k+1)-1<\/sup>Ab
\nwhich is true whenever P(L) is true.
\nHence by the Principle of Mathematical Induction, the result is true for all n \u2208 N .<\/p>\n

Question 2.
\nIf A = \\(\\left[\\begin{array}{lll}
\n1 & 1 & 1 \\\\
\n1 & 1 & 1 \\\\
\n1 & 1 & 1
\n\\end{array}\\right]\\), prove that An<\/sup> = \\(\\left[\\begin{array}{ccc}
\n3^{n-1} & 3^{n-1} & 3^{n-1} \\\\
\n3^{n-1} & 3^{n-1} & 3^{n-1} \\\\
\n3^{n-1} & 3^{n-1} & 3^{n-1}
\n\\end{array}\\right]\\), n \u2208 N
\nSolution:
\n\"NCERT
\nwhich is true
\nwhenever P(k) is true.
\nHence by the Principle of Mathematical Induction P(n) is true for all n \u2208 N.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIf A = \\(\\left[\\begin{array}{ll}
\n3 & -4 \\\\
\n1 & -1
\n\\end{array}\\right]\\), then prove that A\u207f = \\(\\left[\\begin{array}{cc}
\n1+2 n & -4 n \\\\
\nn & 1-2 n
\n\\end{array}\\right]\\), where n is any positive integer.
\nSolution:
\n\"NCERT
\n\u2234 P(k + 1) is true, whenever P(k) is true.
\nHence by the Principle of Mathematical Induction P(n) is true for all n \u2208 N.<\/p>\n

Question 4.
\nIf A and Bare symmetric matrices, prove that AB – BA is a skew symmetric matrix.
\nSolution:
\nA and B are symmetric matrices
\n\u2234 A\u2019 = A and B’ = B<\/p>\n

(i) (AB – BA)\u2019 = (AB)\u2019 – (BA)\u2019
\n= B’A\u2019 – A\u2019B\u2019 since (AB)\u2019 = B\u2019A\u2019
\n= BA – AB since A\u2019 = A and
\nB\u2019 = B
\n= – (AB – BA)
\n\u2234 AB – BA is a skew-symmetric matrix.<\/p>\n

(ii) Let AB = BA
\nTaking transpose on both sides,
\n(AB)\u2019 = (BA)\u2019
\ni.e.,(AB)\u2019 = A\u2019B\u2019 since (BA)\u2019 = A\u2019B\u2019
\ni.e., (AB)\u2019 = AB, since A\u2019 A and B\u2019 = B
\n\u2234 AB \u00a1s a symmetric matrix.
\nLet AB be a symmetric matrix.
\nThen (AB)\u2019 = AB
\n\u21d2 B\u2019A\u2019 = AB
\n\u21d2 BA = AB since B\u2019 = B and A\u2019 = A
\n\u2234 AB = BA<\/p>\n

Question 5.
\nShow that the matrix B’AB is symmetric or skew symmetric according as A is sym\u00acmetric or skew symmetric.
\nSolution:
\nLet A be a symmetric matrix.
\nThen A\u2019 = A
\n\u2234 (B\u2019AB)\u2019 = (B\u2019(AB))\u2019
\n= (AB)\u2019(B\u2019)\u2019
\n= B\u2019A\u2019B
\n= B\u2019AB since A\u2019 = A
\n\u2234 B\u2019AB is a symmetric matrix
\nLet A be a skew-symmetric matrix.
\nThen A\u2019 = – A
\n\u2234 (B\u2019AB)\u2019 (B\u2019(AB))\u2019
\n= (AB)\u2019(B\u2019)\u2019
\n= B\u2019A\u2019B
\n= B\u2019(-A)B since A\u2019= – A
\n= – B\u2019AB
\n\u2234 B\u2019AB is a skew-symmetric matrix.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the values of x, y, z if the matrix A = \\(\\left[\\begin{array}{ccc}
\n0 & 2 y & z \\\\
\nx & y & -z \\\\
\nx & -y & z
\n\\end{array}\\right]\\) satisfied the equation A’A = I.
\nSolution:
\n\"NCERT<\/p>\n

