{"id":49863,"date":"2021-05-19T14:18:03","date_gmt":"2021-05-19T08:48:03","guid":{"rendered":"https:\/\/samacheerguru.com\/?p=49863"},"modified":"2021-05-24T15:10:05","modified_gmt":"2021-05-24T09:40:05","slug":"samacheer-kalvi-11th-maths-guide-chapter-7-ex-7-1-in-tamil","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-11th-maths-guide-chapter-7-ex-7-1-in-tamil\/","title":{"rendered":"Samacheer Kalvi 11th Maths Guide Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.1"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide<\/a> Pdf Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.1 Textbook Questions and Answers, Notes.<\/p>\n

TN Board 11th Maths Solutions Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.1<\/h2>\n

Question 1.<\/p>\n

(i) aij<\/sub> = \\(\\frac{(i-2 j)^{2}}{2}\\) m = 2, n = 3
\n(ii) aij<\/sub> = \\(\\frac{|3 i-4 j|}{4}\\), m = 3, n = 4 \u0b8e\u0ba9 \u0b87\u0bb0\u0bc1\u0b95\u0bcd\u0b95\u0bc1\u0bae\u0bbe\u0bb1\u0bc1: \u0b89\u0bb1\u0bc1\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bca\u0ba3\u0bcd\u0b9f m x n \u0bb5\u0bb0\u0bbf\u0b9a\u0bc8 \u0b89\u0b9f\u0bc8\u0baf A = [aij<\/sub>] \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bc8 \u0b89\u0bb0\u0bc1\u0bb5\u0bbe\u0b95\u0bcd\u0b95\u0bc1\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1:
\n(i) aij<\/sub> = \\(\\frac{(i-2 j)^{2}}{2}\\) m = 2, n = 3
\nA = \\(\\left(\\begin{array}{lll}
\na_{11} & a_{12} & a_{13} \\\\
\na_{21} & a_{22} & a_{23}
\n\\end{array}\\right)\\)<\/p>\n

\"Tamilnadu<\/p>\n

\"Tamilnadu<\/p>\n

(ii) aij<\/sub> = \\(\\frac{|3 i-4 j|}{4}\\), m = 3, n = 4
\nA = \\(\\left(\\begin{array}{llll}
\na_{11} & a_{12} & a_{13} & a_{14} \\\\
\na_{21} & a_{22} & b_{23} & a_{24} \\\\
\na_{31} & a_{32} & a_{33} & a_{34}
\n\\end{array}\\right)\\)<\/p>\n

\"Tamilnadu<\/p>\n

\"Tamilnadu<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\n\\(\\left[\\begin{array}{ccc}
\np^{2}-1 & 0 & -31-q^{3} \\\\
\n7 & r+1 & 9 \\\\
\n-2 & 8 & s-1
\n\\end{array}\\right]=\\left[\\begin{array}{ccc}
\n1 & 0 & -4 \\\\
\n7 & \\frac{3}{2} & 9 \\\\
\n-2 & 8 & -\\pi
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, p, q, r, s \u0b86\u0b95\u0bbf\u0baf\u0bb5\u0bb1\u0bcd\u0bb1\u0bbf\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\np2<\/sup> – 1 = 1
\np2<\/sup> = 1 + 1 = 2 \u21d2 p = \u00b1\u221a2
\n– 31 – q3<\/sup> = -4
\n\u21d2 -q3<\/sup> = -4 + 31 = 27 = 33<\/sup>
\n\u21d2 q = -3
\nr + 1 = \\(\\frac{3}{2}\\)
\n\u21d2 r = \\(\\frac{3}{2}\\) – 1 = \\(\\frac{3-2}{2}=\\frac{1}{2}\\)
\ns – 1 = -\u03c0
\ns = -\u03c0 + 1 = 1 – \u03c0
\n\u2234 p = \u00b1\u221a2, q = -3, r = \\(\\frac{1}{2}\\), s = 1 – \u03c0.<\/p>\n

Question 3.
\n\\(\\left[\\begin{array}{cc}
\n2 x+y & 4 x \\\\
\n5 x-7 & 4 x
\n\\end{array}\\right]=\\left[\\begin{array}{cc}
\n7 & 7 y-13 \\\\
\ny & x+6
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, x + y \u0b90 \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n2x + y = 7 …… (1)
\n7y – 13 = 4x
\n\u21d2 4x – 7y = -13 ………… (2)<\/p>\n

\"Tamilnadu<\/p>\n

y = \\(\\frac{27}{9}\\) = 3
\ny = 3 \u0b8e\u0ba9 (1)-\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f,
\n2x + 3 = 7 \u21d2 2x = 7 – 3 = 4
\nx = 2
\n\u2234 x + y = 2 + 3 = 5<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n2A – B = \\(\\left[\\begin{array}{rrr}
\n6 & -6 & 0 \\\\
\n-4 & 2 & 1
\n\\end{array}\\right]\\) = 0 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd
\nA – 2B = \\(\\left[\\begin{array}{rrr}
\n3 & 2 & 8 \\\\
\n-2 & 1 & -7
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0b9a\u0bcd\u0b9a\u0bae\u0ba9\u0bcd\u0baa\u0bbe\u0b9f\u0bc1\u0b95\u0bb3\u0bc8 \u0ba8\u0bbf\u0bb1\u0bc8\u0bb5\u0bc1 \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0bc1\u0bae\u0bcd, A, B \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1:
\nA – 2B = \\(\\left[\\begin{array}{rrr}
\n3 & 2 & 8 \\\\
\n-2 & 1 & -7
\n\\end{array}\\right]\\) ……(1)<\/p>\n

2A – B =\\(\\left[\\begin{array}{rrr}
\n6 & -6 & 0 \\\\
\n-4 & 2 & 1
\n\\end{array}\\right]\\) ……(2)<\/p>\n

\"Tamilnadu<\/p>\n

\"Tamilnadu<\/p>\n

Question 5.
\nA = \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, A4<\/sup> -\u0b87 \u0b95\u0bbe\u0ba3\u0bcd\u0b95
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1:
\nA = \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\)
\n\u21d2 A4<\/sup> = AAAA
\n= \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\) \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\) \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\) \\(\\left[\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right]\\)<\/p>\n

A2<\/sup> = \\(\\left(\\begin{array}{cc}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right)\\left(\\begin{array}{ll}
\n1 & a \\\\
\n0 & 1
\n\\end{array}\\right)\\)
\n= \\(\\left[\\begin{array}{ll}
\n(1 \\times 1+a \\times 0) & (1 \\times a+a \\times 1) \\\\
\n(0 \\times 1+1 \\times 0) & (0 \\times a+1 \\times 1)
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{cc}
\n(1+0) & (a+a) \\\\
\n(0+0) & (0+1)
\n\\end{array}\\right]=\\left[\\begin{array}{cc}
\n1 & 2 a \\\\
\n0 & 1
\n\\end{array}\\right]\\)<\/p>\n

