{"id":37959,"date":"2023-01-27T10:00:05","date_gmt":"2023-01-27T04:30:05","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=37959"},"modified":"2023-01-28T10:03:12","modified_gmt":"2023-01-28T04:33:12","slug":"plus-two-maths-chapter-wise-questions-answers-chapter-4","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-two-maths-chapter-wise-questions-answers-chapter-4\/","title":{"rendered":"Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants"},"content":{"rendered":"

Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants are part of Plus Two Maths Chapter Wise Questions and Answers<\/a>. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants.<\/p>\n\n\n\n\n\n\n\n\n\n\n
Board<\/strong><\/td>\nSCERT, Kerala<\/td>\n<\/tr>\n
Text Book<\/strong><\/td>\nNCERT Based<\/td>\n<\/tr>\n
Class<\/strong><\/td>\nPlus Two<\/td>\n<\/tr>\n
Subject<\/strong><\/td>\nMaths\u00a0Chapter Wise Questions<\/td>\n<\/tr>\n
Chapter<\/strong><\/td>\nChapter 4<\/td>\n<\/tr>\n
Chapter Name<\/strong><\/td>\nDeterminants<\/td>\n<\/tr>\n
Number of Questions Solved<\/strong><\/td>\n45<\/td>\n<\/tr>\n
Category<\/strong><\/td>\nPlus Two Kerala<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants<\/h2>\n

Plus Two Maths Determinants Three Mark Questions and Answers<\/h3>\n

Question 1.
\nUsing properties of determinants prove \\(\\left|\\begin{array}{ccc}{x} & {y} & {x+y} \\\\{y} & {x+y} & {x} \\\\{x+y} & {x} & {y}\\end{array}\\right|\\) = -2(x3<\/sup> + y3<\/sup>).
\nAnswer:
\n\"Plus
\n= 2(x + y)(-x2<\/sup> + xy – y2<\/sup>) = -2(x3<\/sup> + y3<\/sup>).<\/p>\n

Question 2.
\nIf a, b, c are real numbers and \\(\\left|\\begin{array}{lll}{b+c} & {c+a} & {a+b} \\\\{c+a} & {a+b} & {b+c} \\\\{a+b} & {b+c} & {c+a}\\end{array}\\right|\\) = 0, Show that a = b = c.
\nAnswer:
\n\"Plus
\n2(a + b + c) [(b – c) (c – b) – (b – a) (c – a)] =0 (a+b+c) = 0
\n(a + b + c) = 0 or (b – c) (c – b) = (b – a) (c – a)
\n(a + b + c) = 0 or a = b = c.<\/p>\n

Question 3.
\nSolve using properties of determinants.
\n\\(\\left|\\begin{array}{ccc}{2 x-1} & {x+7} & {x+4} \\\\{x} & {6} & {2} \\\\{x-1} & {x+1} & {3}
\n\\end{array}\\right|\\) = 0
\nAnswer:
\n\"Plus
\n\u21d2 (x – 1) (x2<\/sup> + x – 6x + 6) = 0
\n\u21d2 (x – 1)(x2<\/sup> – 5x + 6) = 0
\n\u21d2 (x – 1) (x – 3) (x – 2) = 0
\n\u21d2 x = 1, x = 3, x = 2.<\/p>\n

Question 4.
\nIf \\(\\left|\\begin{array}{cc}{3} & {x} \\\\{x} & {x}\\end{array}\\right|=\\left|\\begin{array}{cc}{-2} & {2} \\\\{4} & {1}\\end{array}\\right|\\), find the value of x.
\nAnswer:
\n\\(\\left|\\begin{array}{cc}{3} & {x} \\\\{x} & {x}\\end{array}\\right|=\\left|\\begin{array}{cc}{-2} & {2} \\\\{4} & {1}\\end{array}\\right|\\)
\n\u21d2 3x – x2<\/sup> = – 2 – 8
\n\u21d2 x2<\/sup> – 3x – 10 = 0
\n\u21d2 x = 5, -2.<\/p>\n

Question 5.
\nA = \\(\\left[\\begin{array}{ccc}{1} & {-3} & {1} \\\\{2} & {0} & {4} \\\\{1} & {2} & {-2}
\n\\end{array}\\right]\\)<\/p>\n

    \n
  1. Calculate |A| (1)<\/li>\n
  2. Find |adjA| {Hint: using the property A \u00d7 adjA = |A|I} (1)<\/li>\n
  3. Find |3A| (1)<\/li>\n<\/ol>\n

    Answer:
    \n1. |A| = \\(\\left[\\begin{array}{ccc}{1} & {-3} & {1} \\\\{2} & {0} & {4} \\\\{1} & {2} & {-2}
    \n\\end{array}\\right]\\) = – 28.<\/p>\n

    2. A \u00d7 adjA = |A|I
    \n\"Plus<\/p>\n

    3. |3A| = 27 \u00d7 |A| = 27 \u00d7 -28 = -756.<\/p>\n

    Question 6.
    \nUsing properties of determinants proves the following.
    \n\"Plus
    \nAnswer:
    \n\"Plus<\/p>\n

