3<\/sup>).<\/p>\nQuestion 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
\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
\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>\nQuestion 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>\nQuestion 5.
\nA = \\(\\left[\\begin{array}{ccc}{1} & {-3} & {1} \\\\{2} & {0} & {4} \\\\{1} & {2} & {-2}
\n\\end{array}\\right]\\)<\/p>\n
\n- Calculate |A| (1)<\/li>\n
- Find |adjA| {Hint: using the property A \u00d7 adjA = |A|I} (1)<\/li>\n
- 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<\/p>\n
3. |3A| = 27 \u00d7 |A| = 27 \u00d7 -28 = -756.<\/p>\n
Question 6.
\nUsing properties of determinants proves the following.
\n
\nAnswer:
\n<\/p>\n
<\/p>\n
\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>\nPlus 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<\/p>\nQuestion 2.
\n
\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<\/p>\n
<\/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<\/p>\n
\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>\nQuestion 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<\/p>\n<\/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
\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<\/p>\n
\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- Express the system in AX = B form. (1)<\/li>\n
- Find adj A (2)<\/li>\n
- 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<\/p>\n3. The given equations can be expressed in the form AX = B,
\n<\/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<\/p>\nQuestion 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
<\/p>\n
\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
\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>\nQuestion 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
\nHence A(adjA) = (adjA)A = |A|I.<\/p>\nQuestion 11.
\nConsider the following system of equations x + 2y = 4,2x + 5y = 9<\/p>\n
\n- If A = \\(\\left[\\begin{array}{ll}{1} & {2} \\\\{2} & {5}\\end{array}\\right]\\), find |A| (1)<\/li>\n
- Express the above system of equations in the form AX = B (1)<\/li>\n
- Find adj A, A-1<\/sup> (1)<\/li>\n
- 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<\/p>\n
3. Cofactor matrix of A = \\(\\left[\\begin{array}{cc}{5} & {-2} \\\\{-2} & {1}\\end{array}\\right]\\)
\n<\/p>\n
4. We have,
\n
\nx = 2, y = 1.<\/p>\n
Question 12.
\nConsider the point X(-2, -3), B(3, 2), C(-1, -8)<\/p>\n
\n- Find the area of \u2206ABC (2)<\/li>\n
- 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
\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
\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<\/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<\/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<\/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<\/p>\nQuestion 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- Find |A| (1)<\/li>\n
- Find A-1<\/sup> (2)<\/li>\n
- 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<\/p>\n
3. X = A-1<\/sup>B
\n
\n\u21d2 x = 2, y = -3.<\/p>\nPlus 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
\n= a [7a2<\/sup> + 3ab – 6a2<\/sup> – 3ab] = a(a2<\/sup>) = a3<\/sup><\/p>\nQuestion 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<\/p>\n
(ii) Taking \u2018a\u2019 from R1<\/sub>, \u2018b\u2018 from R2<\/sub>,\u2019C\u2019 from R3<\/sub>
\n<\/p>\n<\/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- Find |A| (1)<\/li>\n
- Find adj.A. (2)<\/li>\n
- 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<\/p>\n
3. Given
\n
\ni.e; AX = B \u21d2 X = A-1<\/sup> B
\n<\/p>\nQuestion 4.
\nLet A = \\(\\left[\\begin{array}{ccc}{1} & {-1} & {1} \\\\{2} & {1} & {-3} \\\\{1} & {1} & {1}\\end{array}\\right]\\)<\/p>\n
\n- Is A singular? (1)<\/li>\n
- Find adj A. (2)<\/li>\n
- Obtain A-1<\/sup> (1)<\/li>\n
- 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<\/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>\n4. Given, AX = B \u21d2 X = A-1<\/sup> B
\n
\n\u21d2 x = 2, y = -1, z = 1.<\/p>\nQuestion 5.
\nSolve the following system of linear equations.<\/p>\n
\n- x + y + z = 3, y – z = 0, 2x – y = 1 (6)<\/li>\n
- 5x – 6y + 4z = 15 , 7x + 4y – 3z = 19, 2x + y + 6z = 46 (6)<\/li>\n
- 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<\/p>\n2. 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<\/p>\n<\/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
\n\u21d2 x = -1, y = -2, z = 3.<\/p>\nQuestion 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<\/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
\nSince, |f(x)|= 1<\/p>\n(iii) Yes. From (1) and (2) we have,
\n[f(x)]-1<\/sup> =f(-x).<\/p>\nQuestion 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<\/p>\n
\nC11<\/sub> = 0, C12<\/sub> = 0, C13<\/sub> = 1, C21<\/sub> = 0, C22<\/sub> = -1, C23<\/sub> = -1, C31<\/sub> = 1, C