Question 7.
\nFor what values of x if
\n\\(\\left[\\begin{array}{lll}
\n1 & 2 & 1
\n\\end{array}\\right]\\left[\\begin{array}{lll}
\n1 & 2 & 0 \\\\
\n2 & 0 & 1 \\\\
\n1 & 0 & 2
\n\\end{array}\\right]\\left[\\begin{array}{l}
\n0 \\\\
\n2 \\\\
\nx
\n\\end{array}\\right]\\) = 0
\nSolution:
\n\"NCERT<\/p>\n

Question 8.
\nIf A = \\(\\left[\\begin{array}{cc}
\n3 & 1 \\\\
\n-1 & 2
\n\\end{array}\\right]\\), show that A\u00b2 – 5A + 7I = 0.
\nSolution:
\n\"NCERT<\/p>\n

Question 9.
\nFind x, if
\n\\(\\left[\\begin{array}{lll}
\nx & -5 & -1
\n\\end{array}\\right]\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\left[\\begin{array}{l}
\nx \\\\
\n4 \\\\
\n1
\n\\end{array}\\right]\\) = 0
\nSolution:
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nA manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
\n\"NCERT
\n(i) If unit sale prices of x, y and z are \u20b9 2.50, \u20b9 1.50 and \u20b9 1.00 respectively, find the total revenue in each market with the help of matrix algebra.
\n(ii) If the unit costs of the above three commodities are \u20b9 2.00, \u20b9 1.00 and 50 paise respectively, find the gross profit.
\nSolution:
\n(i) Let
\n\"NCERT
\nrepresents annual sale of products in two markets and the column matrix
\nB = \\(\\left[\\begin{array}{l|l}
\n2.50 \\\\
\n1.50 \\\\
\n1.00
\n\\end{array}\\right] \\begin{aligned}
\n&x \\\\
\n&y \\\\
\n&z
\n\\end{aligned}\\) represents the unit sale price of the commodities x, y, z.
\n\u2234 The revenue collected by each market is given by AB.
\n\"NCERT
\nThe revenue in market I = \u20b9 46,000 and
\nthe revenue in market II = \u20b9 53,000
\nHence gross revenue in two markets
\n= \u20b9 46,000 + 53,000 = \u20b9 99,000<\/p>\n

(ii) Let the column matrix C = \\(\\left[\\begin{array}{c}
\n2.00 \\\\
\n1.00 \\\\
\n0.50
\n\\end{array}\\right]\\)
\nRepre-sent the unit cost price of the commodities x, y and z.
\n\u2234 The cost price of the articles in the two markets is AC.
\n\"NCERT
\nThe cost price of articles in market I = \u20b9 31,000 and the cost price of articles in market II = \u20b9 36,000.
\nHence the gross cost price
\n= \u20b9 31,000 + 36,000 = \u20b9 67,000
\nGross profit = Gross revenue – Gross cost price
\n= \u20b9 99,000 – 67, 000 = \u20b9 32, 000.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nFind the matrix X so that
\nX\\(\\left[\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ccc}
\n-7 & -8 & -9 \\\\
\n2 & 4 & 6
\n\\end{array}\\right]\\)
\nSolution:
\nX\\(\\left[\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ccc}
\n-7 & -8 & -9 \\\\
\n2 & 4 & 6
\n\\end{array}\\right]\\)
\nIt is clear that the order of X is 2 x 2
\nsince the product exists. Let X = \\(\\left[\\begin{array}{ll}
\na & b \\\\
\nc & d
\n\\end{array}\\right]\\).
\n\"NCERT
\nEquating the corresponding elements
\n\u2234 a + 4b = – 7 … (1)
\n2a + 56 = – 8 … (2)
\n3a + 66 = – 9 … (3)
\nc + 4d = 2 … (4)
\n2c + 5d = 4 … (5)
\n3c + 6d = 6 … (6)
\n(1) + (2) \u21d2 3a + 96 = – 15
\n(3) \u21d2 3a + 6b = – 9
\nSubtracting we get 3b = – 6
\n\u21d2 6 = – 2
\n\u2234 (1) \u21d2 a = -7 + 8 = 1
\n(4) + (5) \u21d2 3c + 9d = 6
\n(6) \u21d2 3c + 6d = 6
\nSubtracting 3d = 0 \u21d2 d = 0
\n(4) \u21d2 c = 2
\n\u2234 X = \\(\\left[\\begin{array}{ll}
\na & b \\\\
\nc & d
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{cc}
\n1 & -2 \\\\
\n2 & 0
\n\\end{array}\\right]\\)<\/p>\n