A4<\/sup> = A2<\/sup> A2<\/sup>
\n= \\(\\left[\\begin{array}{cc}
\n1 & 2 a \\\\
\n0 & 1
\n\\end{array}\\right]\\left[\\begin{array}{cc}
\n1 & 2 a \\\\
\n0 & 1
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{rr}
\n((1 \\times 1)+(2 a \\times 0)) & (2 a+2 a) \\\\
\n(0+0) & (0+1)
\n\\end{array}\\right]\\)
\n\u2234 A4<\/sup> = \\(\\left[\\begin{array}{cc}
\n1 & 4 a \\\\
\n0 & 1
\n\\end{array}\\right]\\)<\/p>\n

Question 6.
\nA\u03b1<\/sub> = \\(\\left[\\begin{array}{ll}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd,
\n(i) A\u03b1<\/sub>A\u03b2<\/sub> = A\u03b1+\u03b2<\/sub> \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95.
\n(ii) A\u03b1<\/sub> + A\u03b1<\/sub>T<\/sup> = I \u0b8e\u0ba9\u0bcd\u0bb1 \u0ba8\u0bbf\u0baa\u0ba8\u0bcd\u0ba4\u0ba9\u0bc8\u0baf\u0bc8 \u0ba8\u0bbf\u0bb1\u0bc8\u0bb5\u0bc1 \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0bc1\u0bae\u0bcd \u0b8e\u0ba9\u0bcd \u0b85\u0ba9\u0bc8\u0ba4\u0bcd\u0ba4\u0bc1 \u0bae\u0bc6\u0baf\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0baf\u0bc1\u0bae\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1:
\n(i) A\u03b1<\/sub> = \\(\\left[\\begin{array}{cc}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)<\/p>\n

A\u03b2<\/sub> = \\(\\left[\\begin{array}{cc}
\n\\cos \\beta & -\\sin \\beta \\\\
\n\\sin \\beta & \\cos \\beta
\n\\end{array}\\right]\\)<\/p>\n

\"Tamilnadu<\/p>\n

(ii) A\u03b1<\/sub> + A\u03b1<\/sub>T<\/sup> = I
\nA\u03b1<\/sub> = \\(\\left[\\begin{array}{ll}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)<\/p>\n

A\u03b1<\/sub>T<\/sup> = \\(\\left[\\begin{array}{rr}
\n\\cos \\alpha & \\sin \\alpha \\\\
\n-\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)<\/p>\n

A\u03b1<\/sub> + A\u03b1<\/sub>T<\/sup> = \\(\\left[\\begin{array}{rr}
\n\\cos \\alpha & -\\sin \\alpha \\\\
\n\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]+\\left[\\begin{array}{rr}
\n\\cos \\alpha & \\sin \\alpha \\\\
\n-\\sin \\alpha & \\cos \\alpha
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{cc}
\n2 \\cos \\alpha & 0 \\\\
\n0 & 2 \\cos \\alpha
\n\\end{array}\\right]=\\mathrm{I}=\\left[\\begin{array}{ll}
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right]\\)<\/p>\n

2 cos \u03b1 = 1
\n\u21d2 cos \u03b1 = \\(\\frac{1}{2}\\) = cos \\(\\frac{\\pi}{3}\\)
\n\u21d2 \u03b1 = 2n\u03c0 \u00b1 \\(\\frac{\\pi}{3}\\), n \u2208 Z<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nA = \\(\\left[\\begin{array}{rr}
\n4 & 2 \\\\
\n-1 & x
\n\\end{array}\\right]\\) \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd (A – 2I) (A – 3I) = 0 \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, \u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1:
\n(A – 2I) = \\(\\left[\\begin{array}{cc}
\n4 & 2 \\\\
\n-1 & x
\n\\end{array}\\right]-2\\left[\\begin{array}{ll}
\n2 & 0 \\\\
\n0 & 2
\n\\end{array}\\right]\\)
\n= \\(\\left(\\begin{array}{cc}
\n4 & 2 \\\\
\n-1 & x
\n\\end{array}\\right)-\\left[\\begin{array}{ll}
\n2 & 0 \\\\
\n0 & 2
\n\\end{array}\\right]=\\left[\\begin{array}{cc}
\n2 & 2 \\\\
\n-1 & x-2
\n\\end{array}\\right]\\)<\/p>\n

(A – 3I) = \\(\\left(\\begin{array}{cc}
\n4 & 2 \\\\
\n-1 & x
\n\\end{array}\\right)-3\\left(\\begin{array}{ll}
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right)\\)
\n= \\(\\left(\\begin{array}{cc}
\n4 & 2 \\\\
\n-1 & x
\n\\end{array}\\right)-\\left(\\begin{array}{ll}
\n3 & 0 \\\\
\n0 & 3
\n\\end{array}\\right)=\\left(\\begin{array}{rr}
\n1 & 2 \\\\
\n-1 & x-3
\n\\end{array}\\right)\\)
\n(A – 2I) (A – 3I) = \\(\\left(\\begin{array}{rr}
\n2 & 2 \\\\
\n1 & x-2
\n\\end{array}\\right)\\left(\\begin{array}{cc}
\n1 & 2 \\\\
\n-1 & x-3
\n\\end{array}\\right)\\) = 0 \u0b8e\u0ba9 \u0ba4\u0bb0\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0bc1\u0bb3\u0bcd\u0bb3\u0ba4\u0bc1.<\/p>\n\\(\\left[\\begin{array}{rr}
\n(2-2) & 4+(2 x-6) \\\\
\n1-(x-2) & 2+(x-2)(x-3)
\n\\end{array}\\right]=\\left[\\begin{array}{ll}
\n0 & 0 \\\\
\n0 & 0
\n\\end{array}\\right]\\)\n

4 + 2x – 6 = 0
\n2x – 2 = 0
\nx = 1<\/p>\n

Question 8.
\nA = \\(\\left[\\begin{array}{rrr}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\na & b & -1
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, A2<\/sup> \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b85\u0bb2\u0b95\u0bc1 \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left[\\begin{array}{rrr}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\na & b & -1
\n\\end{array}\\right]\\)<\/p>\n

A2<\/sup> = A.A = \\(\\left[\\begin{array}{rrr}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\na & b & -1
\n\\end{array}\\right]\\) \\(\\left[\\begin{array}{rrr}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\na & b & -1
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{ccc}
\n1+0+0 & 0+0+0 & 0+0+0 \\\\
\n0+0+0 & 0+1+0 & 0+0+0 \\\\
\na+0-a & 0+b-b & 0+0+1
\n\\end{array}\\right]\\)<\/p>\n

= \\(\\left[\\begin{array}{lll}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\) = I \u0b8e\u0ba9\u0bb5\u0bc7 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nA = \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\) \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd A3<\/sup> – 6A2<\/sup> + 7A + kI = 0, \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, k -\u0b90 \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\)<\/p>\n

