{"id":3886,"date":"2023-10-06T06:18:23","date_gmt":"2023-10-06T06:18:23","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3886"},"modified":"2023-10-06T06:20:05","modified_gmt":"2023-10-06T06:20:05","slug":"week-4-ss3-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss3-first-term-mathematics-notes\/","title":{"rendered":"Week 4 – SS3 First Term Mathematics Notes"},"content":{"rendered":"

WEEK 4 DATE\u2026\u2026\u2026..
\n<\/strong>Matrices and determinants: concept, the basic operations of matrices. Identity matrices and equal matrices
\n<\/strong>MATRICES
\n<\/strong>Matrix is a rectangular array of numbers or elements in a row or column.
\n\"\"\/e.g ( a b ), a <\/p>\n

\u00a0 b
\nElements arranged along the horizontal are called ROW<\/strong>. While elements arranged along the vertical is called COLUMN<\/strong>.
\nE,g
\n\"\"\/\"\"\/ 5 7 5 7 row 1
\n\"\"\/\"\"\/\"\"\/ 8 12 8 12 row 2
\nColumn 1 2
\nNOTATION<\/strong>: A matrix is denoted by capital letter and the elements by small letters with reference to the position of the element. The position is defined in terms of the number of rows and columns.The first indicating the row, the second the column, thus:
\n\"\"\/
\n\t\"\"\/ a11<\/sub> a12<\/sub> a13<\/sub>b11 <\/sub>b12<\/sub>
\n\t A = a21<\/sub> a22<\/sub> a23<\/sub>B = b21<\/sub>b22 <\/sub>
\n\t a31<\/sub> a32<\/sub> a33
\n<\/sub>Hence, a21<\/sub>indicates the element in the second row and first column.
\n\t\t<\/sub>\"\"\/EVALUATION:<\/strong> Given the matrix,C<\/strong>= 6 -5 1 -3 write out the elements occupying the following,
\nPositions.C11<\/sub>, C21,<\/sub> C32<\/sub>, C42<\/sub>, C44<\/sub> andC34<\/sub> 2 -4 8 3
\n 4 -7 -6 5
\n -2 9 7 -1
\n\"\"\/Order of a matrix<\/em><\/strong>: A matrix can be identified or described by its order. In describing a matrix, the number of rows is stated first before the number of columns.
\nE.g 6 2 8 is a 2 x 3 matrix, i.e. order 2 by 3.
\n 5 7 3
\n<\/strong>
\n\u00a0BASIC OPERATION OF MATRICES:
\n<\/strong>Addition and subtraction of matrices<\/em><\/strong>: Two or more matrices can be added or subtracted when they are of the same order e.g 2 x 2, 3 x 3 and so on. The sum or subtraction is then determined by adding or subtracting corresponding elements.
\n\"\"\/\"\"\/\"\"\/\"\"\/
\n\tIf A = a11<\/sub> a12<\/sub> B = b11<\/sub> b12<\/sub>A + B = a11<\/sub> + b11<\/sub> a12<\/sub> + b12<\/sub> A \u2013 B = a11<\/sub> \u2013b11<\/sub> a12<\/sub> \u2013b12<\/sub><\/p>\n

\u00a0 a21<\/sub> a22<\/sub> b21<\/sub> b22<\/sub> a21<\/sub>+ b21<\/sub> a22<\/sub> + b22<\/sub>a21<\/sub>– b21<\/sub> a22<\/sub> \u2013 b22<\/sub><\/p>\n

\u00a0
\n\u00a0Example: Given the matrices below, find I A + B II A \u2013 B III B \u2013 A .<\/p>\n

\u00a0\"\"\/\"\"\/ A = 7 6 5 B = 12 -6 4
\n 9 4 8 2 10 1
\n\"\"\/\"\"\/\"\"\/
\n\t I A + B = 19 0 9 II A \u2013 B = -5 12 1 III B \u2013 A = 5 -12 -1
\n 11 14 9 7 -6 7 -7 6 -7<\/p>\n

\u00a0Addition of matrices is commutative, i.e A + B = B + A but matrix subtraction is not commutative,
\nA \u2013 B \u2260 B \u2013 A <\/p>\n

\u00a0MULTIPLICATION OF MATRICES<\/strong>:<\/p>\n

    \n
  1. Scalar multiplication<\/em><\/strong>: This is the multiplication of a matrix by a single number and it is done by multiplying each element in the matrix by the scalar.\n<\/li>\n<\/ol>\n