    \"Plus<\/p>\n

    \"Plus
    \n= 2{-{-c){b{a – c)) – b(-c(c + a))}
    \n= 2{c(ab – cb) + b(c2<\/sup> + ac)}
    \n= 2{abc – c2<\/sup>b + bc2<\/sup> + abc)} = 4abc.<\/p>\n

    Plus Two Maths Determinants Four Mark Questions and Answers<\/h3>\n

    Question 1.
    \n(i) If \\(\\left|\\begin{array}{rrr}{1} & {-3} & {2} \\\\{4} & {-1} & {2} \\\\{3} & {5} & {2}\\end{array}\\right|\\) = 40, then \\(\\left|\\begin{array}{ccc}{1} & {4} & {3} \\\\{-3} & {-1} & {5} \\\\{2} & {2} & {2}\\end{array}\\right|\\) = ?
    \n(a)\u00a0 \u00a00
    \n(b)\u00a0 – 40
    \n(c)\u00a0 40
    \n(d)\u00a0 2 (1)
    \n(ii) \\(\\left|\\begin{array}{rrr}{3} & {-3} & {2} \\\\{12} & {-1} & {2} \\\\{9} & {5} & {2}\\end{array}\\right|\\) = ?
    \n(a) 120
    \n(b)\u00a0 40
    \n(c)\u00a0 – 40
    \n(d)\u00a0 0 (1)
    \n(iii) Show that \u2206 = \\(\\left|\\begin{array}{ccc}{-a^{2}} & {a b} & {a c} \\\\{b a} & {-b^{2}} & {b c} \\\\{a c} & {b c} & {-c^{2}}\\end{array}\\right|\\) = 4a2<\/sup>b2<\/sup>c2<\/sup> (2)
    \nAnswer:
    \n(i) (c) 40<\/p>\n

    (ii) (a)120<\/p>\n

    (iii) \u2206 = abc\\(\\left|\\begin{array}{ccc}{-a} & {a} & {a} \\\\{b} & {-b} & {b} \\\\{c} & {c} & {-c}\\end{array}\\right|\\) take a, b, c from C1<\/sub>, C2<\/sub>, C3<\/sub>
    \n\"Plus<\/p>\n

    Question 2.
    \n\"Plus
    \nAnswer:
    \n(i) \\(\\left|\\begin{array}{ll}{2} & {4} \\\\{5} & {1}\\end{array}\\right|=\\left|\\begin{array}{ll}{2 x} & {4} \\\\{6} & {x}\\end{array}\\right|\\) \u21d2 -18 = 2x2<\/sup> – 24.
    \n\u21d2 2x2<\/sup> = 6 \u21d2 x2<\/sup> = 3 \u21d2 x = \\(\\pm \\sqrt{3}\\).<\/p>\n

    \"Plus<\/p>\n

    \"Plus<\/p>\n

    Question 3.
    \nProve that \\(\\left|\\begin{array}{ccc}{(b+c)^{2}} & {a^{2}} & {a^{2}} \\\\{b^{2}} & {(c+a)^{2}} & {b^{2}} \\\\{c^{2}} & {c^{2}} & {(a+b)^{2}}\\end{array}\\right|\\) = 2abc(a + b + c)3<\/sup>.
    \nAnswer:
    \n\"Plus<\/p>\n

    \"Plus
    \n= (a + b + c)2<\/sup> \u00d7 2ab [(b + c) (c + a) – ab]
    \n= (a + b + c)2<\/sup> \u00d7 2ab [bc + ab + c2<\/sup> + ac – ab)
    \n= (a + b + c)2<\/sup> \u00d7 2abc [a + b + c]
    \n= 2abc (a + b + c)3<\/sup>.<\/p>\n

    Question 4.
    \n(i) Let the value of a determinant is \u2206. Then the value of a determinant obtained by interchanging two rows is
    \n(a) \u2206
    \n(b) -\u2206
    \n(c) 0
    \n(d) 1 (1)
    \n(ii) Show that \\(\\left|\\begin{array}{ccc}{a+b} & {b+c} & {c+a} \\\\{b+c} & {c+a} & {a+b} \\\\{c+a} & {a+b} & {b+c}\\end{array}\\right|=2\\left|\\begin{array}{lll}{a} & {b} & {c} \\\\{b} & {c} & {a} \\\\{c} & {a} & {b}\\end{array}\\right|\\) (3)
    \nAnswer:
    \n(i) (b) -\u2206<\/p>\n

    (ii) Operating C1<\/sub> \u2192 C1<\/sub> + C2<\/sub> + C3<\/sub>, we have
    \n\"Plus<\/p>\n

    \"Plus<\/p>\n

    Question 5.
    \nTest the consistency 3x – y – 2z = 2, 2y – z = -1, 3x – 5y = 3.
    \nAnswer:
    \nThe given system of equations can be put in the matrix form, AX = B, where
    \n\"Plus
    \n|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = 0
    \nC11<\/sub> = -5, C12<\/sub> = -3, C21<\/sub> = -6, C22<\/sub> = 10, C23<\/sub> = 6, C31<\/sub> = 12, C32<\/sub> = 5, C33<\/sub> = 6
    \n\"Plus<\/p>\n