A3<\/sup> = AAA
\nA2<\/sup> = AA = \\(\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{lll}
\n(1+0+4) & (0+0+0) & (2+0+6) \\\\
\n(0+0+2) & (0+4+0) & (0+2+3) \\\\
\n(2+0+6) & (0+0+0) & (4+0+9)
\n\\end{array}\\right]=\\left[\\begin{array}{ccc}
\n5 & 0 & 8 \\\\
\n2 & 4 & 5 \\\\
\n8 & 0 & 13
\n\\end{array}\\right]\\)
\nA3<\/sup> = A2<\/sup>A = \\(\\left[\\begin{array}{ccc}
\n5 & 0 & 8 \\\\
\n2 & 4 & 5 \\\\
\n8 & 0 & 13
\n\\end{array}\\right]\\left[\\begin{array}{lll}
\n1 & 0 & 2 \\\\
\n0 & 2 & 1 \\\\
\n2 & 0 & 3
\n\\end{array}\\right]\\)<\/p>\n

\"Tamilnadu<\/p>\n

Question 10.
\n\u0baa\u0bbf\u0ba9\u0bcd\u0bb5\u0bb0\u0bc1\u0bae\u0bcd \u0ba8\u0bbf\u0baa\u0ba8\u0bcd\u0ba4\u0ba9\u0bc8\u0b95\u0bb3\u0bcd \u0b92\u0bb5\u0bcd\u0bb5\u0bca\u0ba9\u0bcd\u0bb1\u0bc8\u0baf\u0bc1\u0bae\u0bcd \u0ba8\u0bbf\u0bb1\u0bc8\u0bb5\u0bc1 \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0bc1\u0bae\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bc1\u0b95\u0bcd\u0b95\u0bbe\u0ba9 \u0b8e\u0b9f\u0bc1\u0ba4\u0bcd\u0ba4\u0bc1\u0b95\u0bcd\u0b95\u0bbe\u0b9f\u0bcd\u0b9f\u0bc1\u0b95\u0bb3\u0bc8\u0ba4\u0bcd \u0ba4\u0bb0\u0bc1\u0b95.
\n(i) AB \u2260 BA \u0b8e\u0ba9\u0bc1\u0bae\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 A \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd B \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd.
\n(ii) AB = 0 = BA, A # 0 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd B \u2260 0 \u0b8e\u0ba9\u0bc1\u0bae\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 A, B \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd.
\n(ii) AB = 0 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd BA \u2260 0 \u0b8e\u0ba9\u0bc1\u0bae\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 A, B \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nAB \u2260 BA
\n(i) A = \\(\\left[\\begin{array}{rr}
\n1 & 2 \\\\
\n0 & -1
\n\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{rr}
\n2 & 1 \\\\
\n-1 & 0
\n\\end{array}\\right]\\)
\nAB = \\(\\left[\\begin{array}{rr}
\n1 & 2 \\\\
\n0 & -1
\n\\end{array}\\right]\\left[\\begin{array}{rr}
\n2 & 1 \\\\
\n-1 & 0
\n\\end{array}\\right]=\\left[\\begin{array}{cc}
\n(2-2) & (1+0) \\\\
\n(0+1) & (0+0)
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{ll}
\n0 & 1 \\\\
\n1 & 0
\n\\end{array}\\right]\\) …………(1)<\/p>\n

\"Tamilnadu<\/p>\n

(iii) \"Tamilnadu<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nf(x) = \\(\\left[\\begin{array}{rrr}
\n\\cos x & -\\sin x & 0 \\\\
\n\\sin x & \\cos x & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\), \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, f(x)f(y) = f(x + y) \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nf(x) . f(y) = \\(\\left[\\begin{array}{rrr}
\n\\cos x & -\\sin x & 0 \\\\
\n\\sin x & \\cos x & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\) \\(\\left[\\begin{array}{rrr}
\n\\cos y & -\\sin y & 0 \\\\
\n\\sin y & \\cos y & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{c}
\n(\\cos x \\cos y-\\sin x \\sin y+0) \\\\
\n(\\sin x \\cos y+\\cos x \\sin y+0) \\\\
\n(0+0+0)
\n\\end{array}\\right.\\)
\n\\( \\left.\\begin{array}{cc}
\n(-\\cos x \\sin y-\\sin x \\cos y+0) & (0+0+0) \\\\
\n(-\\sin x \\sin y+\\cos x \\cos y+0) & (0+0+0) \\\\
\n(0+0+0) & (0+0+1)
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{ccc}
\n\\cos (x+y) & -\\sin (x+y) & 0 \\\\
\n\\sin (x+y) & \\cos (x+y) & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\)
\n= f(x+y)
\n\u0b8e\u0ba9\u0bb5\u0bc7 f (x) f (y) = f (x+y) \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

Question 12.
\nA \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 A2<\/sup> = A \u0b8e\u0ba9\u0bcd\u0bb1\u0bb5\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 \u0b92\u0bb0\u0bc1 \u0b9a\u0ba4\u0bc1\u0bb0 \u0b85\u0ba3\u0bbf \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, 7A – (I + A)3<\/sup> -\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95. \u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n7A – (I + A)3<\/sup> = 7A [I3<\/sup> + A3<\/sup> + 3A (I + A)]
\n= 7A – [I + A + 3A + 3A]
\n= 7A – [I + 7A]
\n= \"Tamilnadu
\n= -I<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nA = \\(\\left[\\begin{array}{rrr}
\n2 & 0 & -3 \\\\
\n1 & 4 & 5
\n\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{rr}
\n3 & 1 \\\\
\n-1 & 0 \\\\
\n4 & 2
\n\\end{array}\\right]\\)\u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, C = \\(\\left[\\begin{array}{rr}
\n4 & 7 \\\\
\n2 & 1 \\\\
\n1 & -1
\n\\end{array}\\right]\\)
\n[1 -1] \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, A(B + C) = AB + AC, \u0b8e\u0ba9\u0bc1\u0bae\u0bcd \u0baa\u0ba3\u0bcd\u0baa\u0bbf\u0ba9\u0bc8\u0b9a\u0bcd \u0b9a\u0bb0\u0bbf\u0baa\u0bbe\u0bb0\u0bcd\u0b95\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\"Tamilnadu<\/p>\n

= \\(\\left(\\begin{array}{cc}
\n(8+0-3) & (14+0+3) \\\\
\n(4+8+5) & (7+4-5)
\n\\end{array}\\right)=\\left[\\begin{array}{cc}
\n5 & 17 \\\\
\n17 & 6
\n\\end{array}\\right]\\)<\/p>\n

AB + AC = \\(\\left(\\begin{array}{cc}
\n-6 & -4 \\\\
\n19 & 11
\n\\end{array}\\right)+\\left[\\begin{array}{cc}
\n5 & 17 \\\\
\n17 & 6
\n\\end{array}\\right]=\\left[\\begin{array}{cc}
\n-1 & 13 \\\\
\n36 & 17
\n\\end{array}\\right]\\) ……..(2)
\n(1) = (2)
\n\u2234 A(B + C) = AB + AC \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b9a\u0bb0\u0bbf\u0baa\u0bbe\u0bb0\u0bcd\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