    \"\"\/e.g If C = 3 -8 12 find I 2C II -3C
    \n 4 5 7
    \n\"\"\/
    \n\t\"\"\/ I 2C = 6 -16 24 II \u2013 3C = -9 24 -36
    \n 8 10 14 -12 -15 -21<\/p>\n

    \u00a0<\/p>\n

      \n
    1. Multiplication of two matrices<\/em><\/strong>: Two matrices can be multiplied together only when the number of columns in the first is equal to the number of rows in the second matrix.\n<\/li>\n<\/ol>\n

      \"\"\/\"\"\/\"\"\/
      \n\t\t<\/strong>If A= a11<\/sub> a12 <\/sub> B = b1<\/sub>then A. B = a11<\/sub>b1<\/sub> + a12<\/sub>b2<\/sub>
      \n\t a21<\/sub> a22<\/sub> b2<\/sub>a21<\/sub>b1<\/sub> + a22<\/sub>b2
      \n<\/sub>
      \n\u00a0 Matrix by matrix multiplication is not commutative, A.B \u2260 BA
      \n<\/strong>
      \n\u00a0\"\"\/\"\"\/Example: Given the matrices A = 2 3 B = 4 1 find AB.
      \n 5 7 8 9
      \n\"\"\/\"\"\/ 2 3 4 1
      \n AB = x
      \n5 7 8 9<\/p>\n

      \u00a0\"\"\/\"\"\/ AB = 2×4 + 3×8 2 x1 + 3×9 8+24 2 +27
      \n =
      \n 5×4 + 7×8 5×1 + 7×9 20 +56 5+63<\/p>\n

      \u00a0\"\"\/ AB = 32 29<\/p>\n

        \n
      1. 69\n<\/li>\n<\/ol>\n

        \u00a0
        \n\u00a0\"\"\/\"\"\/
        \n\tEVALUATION<\/strong>: 1. Find AC given the matrices A = -3 2 5 C = 4 7 2
        \n0 1 8 3 -5 1
        \n\"\"\/\"\"\/\"\"\/ 9 0 4
        \n2. Given that A = 12 -8 B = 1 4 and C = 9 -5
        \n 3 6 2 8 15 10
        \n Find I. A + B II. B \u2013 C III. 2A \u2013 B + 3C<\/p>\n

        \u00a0TYPES OF MATRICES<\/strong>:
        \nEqual matrices<\/em><\/strong>: Two matrices are said to be equal if corresponding elements are equal and the matrices are of the same order.
        \nSquare matrix<\/em><\/strong>: Is a matrix having the same number of rows and columns. E.g 2 x 2, 3 x 3,and so on
        \nDiagonal matrix:<\/em><\/strong> This is a square matrix with all elements zero except those on the main diagonal.
        \n\"\"\/ 2 0 0
        \n 0 5 0
        \n 0 0 8<\/p>\n

        \u00a0\"\"\/\"\"\/Identity matrix:<\/em><\/strong> It is also called unit matrix and is a diagonal matrix in which the elements on the main diagonal are equal to one (1). It is denoted by I.
        \n<\/strong>2 x 2 ,I =1 0 3 x 3, I = 1 0 0
        \n<\/strong>0 1 0 1 0
        \n<\/strong> 0 0 1
        \n<\/strong>Null matrix<\/em><\/strong>: Is a matrix whose elements are zero. It is denoted by o. <\/strong>i.e
        \n\"\"\/ 0 0 0
        \n 0 0 0
        \n 0 0 0 <\/p>\n

        \u00a0TRANSPOSE OF A MATRIX: <\/em><\/strong>The matrix obtained by interchanging the rows and columns of a matrix is called the transpose matrix. If B<\/strong> is the original matrix, its transpose is denoted by BT<\/sup>. <\/strong>
        \n\t\"\"\/
        \n\t\"\"\/ If A = 2 5 , then AT<\/sup> = 2 -8 7
        \n -8 3 5 3 4
        \n\"\"\/\"\"\/ 7 4
        \nEVALUATION<\/strong>: If A = 1 2 3 and B = 7 10 find I A.B and (A.B)T<\/sup>
        \n\t4 5 6 8 11
        \n9 12 <\/p>\n

        \u00a0
        \n\u00a0WEEKEND ASSIGNMENT\"\"\/\"\"\/
        \n\t\t<\/strong>Given that A = 4 2 3 B = 1 8 9
        \n 5 7 6 3 5 4
        \n\"\"\/\"\"\/
        \n\t1. Find A + B. A. 5 10 12 B. 3 6 12
        \n 8 12 10 2 -2 2<\/p>\n