Question 12.
\nIf A and B are square matrices of the same order such that AB = BA, then prove by induction that AB\u207f = B\u207fA. Further, prove that (AB)\u207f = A\u207fB\u207f for all n \u2208 N.
\nSolution:
\nLet P(n) : AB\u207f = B\u207fA be the given statement.
\nP(1): AB’ = B’A which is true.
\nAssume that P(&) is true, i.e., ABk<\/sup> = Bk<\/sup>A
\nABk+1<\/sup> = (AB)Bk<\/sup> = (BA)Bk<\/sup>
\n= B(ABk<\/sup>)
\n= B(Bk<\/sup>A) since P(k) is true
\n= Bk+1<\/sup>A
\nHence by the Principle of Mathematical In-duction, P(n) is true for all values of n \u2208 N .
\nLet P(n) : (AB)\u207f = A\u207fB\u207f
\nP(1) : (AB)’ = A’B’ \u21d2 AB = AB which is true.
\nAssume that P(k) is true i.e., (AB)k<\/sup> = Ak<\/sup>Bk<\/sup>
\n(AB)k+1<\/sup> = (AB)k<\/sup>.(AB) = Ak<\/sup>.Bk<\/sup>(AB)
\n= Ak<\/sup>(Bk<\/sup>A)B
\n= Ak<\/sup>(ABk<\/sup>)B (since Bk<\/sup> A = ABk<\/sup>)
\n= (Ak<\/sup>.A)(Bk<\/sup>B)
\n= Ak+1<\/sup> Bk+1<\/sup>
\n\u2234 By the Principle of Mathematical Induction P(n) is true for all n \u2208 N.<\/p>\n

\"NCERT<\/p>\n

Choose the correct answer in the following questions.<\/p>\n

Question 13.
\nIf A = \\(\\left[\\begin{array}{cc}
\n\\alpha & \\beta \\\\
\n\\gamma & -\\alpha
\n\\end{array}\\right]\\) is such that A\u00b2 = I, then
\na. 1 + \u03b1\u00b2 + \u03b2\u03b3 = 0
\nb. 1 – \u03b1\u00b2 + \u03b2\u03b3 = 0
\nc. 1 – \u03b1\u00b2 – \u03b2\u03b3 = 0
\nd. 1 + \u03b1\u00b2 – \u03b2\u03b3 = 0
\nSolution:
\nc. 1 – \u03b1\u00b2 – \u03b2\u03b3 = 0
\n\"NCERT<\/p>\n

Question 14.
\nIf the matrix A is both symmetric and skew symmetric, then
\na. A is a diagonal matrix
\nb. A is a zero matrix
\nc. A is a square matrix
\nd. None of these
\nSolution:
\nb. A is a zero matrix
\nFor symmetric matrix aij<\/sub> = aji<\/sub>
\nFor skew symmetric matrix aij<\/sub> = – aji<\/sub>
\nThese conditions are satisfied only if aij<\/sub> = 0 for all i and j.
\nHence the matrix is a zero matrix.<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nIf A is square matrix such that A\u00b2 = A, then (I + A)\u00b3 – 7A is equal to
\na. A
\nb. I – A
\nc. I
\nd. 3A
\nSolution:
\nc. I
\n(I + A)\u00b3 – 7A = I\u00b3 + 3I\u00b2A + 3IA\u00b2 + A\u00b3 – 7A
\n= I + 3A + 3A\u00b2 + A\u00b2A – 7A
\n= I + 3A + 3A + A.A – 7A
\n= I + 6A + A – 7A
\n= I + 7A – 7A = I<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Question 1. Let A = show that (aI + bA)\u207f = a\u207fI + nan-1 bA, where I is the identity matrix …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-3-miscellaneous-exercise\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. 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