2a – b = 0<\/span>
\n– a = -1
\n\u21d2 a = 1
\nWe have, a – b = -1 \u21d2 1 – b = -1 \u21d2 b = 2
\n\u21d2 2a + c = 5 \u21d2 2 + c = 5 \u21d2 c = 3
\n\u21d2 3c + d = 13 \u21d2 9 + d = 13 \u21d2 d = 4.<\/p>\nQuestion 2.
\nSimplify cosx\\(\\left[\\begin{array}{cc}{\\cos x} & {\\sin x} \\\\{-\\sin x} & {\\cos x}\\end{array}\\right]\\) + sinx\\(\\left[\\begin{array}{cc}{\\sin x} & {-\\cos x} \\\\{\\cos x} & {\\sin x}\\end{array}\\right]\\).
\nAnswer:
\n
<\/p>\n
Question 3.
\nSolve the equation for x, y z and t; if
\n\\(2\\left[\\begin{array}{ll}{x} & {z} \\\\{y} & {t}\\end{array}\\right]+3\\left[\\begin{array}{cc}{1} & {-1} \\\\{0} & {2}\\end{array}\\right]=3\\left[\\begin{array}{ll}{3} & {5} \\\\{4} & {6}\\end{array}\\right]\\).
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q3.jpg\")
\n\u21d2 2x + 3 = 9 \u21d2 x = 3
\n\u21d2 2z – 3 = 15 \u21d2 z = 9
\n\u21d2 2y = 12 \u21d2 y = 6
\n\u21d2 2t + 6 = 18 \u21d2 t = 6.<\/p>\n
Question 4.
\nFind A2<\/sup> – 5A + 6I If A = \\(\\left[\\begin{array}{ccc}{2} & {0} & {1} \\\\{2} & {1} & {3} \\\\{1} & {-1} & {0}\\end{array}\\right]\\)
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q4.jpg\")
\nA2<\/sup> – 5A + 6I
\n
<\/p>\nQuestion 5.
\nIf A = \\(\\left[\\begin{array}{cc}{3} & {-2} \\\\{4} & {-2}\\end{array}\\right]\\) find k so that A2<\/sup> = kA – 2I.
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q5.jpg\")
\nGiven A2<\/sup> = kA – 2I
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q5.1.jpg\")
\n1 = 3k – 2
\n\u21d2 k = 1.<\/p>\nQuestion 6.
\nExpress A = \\(\\left[\\begin{array}{ccc}{-1} & {2} & {3} \\\\{5} & {7} & {9} \\\\{-2} & {1} & {1}
\n\\end{array}\\right]\\) as the sum of a symmetric and skew symmetric matrix.
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q6.jpg\")
\nP = 1\/2 (A + AT<\/sup>) is symmetric.
\nQ = 1\/2 (A – AT<\/sup>) is skew symmetric.
\n
<\/p>\nQuestion 7.
\nFind the inverse of the following using elementary transformations.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q7.jpg\")
\nAnswer:
\n(i) Let A = I A
\n
<\/p>\n
(ii) Let A = IA
\n
<\/p>\n
(iii) Let A = IA
\n
<\/p>\n
(iv) Let A = IA
\n
<\/p>\n
Question 8.
\nFind the inverse of the matrix A = \\(\\left[\\begin{array}{cc}{2} & {3} \\\\{-1} & {5}\\end{array}\\right]\\) using row transformation.
\nAnswer:
\nA = \\(\\left[\\begin{array}{cc}{2} & {3} \\\\{-1} & {5}\\end{array}\\right]\\)
\nLet A = IA
\n
<\/p>\n
Question 9.