        \u00a0\"\"\/\"\"\/2. Find A \u2013 B A. -3 -6 -6 B. 3 -6 -6
        \n -2 2 -2 2 2 2
        \n\"\"\/
        \n\t\"\"\/\"\"\/3. Find (A + 2B) T\u00a0\u00a0\u00a0\u00a0<\/sup>A. 6 18 21 B. 6 11 C. 6 18
        \n\u00a0\u00a0\u00a0\u00a0 11 17 14 18 17 21 11
        \n\u00a0\u00a0\u00a0\u00a0 21 14 17 14
        \n4. Find A2<\/sup> \u2013 4I
        \n5. Find BA.<\/p>\n

        \u00a0
        \n\u00a0
        \n\u00a0\"\"\/THEORY
        \n<\/strong>1. Given that\u00a0\u00a0\u00a0\u00a0 P = 1 5 find 2p2<\/sup> \u2013 3p + 5I
        \n\u00a0\u00a0\u00a0\u00a0 -4 2
        \n\"\"\/
        \n\t\"\"\/\"\"\/2. \u00a0\u00a0\u00a0\u00a0 1 3 2 \u00a0\u00a0\u00a0\u00a0 2 4 – 3 3 1 6
        \n If P = 8 -4 4 \u00a0\u00a0\u00a0\u00a0 Q = \u00a0\u00a0\u00a0\u00a0 3 8 4 R = 4 3 2
        \n 7 3 5 \u00a0\u00a0\u00a0\u00a0 -1 3 6 2 – 1 1<\/p>\n

        \u00a0 Find (a) 5P + 2Q (b) 4Q \u2013 2R (iii) 2P + Q + 3R (iv) PR<\/p>\n

        \u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0
        \n\u00a0Completing the square
        \n<\/strong>To make a given expression a perfect square, the quantity to be added is the square of half of the coefficient of x ( or whatever letter is involved).
        \nExamples:
        \nIn each of the following, add the term that makes the given expression into a perfect square , then write the result as the square of a bracketed expression.
        \n1. g2<\/sup> – 4 2<\/sup>\/3 g
        \n2. k2<\/sup> – 11<\/sup>\/3 k
        \n3. m2<\/sup> + 3mn.
        \nSolutions.
        \ng2<\/sup> \u2013 4 2<\/sup>\/3<\/sub> g
        \nthe coefficient of g is \u2013 42<\/sup>\/3 = –14<\/sup>\/3
        \nhalf of –14<\/sup>\/3<\/sub> = \u00bd x –14<\/sup>\/3<\/sub> =-7<\/sup>\/3<\/sub>
        \n\tSquare of half of coefficient of g = (-7<\/sup>\/3)2<\/sup> = 49 ( + 54<\/sup>\/9<\/sub>)
        \n 9
        \n:. 49 must be added to the given expression to make it a perfect square
        \n 9
        \n:. g2<\/sup> -4 2<\/sup>\/3<\/sub> + 49<\/sup>\/9<\/sub> = ( n \u2013 7<\/sup>\/3<\/sub>)
        \nk2<\/sup> – 1 1<\/sup>\/3<\/sub> k. the coefficient of k is \u2026\u2026..11<\/sup>\/3 – –4<\/sup>\/3<\/sub>
        \n\tHalf of –4<\/sup>\/3<\/sub> = \u00bd x –4<\/sup>\/3<\/sub> = –2<\/sup>\/3<\/sub>
        \n\tSquare of half of coefficient of k = ( –2<\/sup>\/3<\/sub>)2<\/sup> = + 4<\/sup>\/9<\/sub>.
        \n:. 4<\/sup>\/9 must be added to the given expression to make it a perfect square
        \n:. K2<\/sup> – 1 1<\/sup>\/3<\/sub> k + 4<\/sup>\/9<\/sub> = ( k \u2013 2<\/sup>\/3<\/sub>)2<\/sup><\/p>\n