Question 14.
\nA\\(\\left[\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6
\n\\end{array}\\right]=\\left[\\begin{array}{rrr}
\n-7 & -8 & -9 \\\\
\n2 & 4 & 6
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0baf \u0b9a\u0bae\u0ba9\u0bcd\u0baa\u0bbe\u0b9f\u0bcd\u0b9f\u0bbf\u0ba9\u0bc8 \u0ba8\u0bbf\u0bb1\u0bc8\u0bb5\u0bc1 \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0bc1\u0bae\u0bcd A \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0baf\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left[\\begin{array}{ll}
\na_{11} & a_{12} \\\\
\na_{21} & a_{22}
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0b95.<\/p>\n\\(\\left[\\begin{array}{ll}
\na_{11} & a_{12} \\\\
\na_{21} & a_{22}
\n\\end{array}\\right]\\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]\\)\n

\"Tamilnadu<\/p>\n

a22<\/sub> = 0 \u0b8e\u0ba9 (3)\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f
\n1a21<\/sub> + 4 \u00d7 0 = 2 ……(3)
\na21<\/sub> = 2 ……(4)
\n\u2234 A = \\(\\left(\\begin{array}{ll}
\na_{11} & a_{12} \\\\
\na_{21} & a_{22}
\n\\end{array}\\right)=\\left(\\begin{array}{rr}
\n1 & -2 \\\\
\n2 & 0
\n\\end{array}\\right)\\)<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nAT<\/sup> = \\(\\left[\\begin{array}{rr}
\n4 & 5 \\\\
\n-1 & 0 \\\\
\n2 & 3
\n\\end{array}\\right]\\) \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd B = \\(\\left[\\begin{array}{rrr}
\n2 & -1 & 1 \\\\
\n7 & 5 & -2
\n\\end{array}\\right]\\) \u0baa\u0bbf\u0ba9\u0bcd\u0bb5\u0bb0\u0bc1\u0bb5\u0ba9\u0bb5\u0bb1\u0bcd\u0bb1\u0bc8\u0b9a\u0bcd \u0b9a\u0bb0\u0bbf\u0baa\u0bbe\u0bb0\u0bcd\u0b95\u0bcd\u0b95.
\n(i) (A + B)T<\/sup> = AT<\/sup> + BT<\/sup> = BT<\/sup> + AT<\/sup>
\n(ii) (A – B)T<\/sup> = AT<\/sup> – BT<\/sup>
\n(iii) (BT<\/sup>)T<\/sup> = B.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n(i) A = \\(\\left[\\begin{array}{ccc}
\n4 & -1 & 2 \\\\
\n5 & 0 & 3
\n\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{ccc}
\n2 & -1 & 1 \\\\
\n7 & 5 & -2
\n\\end{array}\\right]\\)<\/p>\n

(A + B) = \\(\\left[\\begin{array}{lll}
\n(4+2) & (-1-1) & (2+1) \\\\
\n(5+7) & (0+5) & (3-2)
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{ccc}
\n6 & -2 & 3 \\\\
\n12 & 5 & 1
\n\\end{array}\\right]\\)<\/p>\n

(A + B)T<\/sup> = \\(\\left[\\begin{array}{cc}
\n6 & 12 \\\\
\n-2 & 5 \\\\
\n3 & 1
\n\\end{array}\\right]\\) ……….(1)<\/p>\n

BT<\/sup> = \\(\\left[\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right]\\)<\/p>\n

AT<\/sup> + BT<\/sup> = \\(\\left[\\begin{array}{cc}
\n4 & 5 \\\\
\n-1 & 0 \\\\
\n2 & 3
\n\\end{array}\\right]+\\left[\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{cc}
\n6 & 12 \\\\
\n-2 & 5 \\\\
\n3 & 1
\n\\end{array}\\right]\\) ………..(2)<\/p>\n

BT<\/sup> + AT<\/sup> = \\(\\left[\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right]+\\left[\\begin{array}{cc}
\n4 & 5 \\\\
\n-1 & 0 \\\\
\n2 & 3
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{cc}
\n6 & 12 \\\\
\n-2 & 5 \\\\
\n3 & 1
\n\\end{array}\\right]\\) ………(3)
\n(1) = (2) = (3)
\n\u2234 \u0b8e\u0ba9\u0bb5\u0bc7 (A + B)T<\/sup> = AT<\/sup> + BT<\/sup> = BT<\/sup> + AT<\/sup> \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

(ii) ( A- B)T<\/sup> = \\(\\left[\\left(\\begin{array}{rrr}
\n4 & -1 & 2 \\\\
\n5 & 0 & 3
\n\\end{array}\\right)-\\left(\\begin{array}{rrr}
\n2 & -1 & 1 \\\\
\n7 & 5 & -2
\n\\end{array}\\right)\\right]^{\\mathrm{T}}\\)
\n= \\(\\left[\\begin{array}{rrr}
\n(4-2) & -1-(-1) & (2-1) \\\\
\n(5-7) & (0-5) & (3-(-2))
\n\\end{array}\\right]^{\\mathrm{T}}\\)
\n= \\(\\left(\\begin{array}{rrr}
\n2 & 0 & 1 \\\\
\n-2 & -5 & 5
\n\\end{array}\\right)^{\\mathrm{T}}=\\left(\\begin{array}{lr}
\n2 & -2 \\\\
\n0 & -5 \\\\
\n1 & 5
\n\\end{array}\\right)\\) ……….(1)<\/p>\n

AT<\/sup> – BT<\/sup> = \\(\\left[\\begin{array}{cc}
\n4 & 5 \\\\
\n-1 & 0 \\\\
\n2 & 3
\n\\end{array}\\right]-\\left[\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{cc}
\n2 & -2 \\\\
\n0 & -5 \\\\
\n1 & 5
\n\\end{array}\\right]\\) ……….(2)
\n(1) = (2)
\n\u2234 (A – B)T<\/sup> = AT<\/sup> – BT<\/sup> \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

(iii) (BT<\/sup>)T<\/sup> = B
\nBT<\/sup> = \\(\\left(\\begin{array}{rrr}
\n2 & -1 & 1 \\\\
\n7 & 5 & -2
\n\\end{array}\\right)^{\\mathrm{T}}=\\left(\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right)\\)
\n(BT<\/sup>)T<\/sup> = \\(\\left(\\begin{array}{cc}
\n2 & 7 \\\\
\n-1 & 5 \\\\
\n1 & -2
\n\\end{array}\\right)^{\\mathrm{T}}=\\left(\\begin{array}{ccc}
\n2 & -1 & 1 \\\\
\n7 & 5 & -2
\n\\end{array}\\right)\\) = B \u0b8e\u0ba9\u0bb5\u0bc7 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.<\/p>\n