    \"Plus
    \nTherefore the system is inconsistent and has no solutions.<\/p>\n

    Question 6.
    \nConsider the system of equations 2x – 3y = 7 and 3x + 4y = 5<\/p>\n

      \n
    1. Express the system in AX = B form. (1)<\/li>\n
    2. Find adj A (2)<\/li>\n
    3. Solve the system of equations. (1)<\/li>\n<\/ol>\n

      Answer:
      \n1. |A| = \\(\\left|\\begin{array}{cc}{2} & {-3} \\\\{3} & {4}\\end{array}\\right|\\) = 8 + 9 = 17.<\/p>\n

      2. c11<\/sub> = 4, c12<\/sub> = -3, c21<\/sub> = 3, c22<\/sub> = 2,
      \n\"Plus<\/p>\n

      3. The given equations can be expressed in the form AX = B,
      \n\"Plus<\/p>\n

      Question 7.
      \n(i) If A and B are matrices of order 3 such that|A| = -1; |B| = 3, then |3AB| is
      \n(a) -9
      \n(b) -27
      \n(c) -81
      \n(d) 9 (1)
      \n(ii) If A = \\(\\left[\\begin{array}{cc}{1} & {\\tan x} \\\\{-\\tan x} & {1}\\end{array}\\right]\\), Show that AT<\/sup> A-1<\/sup> = \\(\\left[\\begin{array}{cc}{\\cos 2 x} & {-\\sin 2 x} \\\\{\\sin 2 x} & {\\cos 2 x}\\end{array}\\right]\\) (3)
      \nAnswer:
      \n(i) (c) -81 (since |3AB| = 27|A||B|).<\/p>\n

      (ii) |A| = \\(\\left[\\begin{array}{cc}{1} & {\\tan x} \\\\{-\\tan x} & {1}\\end{array}\\right]\\) = sec2<\/sup>x \u2260 0, therefore A is invertible.
      \n\"Plus<\/p>\n

      Question 8.
      \nConsider the determinant \u2206 = \\(\\left|\\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\\\{y} & {y^{2}} & {1+y^{3}} \\\\{z} & {z^{2}} & {1+z^{3}}\\end{array}\\right|\\), Where x, y, z, are different.
      \n(i) Express the above determinant as sum of two determinants. (1)
      \n(ii) Show that if \u2206 = 0, then 1 + xyz = 0. (3)
      \nAnswer:
      \n(i) Given,
      \n\u2206 = \\(\\left|\\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\\\{y} & {y^{2}} & {1+y^{3}} \\\\{z} & {z^{2}} & {1+z^{3}}\\end{array}\\right|=\\left|\\begin{array}{ccc}{x} & {x^{2}} & {1} \\\\{y} & {y^{2}} & {1} \\\\{z} & {z^{2}} & {1}\\end{array}\\right|+\\left|\\begin{array}{ccc}{x} & {x^{2}} & {x^{3}} \\\\{y} & {y^{2}} & {y^{3}} \\\\{z} & {z^{2}} & {z^{3}}\\end{array}\\right|\\)<\/p>\n

      \"Plus<\/p>\n

      \"Plus
      \nGiven, \u2206 = 0 \u21d2 (1 + xyz)(y – x)(z – x)(z – y) = 0 \u21d2 1 + xyz = 0
      \n\u2235 x \u2260 y \u2260 z.<\/p>\n

      Question 9.
      \n(i) The value of the determinant \\(\\left|\\begin{array}{cc}{\\sin 10} & {-\\cos 10} \\\\{\\sin 80} & {\\cos 80}\\end{array}\\right|\\) is
      \n(a) – 1
      \n(b) 1
      \n(c) 0
      \n(d) – 2 (1)
      \n(ii) Using properties of determinants, show that (3)
      \n\\(\\left|\\begin{array}{lll}{a} & {a^{2}} & {b+c} \\\\{b} & {b^{2}} & {c+a} \\\\{c} & {c^{2}} & {a+b}\\end{array}\\right| = (b – c) (c – a) (a – b) (a + b + c)\\)
      \nAnswer:
      \n(i) (b) Since,
      \nsin 10 cos 80 + cos 10 sin 80 = sin (10 + 80) =sin 90 = 1.<\/p>\n

      (ii) Let C3<\/sub> \u2192 C3<\/sub> + C1<\/sub>
      \n\"Plus
      \n= (a + b + c)(b – a)(c – a)(c + a – b – a)
      \n= (a + b + c)(b – a)(c – a)(c – b)
      \n= (b – c)(c – a)(a – b)(a + b + c).<\/p>\n