\n\\(A=\\left[\\begin{array}{ll}{2} & {3} \\\\{4} & {5} \\\\{2} & {1}\\end{array}\\right] B=\\left[\\begin{array}{ccc}{1} & {-2} & {3} \\\\{-4} & {2} & {5}\\end{array}\\right]\\)<\/p>\n
\n- Find AB<\/li>\n
- If C is the matrix obtained from A by the transformation R1<\/sub> \u2192 2R1<\/sub>, find CB<\/li>\n<\/ol>\n
Answer:
\n
<\/p>\n
(ii) Since C is the matrix obtained from A by the transformation R1<\/sub> \u2192 2R1<\/sub>
\n\u21d2 C = \\(\\left[\\begin{array}{ll}{4} & {6} \\\\{4} & {5} \\\\{2} & {1}\\end{array}\\right]\\)
\nThen CB can be obtained by multiplying first row of AB by 2.
\nCB = \\(\\left[\\begin{array}{ccc}{-20} & {-4} & {42} \\\\{-16} & {2} & {37} \\\\{-2} & {-2} & {11}
\n\\end{array}\\right]\\).<\/p>\nQuestion 10.
\nConstruct a 3 \u00d7 4 matrix whose elements are given by<\/p>\n
\n- ay<\/sub> = \\(\\frac{|-3 i+j|}{2}\\) (2)<\/li>\n
- aij<\/sub> = 2i – j (2)<\/li>\n<\/ol>\n
Answer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q10.jpg\")
\na13<\/sub> = 0, a14<\/sub> = \\(\\frac{1}{2}\\), a21<\/sub> = \\(\\frac{5}{2}\\), a22<\/sub> = 2, a23<\/sub> = \\(\\frac{3}{2}\\), a24<\/sub> = 1, a31<\/sub> = 4, a32<\/sub> = \\(\\frac{7}{2}\\), a33<\/sub> = 3, a34<\/sub> = \\(\\frac{5}{2}\\)
\n
<\/p>\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q10.2.jpg\")
\na11<\/sub> = 1, a12<\/sub> = 0, a13<\/sub>= -1, a14<\/sub> = -2, a21<\/sub> = 3, a22<\/sub> = 2, a23<\/sub> = 1, a24<\/sub> = 0, a31<\/sub> = 5, a32<\/sub> = 4, a33<\/sub> = 3, a34<\/sub> = 2
\n
<\/p>\nQuestion 11.
\nExpress the following matrices as the sum of a Symmetric and a Skew Symmetric matrix.
\n(i) \\(\\left[\\begin{array}{ccc}{6} & {-2} & {2} \\\\{-2} & {3} & {-1} \\\\{2} & {-1} & {3}
\n\\end{array}\\right]\\)
\n(ii) \\(\\left[\\begin{array}{ccc}{3} & {3} & {-1} \\\\{-2} & {-2} & {1} \\\\{-4} & {-5} & {2}
\n\\end{array}\\right]\\)
\nAnswer:
\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 12.
\nIf A = \\(\\left[\\begin{array}{ccc}{2} & {4} & {3} \\\\{1} & {0} & {6} \\\\{0} & {-2} & {-3}\\end{array}\\right]\\)<\/p>\n
\n- Find 3A. (1)<\/li>\n
- Find AT<\/sup> (1)<\/li>\n
- Evaluate A + AT<\/sup> , is it symmetric? Justify your answer. (1)<\/li>\n<\/ol>\n
Answer:
\n1. 3A = \\(\\left[\\begin{array}{ccc}{6} & {12} & {9} \\\\{3} & {0} & {18} \\\\{0} & {-6} & {-9}
\n\\end{array}\\right]\\)<\/p>\n
2. AT<\/sup> = \\(\\left[\\begin{array}{ccc}{2} & {1} & {0} \\\\{4} & {0} & {-2} \\\\{3} & {6} & {-3}
\n\\end{array}\\right]\\)<\/p>\n3. A + AT<\/sup>
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-3M-Q12.jpg\")
\nThe elements on both sides of the main diagonal are same. Therefore A + AT <\/sup>is a symmetric matrix.<\/p>\nPlus Two Maths Matrices Four Mark Questions and Answers<\/h3>\n
Question 1.