\"Samacheer<\/p>\n

Question 16.
\n3 \u00d7 4 \u0bb5\u0bb0\u0bbf\u0b9a\u0bc8 \u0b89\u0b9f\u0bc8\u0baf \u0b92\u0bb0\u0bc1 \u0b85\u0ba3\u0bbf A \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd B \u0b8e\u0ba9\u0bcd\u0bb1 \u0b87\u0bb0\u0ba3\u0bcd\u0b9f\u0bc1 \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bc1\u0bae\u0bcd AT<\/sup>B \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd BAT<\/sup> \u0b86\u0b95\u0bbf\u0baf \u0b87\u0bb0\u0ba3\u0bcd\u0b9f\u0bc8\u0baf\u0bc1\u0bae\u0bcd \u0bb5\u0bb0\u0bc8\u0baf\u0bb1\u0bc8 \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0bc1\u0bae\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, B \u0b85\u0ba3\u0bbf\u0baf\u0bbf\u0ba9\u0bcd \u0bb5\u0bb0\u0bbf\u0b9a\u0bc8\u0baf\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = [aij<\/sub>]3 \u00d7 4<\/sub>
\nAT<\/sup> = [aij<\/sub>]4 \u00d7 3<\/sub>
\nB = [bij<\/sub>]x \u00d7 y<\/sub> \u0b8e\u0ba9\u0bcd\u0b95
\nAT<\/sup>B = [aij<\/sub>]4 \u00d7 3<\/sub> [bij<\/sub>]x \u00d7 y<\/sub>
\n\u2234 x = 3
\nBAT<\/sup> = [bij<\/sub>]x \u00d7 y<\/sub> [aij<\/sub>]4 \u00d7 3<\/sub>
\n\u2234 y = 4
\n\u2234 B-\u0ba9\u0bcd \u0bb5\u0bb0\u0bbf\u0b9a\u0bc8 x \u00d7 y = 3 \u00d7 4<\/p>\n

Question 17.
\n\u0baa\u0bbf\u0ba9\u0bcd\u0bb5\u0bb0\u0bc1\u0bae\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bc8 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd ! \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bbf\u0ba9\u0bcd \u0b95\u0bc2\u0b9f\u0bc1\u0ba4\u0bb2\u0bbe\u0b95 \u0b8e\u0bb4\u0bc1\u0ba4\u0bc1\u0b95.
\n(i) \\(\\left[\\begin{array}{ll}
\n4 & -2 \\\\
\n3 & -5
\n\\end{array}\\right]\\) (ii) \\(\\left[\\begin{array}{rrr}
\n3 & 3 & -1 \\\\
\n-2 & -2 & 1 \\\\
\n-4 & -5 & 2
\n\\end{array}\\right]\\).
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n(i) A = \\(\\left[\\begin{array}{ll}
\n4 & -2 \\\\
\n3 & -5
\n\\end{array}\\right]\\) \u21d2 AT<\/sup> = \\(\\left[\\begin{array}{cc}
\n4 & 3 \\\\
\n-2 & -5
\n\\end{array}\\right]\\)<\/p>\n

A + AT<\/sup> = \\(\\left[\\begin{array}{ll}
\n4 & -2 \\\\
\n3 & -5
\n\\end{array}\\right]\\) + \\(\\left[\\begin{array}{cc}
\n4 & 3 \\\\
\n-2 & -5
\n\\end{array}\\right]\\) = \\(\\left[\\begin{array}{rr}
\n8 & 1 \\\\
\n1 & -10
\n\\end{array}\\right]\\)
\nP = \\(\\frac{1}{2}\\)(A + AT<\/sup>) = \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{rr}
\n8 & 1 \\\\
\n1 & -10
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0b95.<\/p>\n

\u0b87\u0baa\u0bcd\u0baa\u0bca\u0bb4\u0bc1\u0ba4\u0bc1 PT<\/sup> = \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{rr}
\n8 & 1 \\\\
\n1 & -10
\n\\end{array}\\right]\\) = P
\n\u0b86\u0b95\u0bc8\u0baf\u0bbe\u0bb2\u0bcd P = \\(\\frac{1}{2}\\) [A + AT<\/sup>] \u0b92\u0bb0\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

Q = \\(\\frac{1}{2}\\) [A – AT<\/sup>] \u0b8e\u0ba9\u0bcd\u0b95.
\n= \\(\\frac{1}{2}\\) \\(\\left(\\left[\\begin{array}{ll}
\n4 & -2 \\\\
\n3 & -5
\n\\end{array}\\right]-\\left[\\begin{array}{cc}
\n4 & 3 \\\\
\n-2 & -5
\n\\end{array}\\right]\\right)\\)
\n= \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{cc}
\n0 & -5 \\\\
\n5 & 0
\n\\end{array}\\right]\\)
\nQT<\/sup> = \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{cc}
\n0 & 5 \\\\
\n-5 & 0
\n\\end{array}\\right]\\) = -Q
\n\u0b86\u0b95\u0bc8\u0baf\u0bbe\u0bb2\u0bcd Q = \\(\\frac{1}{2}\\) [A – AT<\/sup>] \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.
\nA = P + Q
\n= \\(\\frac{1}{2}\\left[\\begin{array}{rr}
\n8 & 1 \\\\
\n1 & -10
\n\\end{array}\\right]+\\frac{1}{2}\\left[\\begin{array}{cc}
\n0 & -5 \\\\
\n5 & 0
\n\\end{array}\\right]\\)
\n\u0b86\u0b95\u0bb5\u0bc7 A \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc8 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bbf\u0ba9\u0bcd \u0b95\u0bc2\u0b9f\u0bc1\u0ba4\u0bb2\u0bbe\u0b95 \u0b8e\u0bb4\u0bc1\u0ba4\u0bb2\u0bbe\u0bae\u0bcd.<\/p>\n

(ii) \"Tamilnadu \u0b92\u0bb0\u0bc1
\nQT<\/sup> = \\(\\frac{1}{2}\\)\\(\\left[\\begin{array}{ccc}
\n0 & -5 & -3 \\\\
\n5 & 0 & -6 \\\\
\n3 & 6 & 0
\n\\end{array}\\right]\\) = -Q
\n\u0b86\u0b95\u0bc8\u0baf\u0bbe\u0bb2\u0bcd Q = \\(\\frac{1}{2}\\) [A – AT<\/sup>] \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.
\n\u21d2 A = P + Q = \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{ccc}
\n6 & 1 & -5 \\\\
\n1 & -4 & -4 \\\\
\n-5 & -4 & 4
\n\\end{array}\\right]\\) + \\(\\frac{1}{2}\\) \\(\\left[\\begin{array}{ccc}
\n0 & 5 & 3 \\\\
\n-5 & 0 & 6 \\\\
\n-3 & -6 & 0
\n\\end{array}\\right]\\)
\n\u0b86\u0b95\u0bc8\u0baf\u0bbe\u0bb2\u0bcd A \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc8 \u0b92\u0bb0\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bbf\u0ba9\u0bcd \u0b95\u0bc2\u0b9f\u0bc1\u0ba4\u0bb2\u0bbe\u0b95 \u0b8e\u0bb4\u0bc1\u0ba4\u0bb2\u0bbe\u0bae\u0bcd.<\/p>\n