      Question 10.
      \n(i) Choose the correct answer from the bracket. Consider a square matrix of order 3. Let C11<\/sub>, C12<\/sub>, C13<\/sub> are cofactors of the elements a11<\/sub>, a12<\/sub>, a13<\/sub> respectively, then a11<\/sub>C11<\/sub> + a12<\/sub>C12<\/sub> + a13<\/sub>C13<\/sub> is (1)
      \n(a) 0
      \n(b) |A|
      \n(c) 1
      \n(d) none of these.
      \n(ii) Verify A(adjA) = (adjA)A = |A|I for the matrix A = \\(\\left[\\begin{array}{ll}{5} & {-2} \\\\{3} & {-2}\\end{array}\\right]\\) that, where I = \\(\\left[\\begin{array}{ll}{1} & {0} \\\\{0} & {1}\\end{array}\\right]\\) (3)
      \nAnswer:
      \n(i) (b) |A|<\/p>\n

      (ii) |A| = \\(\\left|\\begin{array}{cc}{5} & {-2} \\\\{3} & {-2}\\end{array}\\right|\\) = – 4
      \nC11<\/sub> = – 2, C12<\/sub> = – 3, C21<\/sub> = 2, C22<\/sub> = 5
      \n\"Plus
      \nHence A(adjA) = (adjA)A = |A|I.<\/p>\n

      Question 11.
      \nConsider the following system of equations x + 2y = 4,2x + 5y = 9<\/p>\n

        \n
      1. If A = \\(\\left[\\begin{array}{ll}{1} & {2} \\\\{2} & {5}\\end{array}\\right]\\), find |A| (1)<\/li>\n
      2. Express the above system of equations in the form AX = B (1)<\/li>\n
      3. Find adj A, A-1<\/sup> (1)<\/li>\n
      4. Solve the system of equations. (1)<\/li>\n<\/ol>\n

        Answer:
        \n1. |A| = \\(\\left[\\begin{array}{ll}{1} & {2} \\\\{2} & {5}\\end{array}\\right]\\) = 5 – 4 = 1<\/p>\n

        2. The given system of equation can be expressed in the form AX = B.
        \n\"Plus<\/p>\n

        3. Cofactor matrix of A = \\(\\left[\\begin{array}{cc}{5} & {-2} \\\\{-2} & {1}\\end{array}\\right]\\)
        \n\"Plus<\/p>\n

        4. We have,
        \n\"Plus
        \nx = 2, y = 1.<\/p>\n

        Question 12.
        \nConsider the point X(-2, -3), B(3, 2), C(-1, -8)<\/p>\n

          \n
        1. Find the area of \u2206ABC (2)<\/li>\n
        2. Find third vertex of any other triangle with same area and base AB. (2)<\/li>\n<\/ol>\n

          Answer:
          \n1. \\(\\frac{1}{2}\\left|\\begin{array}{ccc}{-2} & {-3} & {1} \\\\{3} & {2} & {1} \\\\{-1} & {-8} & {1}\\end{array}\\right|\\)
          \n\\(\\frac{1}{2}\\) (- 2(2 + 8) + 3(3 + 1) + 1(- 24 + 2)) = – 15
          \nArea of \u2206 ABC = 15.<\/p>\n

          2. The base AB is fixed and the third point is variable. Therefore we can choose any x coordinate and find y coordinate or vice versa.
          \n\"Plus
          \n\u21d2 – 2(2 – y) + 3(3 – 1) + 1(3y – 2) = 30
          \n\u21d2 – 4 + 2y + 6 + 3y – 2 = 30
          \n\u21d2 5y = 30 \u21d2 y – 6
          \nTherefore point is(1, 6).<\/p>\n

          Question 13.
          \nFind the inverse of the following
          \n\"Plus
          \nAnswer:
          \n(i) Let |A| = \\(\\left|\\begin{array}{lll}{1} & {2} & {3} \\\\{0} & {2} & {4} \\\\{0} & {0} & {5}\\end{array}\\right|\\) = 10
          \nC11<\/sub> = 10, C12<\/sub> = 0, C13<\/sub> = 0, C21<\/sub> = – 10, C22<\/sub> = 5, C23<\/sub> = 0, C31<\/sub> = – 2, C32<\/sub> = – 4, C33<\/sub> = 2
          \n\"Plus<\/p>\n

          (ii) Let |A| = \\(\\left|\\begin{array}{ccc}{1} & {0} & {0} \\\\{3} & {3} & {0} \\\\{5} & {2} & {-1}\\end{array}\\right|\\) = -3
          \nC11<\/sub> = -3, C12<\/sub> = 3, C13<\/sub> = -9, C21<\/sub> = 0, C22<\/sub> = -1, C23<\/sub> = -2, C31<\/sub> = 0, C32<\/sub> = 0, C33<\/sub> = 3
          \n\"Plus<\/p>\n

          (iii) Let |A| = \\(\\left|\\begin{array}{ccc}{2} & {1} & {3} \\\\{4} & {-1} & {0} \\\\{-7} & {2} & {1}\\end{array}\\right|\\)
          \n= 2(-1 – 0) -1(4 – 0) + 3(8 – 7) = -3
          \nC11<\/sub> = -1, C12<\/sub> = -4, C13<\/sub> = 1, C21<\/sub> = 5, C22<\/sub> = 23, C23<\/sub> = -11, C31<\/sub> = 3, C32<\/sub> = 12, C33<\/sub> = -6<\/p>\n