\nConsider the following statement: P(n) : An<\/sup> = \\(\\left[\\begin{array}{cc}{1+2 n} & {-4 n} \\\\{n} & {1-2 n}\\end{array}\\right]\\) for all n \u2208 N<\/p>\n\n- Write P (1). (1)<\/li>\n
- If P(k) is true, then show that P( k + 1) is also true. (3)<\/li>\n<\/ol>\n
Answer:
\n1. P(1) : A = \\(\\left[\\begin{array}{cc}{1+2} & {-4} \\\\{1} & {1-2}\\end{array}\\right]=\\left[\\begin{array}{cc}{3} & {-4} \\\\{1} & {-1}\\end{array}\\right]\\)<\/p>\n
2. Assume that P(n) is true n = k
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q1.jpg\")
\nHence P(k+1) is true n \u2208 N.<\/p>\n
Question 2.
\nFind the matrices A and B if 2A + 3B = \\(\\left[\\begin{array}{ccc}{1} & {2} & {-1} \\\\{0{1} & {2} & {4}\\end{array}\\right]\\) and A + 2B = \\(\\left[\\begin{array}{lll}{2} & {0} & {1} \\\\{1} & {1} & {2} \\\\{3} & {1} & {2}\\end{array}\\right]\\).
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q2.jpg\")
\nSolving (1) and (2) \u21d2 2 \u00d7 (2)
\n
<\/p>\n
![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q2.2.jpg\")
\nQuestion 3.<\/p>\n
\n- Construct a 3 \u00d7 3 matrix A = [aij<\/sub>] where aij<\/sub> – 2(i – j) (3)<\/li>\n
- Show that matrix A is skew-symmetric. (1)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
<\/p>\n
2.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q3.1.jpg\")
\nTherefore A is a skew-symmetric matrix.<\/p>\n
Question 4.
\nConsider the following statement P(n ): An<\/sup> = \\(\\left[\\begin{array}{cc}{\\cos n \\theta} & {\\sin n \\theta} \\\\{-\\sin n \\theta} & {\\cos n \\theta}\\end{array}\\right]\\) for all n \u2208 N<\/p>\n\n- Write P(1). (1)<\/li>\n
- If P (k) is true then show that P (k+1) is true (3)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
<\/p>\n
2. Assume that P(n) is true for n = k
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q4.1.jpg\")
\nP(k+1) = Ak+1<\/sup>
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q4.2.jpg\")
\n\u2234 P(k+1) is true. Hence true for all n \u2208 N.<\/p>\nQuestion 5.
\nA = \\(\\left[\\begin{array}{lll}{1} & {2} & {2} \\\\{2} & {1} & {2} \\\\{2} & {2} & {1}\\end{array}\\right]\\), then<\/p>\n
\n- Find 4A and A2<\/sup> (2)<\/li>\n
- Show that A2<\/sup> -4A = 5I3<\/sub> (2)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
<\/p>\n
2.
\n
<\/p>\n
Question 6.
\nLet A = \\(\\left[\\begin{array}{lll}{2} & {1} & {3} \\\\{4} & {1} & {0}\\end{array}\\right]\\) and B= \\(\\left[\\begin{array}{cc}{1} & {-1} \\\\{0} & {2} \\\\{5} & {0}\\end{array}\\right]\\)<\/p>\n
\n- Find AT<\/sup> and BT<\/sup> (1)<\/li>\n
- Find AB (1)<\/li>\n
- Show that (AB)T<\/sup> = BT<\/sup> AT<\/sup> (2)<\/li>\n<\/ol>\n
Answer:
\n1.
\n
<\/p>\n
2.
\n
<\/p>\n
3.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q6.2.jpg\")
\n\u2234 (AB)T<\/sup> = BT<\/sup> AT<\/sup>.<\/p>\nQuestion 7.