\"Samacheer<\/p>\n

Question 18.
\n\\(\\left[\\begin{array}{rr}
\n2 & -1 \\\\
\n1 & 0 \\\\
\n-3 & 4
\n\\end{array}\\right]\\) AT<\/sup> = \\(\\left[\\begin{array}{rrr}
\n-1 & -8 & -10 \\\\
\n1 & 2 & -5 \\\\
\n9 & 22 & 15
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bc1\u0bae\u0bbe\u0bb1\u0bc1\u0bb3\u0bcd\u0bb3 A \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0baf\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 : \\(\\left[\\begin{array}{rr}
\n2 & -1 \\\\
\n1 & 0 \\\\
\n-3 & 4
\n\\end{array}\\right]\\)3 \u00d7 2<\/sub> AT<\/sup> = \\(\\left[\\begin{array}{rrr}
\n-1 & -8 & -10 \\\\
\n1 & 2 & -5 \\\\
\n9 & 22 & 15
\n\\end{array}\\right]\\)3 \u00d7 3<\/sub> \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd
\nAT<\/sup> = \\(\\left[\\begin{array}{lll}
\na_{11} & a_{12} & a_{13} \\\\
\na_{21} & a_{22} & a_{23}
\n\\end{array}\\right]_{2 \\times 3}\\) \u0b8e\u0ba9\u0bcd\u0b95.
\n\\(\\left[\\begin{array}{cc}
\n2 & -1 \\\\
\n1 & 0 \\\\
\n-3 & 4
\n\\end{array}\\right]\\left[\\begin{array}{lll}
\na_{11} & a_{12} & a_{13} \\\\
\na_{21} & a_{22} & a_{23}
\n\\end{array}\\right]=\\left[\\begin{array}{ccc}
\n-1 & -8 & -10 \\\\
\n1 & 2 & -5 \\\\
\n9 & 22 & 15
\n\\end{array}\\right]\\)<\/p>\n

2a11<\/sub> – 1a21<\/sub> = -1 ………(1)
\n1a11<\/sub> + 0a21<\/sub> = 1 ……….(2)
\n1a11<\/sub> + 0 = 1
\na11<\/sub> = 1
\na11<\/sub> = 1 \u0b8e\u0ba9 (1)-\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f
\n2 \u00d7 1 – a21<\/sub> = -1
\n-a21<\/sub> = -1 – 2<\/p>\n

\"Tamilnadu<\/p>\n

2a12<\/sub> – 1a22<\/sub> = -8 ………(3)
\n1a12<\/sub> + 0a22<\/sub> = 2 ………(4)
\na12<\/sub> + 0 = 2
\na12<\/sub> = 2 \u0b8e\u0ba9 (3)-\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f
\n2 \u00d7 2 – a22<\/sub> = -8
\n-a22<\/sub> = -8 – 4 = -12
\na22<\/sub> = 12
\n2a13<\/sub> – 1a23<\/sub> = -10 ……….(5)
\n1a13<\/sub> + 0a23<\/sub> = -5 ………..(6)
\na13<\/sub> + 0 = -5
\na13<\/sub> = -5 \u0b8e\u0ba9 (5)-\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f
\n2 \u00d7 -5 – a23<\/sub> = -10
\n– a23<\/sub> = -10 + 10 =0
\na23<\/sub> = 0
\n\u2234 AT<\/sup> = \\(\\left[\\begin{array}{rrr}
\n1 & 2 & -5 \\\\
\n3 & 12 & 0
\n\\end{array}\\right]\\)
\n\u2234 A = \\(\\left[\\begin{array}{rr}
\n1 & 3 \\\\
\n2 & 12 \\\\
\n-5 & 0
\n\\end{array}\\right]\\)<\/p>\n

Question 19.
\nA = \\(\\left[\\begin{array}{rrr}
\n1 & 2 & 2 \\\\
\n2 & 1 & -2 \\\\
\nx & 2 & y
\n\\end{array}\\right]\\) \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd AAT<\/sup> = 9I \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, x, y -\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nAAT<\/sup> = 9I
\n\\(\\left[\\begin{array}{ccc}
\n1 & 2 & 2 \\\\
\n2 & 1 & -2 \\\\
\nx & 2 & y
\n\\end{array}\\right]\\left[\\begin{array}{ccc}
\n1 & 2 & x \\\\
\n2 & 1 & 2 \\\\
\n2 & -2 & y
\n\\end{array}\\right]=9\\left[\\begin{array}{lll}
\n1 & 0 & 0 \\\\
\n0 & 1 & 0 \\\\
\n0 & 0 & 1
\n\\end{array}\\right]\\)
\nx + 4 + 2y = 0
\nx + 2y = -4 ………(1)
\nx + 4 + 2y = 0
\nx + 2y = -4 ……….(2)
\n2x + 2 – 2y = 0<\/p>\n

\"Tamilnadu<\/p>\n

y = -1 \u0b8e\u0ba9 (1)-\u0bb2\u0bcd \u0baa\u0bbf\u0bb0\u0ba4\u0bbf\u0baf\u0bbf\u0b9f
\nx + 2 \u00d7 -1 = – 4
\nx – 2 = -4
\nx = -4 + 2 ……….(2)
\nx = -2
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 : –
\nx = -2
\ny = -1<\/p>\n

\"Samacheer<\/p>\n

Question 20.
\n(i) x \u0ba9\u0bcd \u0b8e\u0bae\u0bcd\u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bcd\u0b95\u0bc1 , A = \\(\\left[\\begin{array}{rrr}
\n0 & 1 & -2 \\\\
\n-1 & 0 & x^{3} \\\\
\n2 & -3 & 0
\n\\end{array}\\right]\\) \u0baf\u0bc8 \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd?
\n(ii) \\(\\left[\\begin{array}{ccc}
\n0 & p & 3 \\\\
\n2 & q^{2} & -1 \\\\
\nr & 1 & 0
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, p, q, r-\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left[\\begin{array}{ccc}
\n0 & 1 & -2 \\\\
\n-1 & 0 & x^{3} \\\\
\n2 & -3 & 0
\n\\end{array}\\right]\\)<\/p>\n