          \"Plus<\/p>\n

          (iv) Let |A| = \\(\\left|\\begin{array}{ccc}{1} & {-1} & {2} \\\\{0} & {2} & {-3} \\\\{3} & {-2} & {4}\\end{array}\\right|\\)
          \n= 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = -1
          \nC11<\/sub> = 2, C12<\/sub> = -9, C13<\/sub> = -6, C21<\/sub> = 0, C22<\/sub> = -2, C23<\/sub> = -1, C31<\/sub> = 3, C32<\/sub> = 3, C33<\/sub> = 2
          \n\"Plus<\/p>\n

          Question 14.
          \nConsider the system of equations 5x + 2y = 4, 7x + 3y = 5. If A = \\(\\left[\\begin{array}{ll}{5} & {2} \\\\{7} & {3}\\end{array}\\right]\\), X = \\(\\left[\\begin{array}{l}{\\mathrm{r}} \\\\{y}\\end{array}\\right]\\) and B = \\(\\left[\\begin{array}{l}{4} \\\\{5}\\end{array}\\right]\\)<\/p>\n

            \n
          1. Find |A| (1)<\/li>\n
          2. Find A-1<\/sup> (2)<\/li>\n
          3. Solve the above system of equations. (1)<\/li>\n<\/ol>\n

            Answer:
            \n1. |A| = \\(\\left|\\begin{array}{ll}{5} & {2} \\\\{7} & {3}\\end{array}\\right|\\) = 15 – 14 = 1.<\/p>\n

            2. Given, A = \\(\\left[\\begin{array}{ll}{5} & {2} \\\\{7} & {3}\\end{array}\\right]\\)
            \n\"Plus<\/p>\n

            3. X = A-1<\/sup>B
            \n\"Plus
            \n\u21d2 x = 2, y = -3.<\/p>\n

            Plus Two Maths Determinants Six Mark Questions and Answers<\/h3>\n

            Question 1.
            \n(i) Let A be a square matrix of order \u2018n\u2019 then |KA| = …….. (1)
            \n(ii) Find x if \\(\\left|\\begin{array}{cc}{x} & {2} \\\\{18} & {x}\\end{array}\\right|=\\left|\\begin{array}{cc}{6} & {2} \\\\{18} & {6}\\end{array}\\right|\\) (2)
            \n(iii) Choose the correct answer from the bracket. The value of the determinant \\(\\left|\\begin{array}{ccc}{0} & {p-q} & {p-r} \\\\{q-p} & {0} & {q-r} \\\\{r-p} & {r-q} & {0}
            \n\\end{array}\\right|\\) is ….. (1)
            \n(iv) Consider \\(\\left|\\begin{array}{ccc}{a} & {a+b} & {a+b+c} \\\\{2 a} & {3 a+2 b} & {4 a+3 b+2 c} \\\\{3 a} & {6 a+3 b} & {10 a+6 b+3 c}\\end{array}\\right|\\) (2)
            \nAnswer:
            \n(i) If A be a square matrix of order n, then |KA| = Kn<\/sup>|A|<\/p>\n

            (ii) \\(\\left|\\begin{array}{cc}{x} & {2} \\\\{18} & {x}\\end{array}\\right|=\\left|\\begin{array}{cc}{6} & {2} \\\\{18} & {6}\\end{array}\\right|\\) \u21d2 x2<\/sup> – 36 = 0
            \n\u21d2 x2<\/sup> = 36 \u21d2 x = \u00b16.<\/p>\n

            (iii) (c) 0 (since the given determinant is the determinant of a third order skew symmetric matrix)<\/p>\n

            \"Plus
            \n= a [7a2<\/sup> + 3ab – 6a2<\/sup> – 3ab] = a(a2<\/sup>) = a3<\/sup><\/p>\n

            Question 2.
            \n(i) Let \\(\\left|\\begin{array}{lll}{1} & {3} & {2} \\\\{2} & {0} & {1} \\\\{3} & {4} & {3}
            \n\\end{array}\\right|\\) = 3, then what is the value of \\(\\left|\\begin{array}{lll}{1} & {3} & {2} \\\\{4} & {0} & {2} \\\\{3} & {4} & {3}\\end{array}\\right|\\) = ? and\\(\\left|\\begin{array}{lll}{6} & {7} & {6} \\\\{2} & {0} & {1} \\\\{3} & {4} & {3}\\end{array}\\right| \\) = ? (2)
            \n(Hint: Use the properties of determinants)
            \n(ii) Using properties of determinants show that (4)
            \n\\(\\left|\\begin{array}{ccc}{1+a} & {1} & {1} \\\\{1} & {1+b} & {1} \\\\{1} & {1} & {1+c}\\end{array}\\right|=a b c\\left(1+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)\\)
            \nAnswer:
            \n\"Plus<\/p>\n