\nA = \\(\\left[\\begin{array}{ccc}{1} & {-3} & {1} \\\\{2} & {0} & {4} \\\\{1} & {2} & {-2}\\end{array}\\right]\\) Express A as the sum of a symmetric and skew symmetric matrix.
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-4M-Q7.jpg\")
\n\\(\\frac{1}{2}\\) (A + AT<\/sup>) + \\(\\frac{1}{2}\\) (A – AT<\/sup>)
\n
<\/p>\nQuestion 8.<\/p>\n
\n- Consider a 2 \u00d7 2 matrix A = [aij<\/sub>], where aij<\/sub> = \\(\\frac{(i+j)^{2}}{2}\\)<\/li>\n
- Write the transpose of A. (2)<\/li>\n
- Show that A is symmetric. (2)<\/li>\n<\/ol>\n
Answer:
\n1. A = \\(\\left[\\begin{array}{ll}{2} & {\\frac{9}{2}} \\\\{\\frac{9}{2}} & {8}\\end{array}\\right]\\)<\/p>\n
2. AT<\/sup> = \\(\\left[\\begin{array}{ll}{2} & {\\frac{9}{2}} \\\\{\\frac{9}{2}} & {8}\\end{array}\\right]\\)<\/p>\n3. AT<\/sup> = A therefore symmetric matrix.<\/p>\nQuestion 9.
\nA = \\(\\left[\\begin{array}{ll}{6} & {5} \\\\{7} & {6}\\end{array}\\right]\\) is a matrix<\/p>\n
\n- What is the order of A. (1)<\/li>\n
- Find A2<\/sup> and 12 A. (2)<\/li>\n
- If f(x) = xT<\/sup> – 12x +1; find f(A). (1)<\/li>\n<\/ol>\n
Answer:
\n1. Order of A is 2 \u00d7 2.<\/p>\n
2.
\n
<\/p>\n
3. f(x) = x2<\/sup> – 12x + 1 \u21d2 f(A) = A2<\/sup> – 12A + I
\n
<\/p>\nPlus Two Maths Matrices Six Mark Questions and Answers<\/h3>\n
Question 1.
\nLet A = \\(\\left[\\begin{array}{ll}{2} & {4} \\\\{3} & {2}\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{cc}{1} & {3} \\\\{-2} & {5}\\end{array}\\right]\\), C = \\(\\left[\\begin{array}{rr}{-2} & {5} \\\\{3} & {4}\\end{array}\\right]\\)
\nFind each of the following
\n(i) A + B; A – B
\n(ii) 3A – C
\n(iii) AB
\n(iv) BA
\nAnswer:
\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 2.
\nLet A = \\(\\left[\\begin{array}{ll}{1} & {2} \\\\{3} & {4}\\end{array}\\right]\\); B = \\(\\left[\\begin{array}{ll}{2} & {1} \\\\{4} & {5}\\end{array}\\right]\\); C = \\(\\left[\\begin{array}{ccc}{1} & {-1} \\\\{0} & {2}\\end{array}\\right]\\)
\n(i) Find A + B and A – B (2)
\n(ii) Show that (A + B) + C = A + (B + C) (2)
\n(iii) Find AB and BA
\nAnswer:
\n
<\/p>\n
![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q2.1.jpg\")
\n\u2234 (A + B) + C = A + (B + C)
\n
<\/p>\n
Question 3.
\nA = \\(\\left[\\begin{array}{ccc}{-1} & {0} & {2} \\\\{4} & {0} & {-3}\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{cc}{0} & {2} \\\\{-1} & {3} \\\\{0} & {4}\\end{array}\\right]\\)<\/p>\n
\n- What is the order of matrix AB ? (1)<\/li>\n
- Find AT<\/sup>, BT<\/sup> (2)<\/li>\n
- Verify (AB)T<\/sup> = BT<\/sup> AT<\/sup> (3)<\/li>\n<\/ol>\n
Answer:
\n1. Order of AB is 2 \u00d7 2. Since order of A is 2 \u00d7 3 and B is 3 \u00d7 2.<\/p>\n
2.
