Linear Algebra

Detailed SolutionIt is given that X1, X2, … XM are M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X1, X2 … XM, -X1, -X2 … -XM is A. 2M B. M + 1 C. M D. dependent on the choice of X1, X2, … XM

Detailed SolutionThe rank of the matrix \[\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ – 1}&0 \\ 1&1&1 \end{array}} \right]\] is A. 0 B. 1 C. 2 D. 3

Detailed SolutionThe eigen values of a symmetric matrix are all A. complex with non-zero positive imaginary part B. complex with non-zero negative imaginary part C. real D. pure imaginary

Detailed SolutionWhich one of the following equations is a correct identity for arbitrary 3 × 3 real matrices P, Q and R? A. P(Q + R) = PQ + RP B. (P – Q)2 = P2 – 2PQ + Q2 C. det (P + Q) = det P + det Q D. (P + Q)2 = P2 + PQ + QP + Q2

Detailed SolutionLet X be a square matrix. Consider the following two statements on X. I. X is invertible. II. Determinant of X is non-zero. Which one of the following is TRUE? A. I implies II; II does not imply I B. II implies I; I does not imply II C. I and II are equivalent statements D. I does not imply II; II does not imply I

” option2=”\[ – \frac{3}{5}\]” option3=”\[\frac{3}{5}\]” option4=”\[\frac{4}{5}\]” correct=”option3″]

Detailed SolutionFor a matrix \[\left[ {\text{M}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}} \\ {\text{x}}&{\frac{3}{5}} \end{array}} \right],\] the transpose of the matrix is equal to the inverse of the matrix, [M]T = [M]-1. The value of x is given by A. \[ – \frac{4}{5}\] B. \[ – \frac{3}{5}\] C. \[\frac{3}{5}\] D. \[\frac{4}{5}\]

Detailed SolutionThe value of q for which the following set of linear equation 2x + 3y = 0; 6x + qy = 0 can have non-trivial solution is A. 2 B. 7 C. 9 D. 11

” option2=”\[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} – 3{{\text{e}}_2} + \frac{{11}}{5}{{\text{e}}_3}\]” option3=”\[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} + 3{{\text{e}}_2} + \frac{{11}}{5}{{\text{e}}_3}\]” option4=”\[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} + 3{{\text{e}}_2} – \frac{{11}}{5}{{\text{e}}_3}\]” correct=”option1″]

Detailed SolutionIf the vectors e1 = (1, 0, 2), e2 = (0, 1, 0) and e3 = (-2, 0, 1) form an orthogonal basis of the three-dimensional real space R3, then the vector u = (4, 3, -3) \[ \in \] R3 can be expressed as A. \[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} – 3{{\text{e}}_2} – \frac{{11}}{5}{{\text{e}}_3}\] B. \[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} – 3{{\text{e}}_2} + \frac{{11}}{5}{{\text{e}}_3}\] C. \[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} + 3{{\text{e}}_2} + \frac{{11}}{5}{{\text{e}}_3}\] D. \[{\text{u}} = – \frac{2}{5}{{\text{e}}_1} + 3{{\text{e}}_2} – \frac{{11}}{5}{{\text{e}}_3}\]

” option2=”\[\frac{{{\text{n}}\left( {{\text{n}} – 1} \right)}}{2}\]” option3=”\[\frac{{{\text{n}}\left( {{\text{n}} + 1} \right)\left( {2{\text{n}} + 1} \right)}}{6}\]” option4=”\[{{\text{n}}^2}\]” correct=”option4″]

Detailed SolutionA real n × n matrix A = {aij} is defined as follows: aij = i, if i = j, otherwise 0 The summation of all n eigen values of A is A. \[\frac{{{\text{n}}\left( {{\text{n}} + 1} \right)}}{2}\] B. \[\frac{{{\text{n}}\left( {{\text{n}} – 1} \right)}}{2}\] C. \[\frac{{{\text{n}}\left( {{\text{n}} + 1} \right)\left( {2{\text{n}} + 1} \right)}}{6}\] D. \[{{\text{n}}^2}\]

Detailed SolutionThe value of x3 obtained by solving following system of linear equation is x1 + 2×2 – 2×3 = 4 2×1 + x2 + x3 = -2 -x1 + x2 – x3 = 2 A. -12 B. -2 C. 0 D. 12