            (ii) Taking \u2018a\u2019 from R1<\/sub>, \u2018b\u2018 from R2<\/sub>,\u2019C\u2019 from R3<\/sub>
            \n\"Plus<\/p>\n

            \"Plus<\/p>\n

            Question 3.
            \nIf A = \\(\\left[\\begin{array}{ccc}{2} & {-3} & {5} \\\\{3} & {2} & {-4} \\\\{1} & {1} & {-2}\\end{array}\\right]\\)<\/p>\n

              \n
            1. Find |A| (1)<\/li>\n
            2. Find adj.A. (2)<\/li>\n
            3. Solve 2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3 (3)<\/li>\n<\/ol>\n

              Answer:
              \n1. A = \\(\\left[\\begin{array}{ccc}{2} & {-3} & {5} \\\\{3} & {2} & {-4} \\\\{1} & {1} & {-2}\\end{array}\\right]\\)
              \n|A| = 2 \u00d7 0 + 3x – 2 + 5 = -1.<\/p>\n

              2. Co.factor A
              \n\"Plus<\/p>\n

              3. Given
              \n\"Plus
              \ni.e; AX = B \u21d2 X = A-1<\/sup> B
              \n\"Plus<\/p>\n

              Question 4.
              \nLet A = \\(\\left[\\begin{array}{ccc}{1} & {-1} & {1} \\\\{2} & {1} & {-3} \\\\{1} & {1} & {1}\\end{array}\\right]\\)<\/p>\n

                \n
              1. Is A singular? (1)<\/li>\n
              2. Find adj A. (2)<\/li>\n
              3. Obtain A-1<\/sup> (1)<\/li>\n
              4. Using A-1<\/sup> solve the system of equations x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2 (2)<\/li>\n<\/ol>\n

                Answer:
                \n1. A = \\(\\left[\\begin{array}{ccc}{1} & {-1} & {1} \\\\{2} & {1} & {-3} \\\\{1} & {1} & {1}\\end{array}\\right]\\)
                \n\u21d2 |A| = 4 + 5 + 1 = 10 \u2260 0
                \nA is non singular matrix.<\/p>\n

                2. Cofactor A
                \n\"Plus<\/p>\n

                3. A-1<\/sup> = \\(\\frac{1}{10}\\) \\(\\left[\\begin{array}{ccc}{4} & {2} & {2} \\\\{-5} & {0} & {5} \\\\{1} & {-2} {3}\\end{array}\\right]\\)<\/p>\n

                4. Given, AX = B \u21d2 X = A-1<\/sup> B
                \n\"Plus
                \n\u21d2 x = 2, y = -1, z = 1.<\/p>\n

                Question 5.
                \nSolve the following system of linear equations.<\/p>\n

                  \n
                1. x + y + z = 3, y – z = 0, 2x – y = 1 (6)<\/li>\n
                2. 5x – 6y + 4z = 15 , 7x + 4y – 3z = 19, 2x + y + 6z = 46 (6)<\/li>\n
                3. x + 2y + 5z = 10, x – y – z = -2, 2x + 3_y-2 = -11 (6)<\/li>\n<\/ol>\n

                  Answer:
                  \n1. Let AX = B
                  \nWhere A = \\(\\left[\\begin{array}{ccc}{1} & {1} & {1} \\\\{0} & {1} & {-1} \\\\{2} & {-1} & {0}\\end{array}\\right], X=\\left[\\begin{array}{c}{x} \\\\{y} \\\\{z}\\end{array}\\right], B=\\left[\\begin{array}{l}{3} \\\\{0} \\\\{1}\\end{array}\\right]\\)
                  \n|A| = 1(0 – 1) – 1(0 + 2) + 1(0 – 2) = -5
                  \nC11<\/sub> = -1, C12<\/sub> = -2, C13<\/sub> = -2, C21<\/sub> = -1, C22<\/sub> = 3, C23<\/sub> = 3, C31<\/sub> = -2, C32<\/sub> = 1, C33<\/sub> = 1
                  \n\"Plus<\/p>\n

                  2. Let AX = B,
                  \nWhere A = \\(\\left[\\begin{array}{ccc}{5} & {-6} & {4} \\\\{7} & {4} & {-3} \\\\{2} & {1} & {6}\\end{array}\\right], X=\\left[\\begin{array}{c}{x} \\\\{y} \\\\{z}\\end{array}\\right],B=\\left[\\begin{array}{c}{15} \\\\{19} \\\\{46}\\end{array}\\right]\\)
                  \n|A| = 5(24 + 3) + 6(42 + 6) + 4(7 – 8) = 419
                  \nC11<\/sub> = 27, C12<\/sub> = -48, C13<\/sub> = -1, C21<\/sub> = -1, C22<\/sub> = 22, C23<\/sub> = -17, C31<\/sub> = 2, C32<\/sub> = 43, C33<\/sub> = 62
                  \n\"Plus<\/p>\n