\n
<\/p>\n
3.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q3.1.jpg\")
\n(AB)T<\/sup> = BT<\/sup> AT<\/sup>.<\/p>\nQuestion 4.
\nLet A = \\(\\left[\\begin{array}{rrr}{1} & {2} & {-3} \\\\{2} & {1} & {-1}\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{ll}{2} & {3} \\\\{5} & {4} \\\\{1} & {6}\\end{array}\\right]\\)
\n(i) FindAB. (1)
\n(ii) Find AT<\/sup>, BT<\/sup> & (AB)T<\/sup> (3)
\n(iii) Verify that (AB)T<\/sup> = BT<\/sup> AT<\/sup> (2)
\nAnswer:
\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 5.
\nIf A = \\(\\left[\\begin{array}{c}{-2} \\\\{4} \\\\{5}\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{lll}{1} & {3} & {6}\\end{array}\\right]\\)
\n(i) Find AT<\/sup>, BT<\/sup> (1)
\n(ii) Find (AB)T<\/sup> (2)
\n(iii) Verify (AB)T<\/sup> = BT<\/sup> AT<\/sup> (3)
\nAnswer:
\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 6.
\nLet A = \\(\\left[\\begin{array}{cc}{3} & {1} \\\\{-1} & {2}\\end{array}\\right]\\)
\n(i) Find A2<\/sup> (1)
\n(ii) Show that A2<\/sup> – 5A + 7I = 0 (1)
\n(iii) Using this result find A-1<\/sup> (2)
\n(iv) Slove the following equation using matrix: 3x + y = 1, – x + 2y = 2.
\nAnswer:
\n
<\/p>\n
<\/p>\n
(iii) A2<\/sup> – 5A + 7I = 0 \u21d2 A2<\/sup> – 5A = -7I,
\nmultiplying by A-1<\/sup> on both sides,
\n\u21d2 A – 5I = -7 A-1<\/sup>
\n
<\/p>\n(iv) The equation can be represented in matrix form as follows, AX = B \u21d2 X = A-1<\/sup>B
\n
<\/p>\nQuestion 7.
\nA = \\(\\left[\\begin{array}{ccc}{1} & {2} & {3} \\\\{3} & {-2} & {1} \\\\{4} & {2} & {1}
\n\\end{array}\\right]\\)
\n(i) Show that A3<\/sup> – 23A – 40I = 0 (3)
\n(ii) Hence find A-1<\/sup> (3)
\nAnswer:
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q7.jpg\")
\nA3<\/sup> – 23A – 40I = 0
\n
<\/p>\n(ii) A-1<\/sup>A3<\/sup> – 23 A-1<\/sup>A – 40A-1<\/sup>I = 0
\n\u21d2 A2<\/sup> – 23I – 40A-1<\/sup> = 0
\n
<\/p>\nQuestion 8.
\nA is a third order square matrix and \\(a_{i j}=\\left\\{\\begin{aligned}-i+2 j & \\text { if } i=j \\\\i \\times j & \\text { if } i \\neq j\\end{aligned} \\text { and } B=\\left[\\begin{array}{lll}{2} & {1} & {1} \\\\{1} & {1} & {5} \\\\{1} & {5} & {2}\\end{array}\\right]\\right.\\)<\/p>\n
\n- Construct the matrix A. (1)<\/li>\n
- Interpret the matrix A. (1)<\/li>\n
- Find AB – BA. (3)<\/li>\n
- Interpret the matrix AB – BA. (1)<\/li>\n<\/ol>\n
Answer:
\n1. a11<\/sub> = 1, a12<\/sub> = 2, a13<\/sub> = 3, a21<\/sub> = 2, a22<\/sub> = 2, a23<\/sub> = 6, a31<\/sub> = 3, a32<\/sub> = 6, a33<\/sub> = 3
\nA = \\(\\left[\\begin{array}{lll}{1} & {2} & {3} \\\\{2} & {2} & {6} \\\\{3} & {6} & {3}\\end{array}\\right]\\)<\/p>\n2. Now,
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q8.jpg\")
\nTherefore A is symmetric matrix.<\/p>\n
3.