        \u00a0<\/p>\n

          \n
        1. m c xx<\/sup> + 3mn\n<\/li>\n<\/ol>\n

          the coefficient of m is 3n .
          \nhalf of + 3n =1<\/sup>\/2<\/sub> x + 3n = + 3n
          \n\t 2.
          \nEVALUATION.
          \nIn each of the following add the term that makes the given expression into a perfect square . Write the result as the square of a bracketed expression
          \n1b2<\/sup> – 4<\/sup>\/5b
          \n2.u2<\/sup> – 1 3<\/sup>\/5 u
          \nWEEKEND ASSIGNMENT
          \n1.Find the value of k such that x2 + 5x +k is a perfect square.
          \n (a) 2 \u00bd \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 4 \u00bc ( c) 6 \u00bc ( d) 25 (e) 100.
          \n2. What must be addedto n2<\/sup> + 1 1\/s n to make it a perfect square?
          \n (a) 3<\/sup>\/5<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 1 1<\/sup>\/5<\/sub> ( c) 16<\/sup>\/ 9<\/sub>\u00a0\u00a0\u00a0\u00a0 ( d) 9<\/sup>\/16<\/sub> ( e ) 1 \u00bc
          \n3. Solve the equation
          \n ( x + 1 \u00bc )2<\/sup> = 1 9<\/sup>\/16<\/sub>
          \n\t( a) 2 \u00bd, 0 ) (b) (0,2) ( c ) ( 0, – ) ( d ) ( 0, -2 \u00bd ) (e) ( 3, 1 \u00bd ).
          \n4.Solve the equation
          \n ( x +1<\/sup>\/3<\/sub> )2<\/sup> =4<\/sup>\/9<\/sub>
          \n\t( – 1<\/sup>\/3<\/sub>, 1) (b) (1, 1<\/sup>\/3<\/sub> ) ( c) (2, 2<\/sup>\/3<\/sub> ) (d ) (-2<\/sup>\/3<\/sub>, 3) (e )(1<\/sup>\/3<\/sub>, -1).
          \n5. What must be added to v2 \u2013 3\/4 v to make it a perfect square?
          \n\u00a0\u00a0\u00a0\u00a0(a) 9<\/sup>\/64<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 3<\/sup>\/8<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0( c) –3<\/sup>\/8<\/sub>\u00a0\u00a0\u00a0\u00a0(d) 7 1<\/sup>\/9<\/sub> (e) 2<\/sup>\/9<\/sub>
          \n\tTheory
          \nIn each of the following, add the term that makes the given expression into a perfect square. Then write the result as the square of a bracketed expression:<\/p>\n

          \u00a0ii. u2<\/sup> \u2013 1 3<\/sup>\/5u
          \n2. a2<\/sup> \u2013 6ad<\/p>\n

          \u00a0
          \n\u00a0TOPIC: SOLUTION OF QUADRATIC EQUATION & SYMMETRIC PROPERTIES OF THE ROOT OF QUATION EQUATION
          \n<\/strong>CONTENT
          \n<\/strong><\/p>\n

            \n
          • \n
            Method of Factorization
            \n\t\t\t\t<\/strong><\/div>\n
              \n
            • \n
              Completing the square method.\n<\/div>\n<\/li>\n
            • \n
              Quadratic formula\n<\/div>\n<\/li>\n
            • \n
              Sum & Product of Roots of a Quadratic Equation\n<\/div>\n<\/li>\n
            • \n
              Symmetric Properties of Roots\n<\/div>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n

              Method of Factorization
              \n<\/strong>A quadratic equation is an expression of the form ax2<\/sup> + bx + c = 0 in which a, b & c are numerals; and also the highest power of x is 2 & that the power of x will neither be fractions nor negatives. Quadratic equations can be solved using the method of factorization, completing the square, quadratic formula& graphical method
              \nSteps in solving quadratic equation: (1)examine the middle term whose power of x is 1. (2) Find the product of the first & last term. (3) Find two terms whose sum is equal to the middle term & product is equal to the value of the product of the first & last term (4) Replace the middle term by two the two terms in step 3. (5) Factorize the first two & last two terms (6) equate the linear factors to zero to find the value of x.
              \nExample \u2013 Solve by factorization: X2<\/sup> + 7X + 10 = 0
              \nSolution
              \nX2<\/sup> + 7X + 10 = 0
              \nX2<\/sup> +2X + 5X + 10 = 0
              \nX(X + 2) + 5(X + 2) = 0
              \n(X+2) (X + 5) = 0
              \nX + 2 = 0
              \nX = -2 OR
              \nX + 5 = 0
              \nX = -5
              \nHence X = -2 or -5<\/p>\n","protected":false},"excerpt":{"rendered":"

              WEEK 4 DATE\u2026\u2026\u2026.. Matrices and determinants: concept, the basic operations of matrices. Identity matrices and…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,302],"tags":[],"class_list":["post-3886","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3886","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3886"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3886\/revisions"}],"predecessor-version":[{"id":3887,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3886\/revisions\/3887"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3886"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3886"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3886"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}