                  \"Plus<\/p>\n

                  3. Let AX = B
                  \n\\(\\text { Where } A=\\left[\\begin{array}{ccc}{1} & {2} & {5} \\\\{1} & {-1} & {-1} \\\\{2} & {3} & {-1}
                  \n\\end{array}\\right], X=\\left[\\begin{array}{c}{x} \\\\{y} \\\\{z}\\end{array}\\right], B=\\left[\\begin{array}{c}{10} \\\\{-2} \\\\{-11}\\end{array}\\right]\\)
                  \n|A| = 1(4) – 2(1) + 5(5) = 27
                  \nC11<\/sub> = 4, C12<\/sub> = -1, C13<\/sub> = 5, C21<\/sub> = 17, C22<\/sub> = -11, C23<\/sub> = 1, C31<\/sub> = 3, C32<\/sub> = 6, C33<\/sub> = -3
                  \n\"Plus
                  \n\u21d2 x = -1, y = -2, z = 3.<\/p>\n

                  Question 6.
                  \nIf f(x) = \\(\\left[\\begin{array}{ccc}{\\cos x} & {-\\sin x} & {0} \\\\{\\sin x} & {\\cos x} & {0} \\\\{0} & {0} & {1}\\end{array}\\right]\\)
                  \n(i) Find f(-x) (2)
                  \n(ii) Find (f(x)]-1<\/sup> (2)
                  \n(iii) Is |f(x)]-1<\/sup> = f(-x)? (2)
                  \nAnswer:
                  \n\"Plus<\/p>\n

                  (ii) |f(x)| = \\(\\left[\\begin{array}{ccc}{\\cos x} & {-\\sin x} & {0} \\\\{\\sin x} & {\\cos x} & {0} \\\\{0} & {0} & {1}\\end{array}\\right]\\) = cos x (cos x) + sin x (sin x) = 1 \u2260 0
                  \nTherefore , [f(x)]-1<\/sup> exists.
                  \nThe cofactors are as follows.
                  \nC11<\/sub> = cos x, C12<\/sub> = -sin x, C13<\/sub> = 0, C21<\/sub> = sin x, C22<\/sub> = cos x, C23<\/sub> = 0, C31<\/sub> = 0, C32<\/sub> = 0, C33<\/sub> = 1
                  \n\"Plus
                  \nSince, |f(x)|= 1<\/p>\n

                  (iii) Yes. From (1) and (2) we have,
                  \n[f(x)]-1<\/sup> =f(-x).<\/p>\n

                  Question 7.
                  \n(i) Choose the correct answer from the bracket. If A = \\(\\left[\\begin{array}{cc}{2} & {3} \\\\{1} & {-2}\\end{array}\\right]\\) and A-1<\/sup> = kA, then the value of \u2018k\u2019 is
                  \n(a) 7
                  \n(b) -7
                  \n(c) \\(\\frac{1}{7}\\)
                  \n(d)\\(-\\frac{1}{7}\\) (1)
                  \n(ii) If A = \\(\\left[\\begin{array}{ccc}{1} & {-1} & {1} \\\\{2} & {-1} & {0} \\\\{1} & {0} & {0}
                  \n\\end{array}\\right]\\),
                  \n(a) A2<\/sup> (2)
                  \n(b) Show that A2<\/sup> = A-1<\/sup> (3)
                  \nAnswer:
                  \n\"Plus<\/p>\n

                  \"Plus
                  \nC11<\/sub> = 0, C12<\/sub> = 0, C13<\/sub> = 1, C21<\/sub> = 0, C22<\/sub> = -1, C23<\/sub> = -1, C31<\/sub> = 1, C32<\/sub> = 2, C33<\/sub> = 1
                  \n\"Plus<\/p>\n

                  Question 8.
                  \n\u2018Arjun\u2019 purchased 3 pens, 2 purses, and 1 instrument box and pays Rs. 410. From the same Shop \u2018Deeraj\u2019 purchases 2 pens, 1 purse, and 2 instrument boxes and pays Rs.290, while \u2018Sindhu\u2019 purchases 2pens, 2 purses, 2 instrument boxes and pays Rs. 440.<\/p>\n

                    \n
                  1. Translate the equation into system of linear equations. (2)<\/li>\n
                  2. The cost of one pen, one purse and one instrument box using matrix method. (4)<\/li>\n<\/ol>\n

                    Answer:
                    \n1. Let The price of one pen is Rs.x, one purse is Rs.y and one instrument box be Rs.z
                    \n3x + 2y + z = 410; 2x + y + 2z =290; 2x + 2y + 2z = 440(1) 2 mts.<\/p>\n

                    2. The system can be represented by the matrix equation AX = B
                    \n\"Plus<\/p>\n

                    \"Plus
                    \nC11<\/sub> = -2, C12<\/sub> = 0, C13<\/sub> = 2, C21<\/sub> = -2, C22<\/sub> = 4, C23<\/sub> = -2, C31<\/sub> = 3, C32<\/sub> = -4, C33<\/sub> = -1
                    \n\"Plus
                    \nHence the cost one pen is Rs.20, one purse is Rs. 150 and one instrument box is Rs. 50.<\/p>\n