\u21d2 AT<\/sup> = \\(\\left[\\begin{array}{ccc}
\n0 & -1 & 2 \\\\
\n1 & 0 & -3 \\\\
\n-2 & x^{3} & 0
\n\\end{array}\\right]\\)
\nAT<\/sup> = -A
\n\u21d2 \\(\\left[\\begin{array}{ccc}
\n0 & -1 & 2 \\\\
\n1 & 0 & -3 \\\\
\n-2 & x^{3} & 0
\n\\end{array}\\right]=\\left[\\begin{array}{ccc}
\n0 & -1 & 2 \\\\
\n1 & 0 & -x^{3} \\\\
\n-2 & 3 & 0
\n\\end{array}\\right]\\)
\n\u2234 x3<\/sup> = 3
\n\u21d2 x = 3\\(\\frac{1}{3}\\)<\/sup><\/p>\n

(ii) A = \\(\\left[\\begin{array}{ccc}
\n0 & p & 3 \\\\
\n2 & q^{2} & -1 \\\\
\nr & 1 & 0
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0b95.
\n\u2234 AT<\/sup> = \\(\\left[\\begin{array}{ccc}
\n0 & 2 & r \\\\
\np & q^{2} & 1 \\\\
\n3 & -1 & 0
\n\\end{array}\\right]\\)
\nAT<\/sup> = -A
\n\u21d2 \\(\\left[\\begin{array}{ccc}
\n0 & 2 & r \\\\
\np & q^{2} & 1 \\\\
\n3 & -1 & 0
\n\\end{array}\\right]=\\left[\\begin{array}{rrr}
\n0 & -p & -3 \\\\
\n-2 & -q^{2} & 1 \\\\
\n-r & -1 & 0
\n\\end{array}\\right]\\)
\np = -2
\nq2<\/sup> = -q2<\/sup>
\n\u21d2 2q2<\/sup> = 0
\nq = 0
\nr = -3<\/p>\n

Question 21.
\naij<\/sub> = i – j \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, A =[ aij<\/sub>]3 \u00d7 3<\/sub>, \u0b8e\u0ba9\u0bcd\u0bb1 \u0b85\u0ba3\u0bbf\u0baf\u0bc8 \u0b89\u0bb0\u0bc1\u0bb5\u0bbe\u0b95\u0bcd\u0b95\u0bc1\u0b95. \u0bae\u0bc7\u0bb2\u0bc1\u0bae\u0bcd, A \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe \u0b85\u0bb2\u0bcd\u0bb2\u0ba4\u0bc1 \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe \u0b8e\u0ba9\u0b95\u0bcd \u0b95\u0bc2\u0bb1\u0bc1\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left[\\begin{array}{lll}
\na_{11} & a_{12} & a_{13} \\\\
\na_{21} & a_{22} & a_{23} \\\\
\na_{31} & a_{32} & a_{33}
\n\\end{array}\\right]\\)
\naij<\/sub> = i – j
\n\u2234 a11<\/sub> = 1 – 1 = 0
\na12<\/sub> = 1 – 2 = -1
\na13<\/sub> = 1 – 3 = -2
\na21<\/sub> = 2 – 1 = 1
\na22<\/sub> = 2 – 2 = 0
\na23<\/sub> = 2 – 3 = -1
\na31<\/sub> = 3 – 1 = 2
\na32<\/sub> = 3 – 2 = 1
\na33<\/sub> = 3 – 3 = 0
\n\u2234 A = \\(\\left[\\begin{array}{ccc}
\n0 & -1 & -2 \\\\
\n1 & 0 & -1 \\\\
\n2 & 1 & 0
\n\\end{array}\\right]\\)<\/p>\n

\u21d2 AT<\/sup> = \\(\\left[\\begin{array}{ccc}
\n0 & 1 & 2 \\\\
\n-1 & 0 & 1 \\\\
\n-2 & -1 & 0
\n\\end{array}\\right]\\) = -A
\n\u2234 A \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b93\u0bb0\u0bcd \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

\"Samacheer<\/p>\n

Question 22.
\nA, B \u0b8e\u0ba9\u0bcd\u0baa\u0ba9 \u0b87\u0bb0\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bbf\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0bcd\u0b95. AB = BA \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, AB \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95. \u0bae\u0bc7\u0bb2\u0bc1\u0bae\u0bcd \u0b87\u0ba4\u0ba9\u0bcd \u0bae\u0bb1\u0bc1\u0ba4\u0bb2\u0bc8\u0baf\u0bc1\u0bae\u0bcd \u0b89\u0ba3\u0bcd\u0bae\u0bc8 \u0b8e\u0ba9 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0ba4\u0bb0\u0bb5\u0bc1 :
\nA = AT<\/sup>, B = BT<\/sup>,
\nAB = BA \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95 \u0bb5\u0bc7\u0ba3\u0bcd\u0b9f\u0bbf\u0baf\u0ba4\u0bc1 :
\n(AB)T<\/sup> = AB
\n\u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0ba3\u0bae\u0bcd :
\n(AB)T<\/sup> = BT<\/sup> . AT<\/sup>
\n[ \u2234 \u0ba8\u0bbf\u0bb0\u0bc8 \u0ba8\u0bbf\u0bb0\u0bb2\u0bcd \u0bae\u0bbe\u0bb1\u0bcd\u0bb1\u0bc1 \u0bb5\u0bbf\u0ba4\u0bbf\u0baf\u0bbf\u0ba9\u0bcd \u0baa\u0bbf\u0ba9\u0bcd \u0ba4\u0bbf\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bc8 \u0bb5\u0bbf\u0ba4\u0bbf]
\n= B . A [\u2235 BT<\/sup> = B, A\u2235 = A]
\n= AB [AB = BA \u0ba4\u0bb0\u0bb5\u0bc1]
\n\u2234 (AB)T<\/sup> = AB
\n\u2234 AB \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b93\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0bbf\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0ba4\u0bc1.
\n\u0bae\u0bb1\u0bc1\u0ba4\u0bb2\u0bc8 :
\n\u0ba4\u0bb0\u0bb5\u0bc1 :
\nA = AT<\/sup>, B = BT<\/sup>
\n(AB)T<\/sup> = AB
\n\u0ba8\u0bbf\u0bb0\u0bc1\u0baa\u0ba3\u0bae\u0bcd :
\n(AB)T<\/sup> = BT<\/sup> AT<\/sup> = BA
\n\u2234 AB = BA<\/p>\n

Question 23.
\nA, B \u0b8e\u0ba9\u0bcd\u0baa\u0ba9 \u0b9a\u0bae\u0bb5\u0bb0\u0bbf\u0b9a\u0bc8\u0baf\u0bc1\u0bb3\u0bcd\u0bb3 \u0b87\u0bb0\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, \u0baa\u0bbf\u0ba9\u0bcd\u0bb5\u0bb0\u0bc1\u0bb5\u0ba9\u0bb5\u0bb1\u0bcd\u0bb1\u0bc8 \u0ba8\u0bbf\u0bb1\u0bc1\u0bb5\u0bc1\u0b95.
\n(i) AB + BA \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd
\n(ii) AB – BA \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0ba4\u0bb0\u0bb5\u0bc1 : AT<\/sup> = A, BT<\/sup> = B
\n(i) (AB + BA)T<\/sup> = (AB)T<\/sup> + (BA)T<\/sup>
\n= BT<\/sup>AT<\/sup> + AT<\/sup>BT<\/sup>
\n= BA + AB
\n= AB + BA
\n\u2234 AB + BA \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