\n
<\/p>\n
<\/p>\n
<\/p>\n
4.
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q8.4.jpg\")
\n= -(AB – BA)
\n\u2234 skew symmetric matrix.<\/p>\n
Question 9.
\nFind x and y if
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q9.jpg\")
\nAnswer:
\n
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Question 10.
\nGiven that A + B = \\(\\left[\\begin{array}{ll}{2} & {5} \\\\{7} & {8}\\end{array}\\right]\\) and A – B = \\(\\left[\\begin{array}{ll}{6} & {8} \\\\{4} & {3}\\end{array}\\right]\\)<\/p>\n
\n- Find 2A. (1)<\/li>\n
- Find A2<\/sup> – B2<\/sup>. (3)<\/li>\n
- Is it equal to (A + B) (A – B)? Give reason (2)<\/li>\n<\/ol>\n
Answer:
\n1. 2A = A + B + A – B
\n
<\/p>\n
2.
\n
<\/p>\n
3. (A + B)(A – B)
\n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q10.2.jpg\")
\n(A + B)(A – B) = A2<\/sup> + AB – BA – B2<\/sup>
\n\u2260 A2<\/sup> – B2<\/sup>
\n\u2235 AB \u2260 BA.<\/p>\nQuestion 11.
\n(i) Consider A = \\(\\left[\\begin{array}{lll}{1} & {x} & {1}\\end{array}\\right]\\), B = \\(\\left[\\begin{array}{ccc}{1} & {3} & {2} \\\\{2} & {5} & {1} \\\\{15} & {3} & {2}
\n\\end{array}\\right]\\), C = \\(\\left[\\begin{array}{l}{1} \\\\{2} \\\\{x}\\end{array}\\right]\\) (2)<\/p>\n
\n\n\nA – Matrix<\/td>\n | B – Order<\/td>\n<\/tr>\n |
\nA<\/td>\n | 3 \u00d7 1<\/td>\n<\/tr>\n |
\nB<\/td>\n | 1 \u00d7 1<\/td>\n<\/tr>\n |
\nBC<\/td>\n | 2 \u00d7 2<\/td>\n<\/tr>\n |
\nABC<\/td>\n | 3 \u00d7 3<\/td>\n<\/tr>\n |
\n<\/td>\n | 1 \u00d7 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n (ii) Find x if ABC = 0 (4) \nAnswer: \n(i)<\/p>\n \n\n\nA – Matrix<\/td>\n | B – Order<\/td>\n<\/tr>\n | \nA<\/td>\n | 1 \u00d7 3<\/td>\n<\/tr>\n | \nB<\/td>\n | 3 \u00d7 3<\/td>\n<\/tr>\n | \nBC<\/td>\n | 3 \u00d7 1<\/td>\n<\/tr>\n | \nABC<\/td>\n | 1 \u00d7 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n (ii) Given, ABC = 0 \n![\"Plus](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/02\/Plus-Two-Maths-Chapter-Wise-Questions-and-Answers-Chapter-3-Matrices-6M-Q11.jpg\") \n\u21d2 x2<\/sup> + 16x + 28 = 0 \n\u21d2 (x + 14)(x + 2) = 0 \n\u21d2 x = -14, -2.<\/p>\nWe hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 3 Matrices will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 3 Matrices, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":" Plus Two Maths Chapter Wise Questions and Answers Chapter 3 Matrices are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 3 Matrices. Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Maths\u00a0Chapter Wise Questions Chapter Chapter 3 Chapter … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[42728],"tags":[],"yoast_head":"\nPlus Two Maths Chapter Wise Questions and Answers Chapter 3 Matrices - CBSE Library<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n | |