Assume for simplicity that this is regionII. Now consider B + C(ii) Consider 2B + 3A2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix. S = R. \end{array} \)For any point on the Real number line, this function is defined. Construct a 3×4 matrix A = [ai j] whose elements ai j are given by:
(i) ai j = i + j(ii) ai j = i – j(iii) ai j = 2i(iv) ai j = j(v) ai j = ½ |-3i + j|Solution:(i) Given ai j = i + jLet A = [ai j]2×3So, the elements in a 3 × 4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 + 1 = 2a12 = 1 + 2 = 3a13 = 1 + 3 = 4a14 = 1 + 4 = 5a21 = 2 + 1 = 3a22 = 2 + 2 = 4a23 = 2 + 3 = 5a24 = 2 + 4 = 6a31 = 3 + 1 = 4a32 = 3 + 2 = 5a33 = 3 + 3 = 6a34 = 3 + 4 = 7Substituting these values in matrix A we get,A =
(ii) Given ai j = i – jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1 – 1 = 0a12 = 1 – 2 = – 1a13 = 1 – 3 = – 2a14 = 1 – 4 = – 3a21 = 2 – 1 = 1a22 = 2 – 2 Going Here 0a23 = 2 – 3 = – 1a24 = 2 – 4 = – 2a31 = 3 – 1 = 2a32 = 3 – 2 = 1a33 = 3 – 3 = 0a34 = 3 – 4 = – 1Substituting these values in matrix A we get,A =
(iii) Given ai j = 2iLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 2×1 = 2a12 = 2×1 = 2a13 = 2×1 = 2a14 = 2×1 = 2a21 = 2×2 = 4a22 = 2×2 = 4a23 = 2×2 = 4a24 = 2×2 = 4a31 = 2×3 = 6a32 = 2×3 = 6a33 = 2×3 = 6a34 = 2×3 = 6Substituting these values in matrix A we get,A =
(iv) Given ai j = jLet A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 = 1a12 = 2a13 = 3a14 = 4a21 = 1a22 = 2a23 = 3a24 = 4a31 = 1a32 = 2a33 = 3a34 = 4Substituting these values in matrix A we get,A =
(vi) have a peek at these guys ai j = ½ |-3i + j|Let A = [ai j]2×3So, the elements in a 3×4 matrix area11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34A =
a11 =
a12 =
a13 =
a14 =
a21 =
a22 =
a23 =
a24 =
a31 =
a32 =
a33 =
a34 =
Substituting these values in matrix A we get,A =
Multiplying by negative sign we get,7. Get excellent practice papers and Solved examples to grasp the concept and check for speed and make you ready for big day.
Warning: Minimum Chi Square Method
dv/dxi) y = (ax2 + bx) let v = (ax2 + bx), so y = vdy/dx = 1/2 (ax 2 + bx )-1/2 . | We say that f is concave up on I if and only if for every x, y 2 I and every t 2 [0, 1] we havef (tx + (1 t)y) t f (x) + (1 t) f (y). NCERT help provides best NCERT solution, Text books, Model test papers, Last Year Question papers, Latest Syllabus in Hindi and English for Class 6, 7, 8, 9, 10, 11 and 12. Students should download BYJU’S RD Sharma Solutions for Class 12 Maths Chapter 5 due to the reasons listed below:Your Mobile number and Email id will not be published.
Like ? Then You’ll Love This Testing Equivalence Using CI
.