(ii) (AB – BA)T<\/sup> = (AB)T<\/sup> – (BA)T<\/sup>
\n= BT<\/sup>AT<\/sup> – AT<\/sup>BT<\/sup>
\n= BA – AB
\n= -(AB – BA)
\n\u2234 AB \u2013 BA \u0b8e\u0ba9\u0bcd\u0baa\u0ba4\u0bc1 \u0b8e\u0ba4\u0bbf\u0bb0\u0bcd \u0b9a\u0bae\u0b9a\u0bcd\u0b9a\u0bc0\u0bb0\u0bcd \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

\"Samacheer<\/p>\n

Question 24.
\n\u0b93\u0bb0\u0bcd \u0b85\u0b99\u0bcd\u0b95\u0bbe\u0b9f\u0bbf\u0baf\u0bbf\u0bb2\u0bcd \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf, \u0b89\u0bb2\u0bb0\u0bcd \u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0 \u0bc1\u0baa\u0bcd\u0baa\u0bc1 \u0b86\u0b95\u0bbf\u0baf\u0bb5\u0bb1\u0bcd\u0bb1\u0bc8\u0b95\u0bcd \u0b95\u0bca\u0ba3\u0bcd\u0b9f\u0bc1 \u0bae\u0bc2\u0ba9\u0bcd\u0bb1\u0bc1 \u0bb5\u0bbf\u0ba4\u0bae\u0bbe\u0ba9 \u0baa\u0bb0\u0bbf\u0b9a\u0bc1\u0baa\u0bcd \u0baa\u0bc8\u0b95\u0bb3\u0bcd \u0ba4\u0baf\u0bbe\u0bb0\u0bcd \u0b9a\u0bc6\u0baf\u0bcd\u0baf\u0baa\u0bcd\u0baa\u0b9f\u0bc1\u0b95\u0bbf\u0ba9\u0bcd\u0bb1\u0ba9. \u0baa\u0bc8 -\u0bb2\u0bcd 100 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf, 100 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0b89\u0bb2\u0bb0\u0bcd \u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd 50 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bc1\u0bae\u0bcd, \u0baa\u0bc8 II-\u0bb2\u0bcd 200 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf, 100 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0b89\u0bb2\u0bb0\u0bcd \u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd 100 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bc1\u0bae\u0bcd, \u0baa\u0bc8 III-\u0bb2\u0bcd 250 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf, 250 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0b89\u0bb2\u0bb0\u0bcd\u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8 \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd 150 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bc1\u0bae\u0bcd \u0b89\u0bb3\u0bcd\u0bb3\u0ba9. .50 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf\u0baf\u0bbf\u0ba9\u0bcd \u0bb5\u0bbf\u0bb2\u0bc8 \u20b950\/-, 50 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0b89\u0bb2\u0bb0\u0bcd \u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8\u0baf\u0bbf\u0ba9\u0bcd \u0bb5\u0bbf\u0bb2\u0bc8 \u20b910\/- \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd 50 \u0b95\u0bbf\u0bb0\u0bbe\u0bae\u0bcd \u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bbf\u0ba9\u0bcd \u0bb5\u0bbf\u0bb2\u0bc8 \u20b960\/- \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, \u0b92\u0bb5\u0bcd\u0bb5\u0bca\u0bb0\u0bc1 \u0baa\u0bb0\u0bbf\u0b9a\u0bc1\u0baa\u0bcd \u0baa\u0bc8\u0baf\u0bbf\u0ba9\u0bcd \u0bb5\u0bbf\u0bb2\u0bc8\u0baf\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0baa\u0bb0\u0bbf\u0b9a\u0bc1\u0baa\u0bcd \u0baa\u0bc8\u0b95\u0bb3\u0bc8 \u0b93\u0bb0\u0bcd \u0b85\u0ba3\u0bbf \u0bb5\u0b9f\u0bbf\u0bb5\u0bbf\u0bb2\u0bcd \u0b8e\u0bb4\u0bc1\u0ba4<\/p>\n

A = \u0bae\u0bc1\u0ba8\u0bcd\u0ba4\u0bbf\u0bb0\u0bbf
\n\u0b89\u0bb2\u0bb0\u0bcd\u0ba8\u0bcd\u0ba4 \u0ba4\u0bbf\u0bb0\u0bbe\u0b9f\u0bcd\u0b9a\u0bc8
\n\u0baa\u0bbe\u0ba4\u0bbe\u0bae\u0bcd \u0baa\u0bb0\u0bc1\u0baa\u0bcd\u0baa\u0bc1 \\(\\left[\\begin{array}{ccc}
\n\\mathrm{I} & \\mathrm{II} & \\mathrm{III} \\\\
\n100 & 200 & 250 \\\\
\n100 & 100 & 250 \\\\
\n50 & 100 & 150
\n\\end{array}\\right]\\)
\nk = \\(\\frac{1}{50}\\)
\n\u0bb5\u0bbf\u0bb2\u0bc8 B = [50 10 60]
\n\u0baa\u0bb0\u0bbf\u0b9a\u0bc1\u0baa\u0bcd \u0baa\u0bc8\u0b95\u0bb3\u0bbf\u0ba9\u0bcd \u0bb5\u0bbf\u0bb2\u0bc8 = B (kA)<\/p>\n

= [50 10 60] \\(\\left(\\frac{1}{50}\\left[\\begin{array}{ccc}
\n\\mathrm{I} & \\mathrm{II} & \\mathrm{III} \\\\
\n100 & 200 & 250 \\\\
\n100 & 100 & 250 \\\\
\n50 & 100 & 150
\n\\end{array}\\right]\\right)\\)
\nBA = [50 10 60 \\(\\left[\\begin{array}{lll}
\n\\mathrm{I} & \\mathrm{II} & \\mathrm{II} \\\\
\n2 & 4 & 5 \\\\
\n2 & 2 & 5 \\\\
\n1 & 2 & 3
\n\\end{array}\\right]\\)
\n= \\(\\left[\\begin{array}{c}
\n100+20+60 \\\\
\n200+20+120 \\\\
\n250+50+180
\n\\end{array}\\right]=\\left[\\begin{array}{c}
\n180 \\\\
\n340 \\\\
\n480
\n\\end{array}\\right]\\)
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0baa\u0bc8 I = \u20b9180
\n\u0baa\u0bc8 II = \u20b9340
\n\u0baa\u0bc8 III = \u20b9 480 \u0b86\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.1 Textbook Questions and Answers, Notes. TN Board 11th Maths Solutions Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.1 Question 1. (i) aij = m = 2, n = 3 (ii) aij = , m = 3, […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/49863"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=49863"}],"version-history":[{"count":0,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/49863\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=49863"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=49863"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=49863"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}