                    Question 9.
                    \nIf A = \\(\\left[\\begin{array}{ccc}{2} & {-3} & {5} \\\\{3} & {2} & {-4} \\\\{1} & {1} & {-2}\\end{array}\\right]\\)<\/p>\n

                      \n
                    1. Find A-1<\/sup> (3)<\/li>\n
                    2. Using it solve the system of equations 2x – 3y + 5z = 16, 3x + 2y – 4z = -4, x + y – 2z = -3 (3)<\/li>\n<\/ol>\n

                      Answer:
                      \n1. A = \\(\\left[\\begin{array}{ccc}{2} & {-3} & {5} \\\\{3} & {2} & {-4} \\\\{1} & {1} & {-2}\\end{array}\\right]\\)
                      \n\u21d2 |A| = 0 + 3x – 2 + 5 = -1
                      \n\"Plus<\/p>\n

                      2. Given AX = B
                      \n\u21d2 X = A-1<\/sup>B
                      \n\"Plus
                      \n\u21d2 x = 2, y = 1, z = 3.<\/p>\n

                      Question 10.
                      \nConsider the following system of equations x + y + 3z = 5, x + 3y – 3z = 1, -2x – 4y – 4z = -10
                      \n(i) Convert the given system in the form AX = B (1)
                      \n(ii) Find A-1<\/sup> (3)
                      \n(iii) Hence solve the system of equations. (2)
                      \nAnswer:
                      \n\"Plus<\/p>\n

                      (ii) i.e; AX = B, \u21d2 X = A-1<\/sup> B \u21d2 |A| = -24 + 10 + 6 = -8
                      \n\"Plus<\/p>\n

                      (iii) X = A-1<\/sup>B
                      \n\"Plus
                      \n= \\(-\\frac{1}{8}\\) \\(\\left[\\begin{array}{l}{-8} \\\\{-8} \\\\{-8}\\end{array}\\right]\\)
                      \n\u21d2 x = 1, y = 1, z = 1.<\/p>\n

                      Question 11.
                      \nSolve the following system by equations by matrix method x + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = -11.
                      \nAnswer:
                      \nx + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = 11
                      \n\"Plus
                      \n\u21d2 x = -1, y = -2, z = 3.<\/p>\n

                      Question 12.
                      \nIf A = \\(\\left[\\begin{array}{ccc}{3} & {-2} & {3} \\\\{2} & {1} & {-1} \\\\{4} & {-3} & {2}\\end{array}\\right]\\)<\/p>\n

                        \n
                      1. Find |A| (1)<\/li>\n
                      2. Find A-1<\/sup> (3)<\/li>\n
                      3. Solve the linear equations 3x – 2y + 3z = 8; 2x + y – z = 1; 4x – 3y + 2z = 4 (2)<\/li>\n<\/ol>\n

                        Answer:
                        \n1. |A| = \\(\\left[\\begin{array}{ccc}{3} & {-2} & {3} \\\\{2} & {1} & {-1} \\\\{4} & {-3} & {2}\\end{array}\\right]\\)
                        \n= 3(2 – 3) + 2(4 + 4) + 3(- 6 – 4) = -17.<\/p>\n

                        2. |A| \u2260 0, hence its inverse exists.
                        \nA-1<\/sup> = \\(\\frac{1}{|A|}\\)adj A
                        \nC11<\/sub> = -1, C12<\/sub> = -8, C13<\/sub> = -10, C21<\/sub> = -5, C22<\/sub> = -6, C23<\/sub> = 1, C31<\/sub> = -1, C32<\/sub> = 9, C33<\/sub> = 7
                        \n\"Plus<\/p>\n

                        3. The given system of linear equations is of the form
                        \n\"Plus
                        \n\u2234 We have, x = 1, y = 2, z = 3.<\/p>\n

                        Question 13.
                        \nif \\(\\left[\\begin{array}{cc}{2} & {5} \\\\{-3} & {7}\\end{array}\\right] \\times A=\\left[\\begin{array}{cc}{17} & {-1} \\\\{47} & {-13}\\end{array}\\right]\\) then
                        \n(i) Find the 2 \u00d7 2 matrix A. (3)
                        \n(ii) Find A2<\/sup>. (1)
                        \n(iii) Show that A2<\/sup> + 5A – 6I = 0, where I is the identity matrix of order 2. (2)
                        \nAnswer:
                        \n\"Plus<\/p>\n

                        \"Plus<\/p>\n

                        \"Plus<\/p>\n

                        \"Plus<\/p>\n

                        We hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants, 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 4 Determinants are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants. Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Maths\u00a0Chapter Wise Questions Chapter Chapter 4 Chapter […]<\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\nPlus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants - A Plus Topper<\/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:\/\/www.aplustopper.com\/plus-two-maths-chapter-wise-questions-answers-chapter-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants\" \/>\n<meta property=\"og:description\" content=\"Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants are part of Plus Two Maths Chapter Wise Questions and Answers. 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