This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) focuses on “Matrix Algebra”.
1. What is a matrix?
a) Group of elements
b) Array of elements
c) Group of columns and rows
d) Array of numbers
View Answer
Explanation: A matrix is an array of elements. The matrix A is denoted as [A]. An element located in the ith row and j th column is denoted as aij. A matrix is a collection of numbers arranged into a fixed number of rows and columns.
2. Which of the following is a row vector?
a) \(\left[ \begin{array}{c c r } 0 & 1 & 2 \end{array}\right]\)
b) \(\left[ \begin{array}{c c r r} 0\\ 1 \\ 2\end{array}\right]\)
c) \(\left[ \begin{array}{c r} 0 & 1\\ 1 & 2\end{array}\right]\)
d) \(\left[ \begin{array}{c c r r} 0 & 3 & 6 \\ 1 & 8 & 4 \\ 0 & 5 & 7\end{array}\right]\)
View Answer
Explanation: A matrix of (1 x n) dimension is called row vector. A matrix of (n x1) dimension is called column vector.
For example
d=[ 1 2 3 4] is a row vector.
c = \(\left[ \begin{array}{c c r r} 0\\ 1 \\ 2\end{array}\right]\) is a column vector.
3. (A B C)T = _______
a) (C B A)T
b) BT CT AT
c) CT BT AT
d) AT BT CT
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Explanation: A matrix which is formed by turning all the rows of given matrix into columns and vice versa is called a transpose of matrix. The transpose of a product is given as the product of the transposes in the reverse order.
(A B C)T = CT BT AT.
4. The derivative of Ax with respect to variable xp is given by __________
a) \(\frac{d}{dx}\)(A x)=xp
b) \(\frac{d}{dx}\)(xp)=A x
c) ∫ A x=xp
d) ∫xp=Ax
View Answer
Explanation: Let A be an (n x n) matrix of constants and x = [x1 x2 x3…… xn]T be column vector of n variables. Then, (formula) derivative of A x with respect to variable xp is given by
\(\frac{d}{dx}\) (A x)=xp.
5. A symmetric matrix is called ____________, if all its Eigen values are strictly positive i.e., greater than zero.
a) Negative definite
b) Positive definite
c) Co- definite
d) Alternative definite
View Answer
Explanation: If all Eigen values of symmetric matrix are positive then the matrix is called as positive definite matrix. A symmetric matrix A of dimension (n x n) is positive definite if, for any non zero vector x = [x1 x2 x3…… xn]T. That is xT Ax > 0.
6. A A-1=A-1A is a condition for ________
a) Singular matrix
b) Nonsingular matrix
c) Matrix inversion
d) Ad joint of matrix
View Answer
Explanation: If det A not equal to zero, then A has an inverse, denoted by A-1. The inverse satisfies the relation
A A-1 =A-1A= I
7. A positive definite symmetric matrix A can be decomposed into form A=LLT this decomposition is called ________
a) Cholesky
b) Rayleighs
c) Galerkins
d) Potential energy
View Answer
Explanation: L is the lower triangular matrix, and its transpose LT is upper triangular matrix. This is called Cholesky decomposition. It is a decomposition of a positive definite matrix into a product of lower triangular matrix and its conjugate transpose.
8. Det(A-λI)=0 is a ________
a) Characteristic equation
b) Matrix equation
c) Inversion of matrix
d) Cholesky’s equation
View Answer
Explanation: A non zero solution will occurs when (A-ɅI) is a singular matrix or det(A-ɅI)=0 it is a characteristic equation. A characteristic equation is the equation which is solved to find the Eigen values, also called the characteristic polynomials.
9. \(\left[ \begin{array}{c c c r r r r}
2 & -1 & 6 & 3 \\ 0 & 14 & 8 & 0 \\ 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 3\end{array}\right]\) is a _____
a) Principle diagonal matrix
b) Upper triangular matrix
c) Lower triangular matrix
d) Singular matrix
View Answer
Explanation: An upper triangular or right triangular matrix is one whose elements below the principal diagonal elements are zero. The sum or product or inverse of any two upper triangular matrixes is an upper triangular matrix.
10. A=\(\left[ \begin{array}{c c r r}
3 & 2 & 1 \\ 4 & 5 & -8 \\ 10 & 0 & 5\end{array}\right]\)Then det (A) =
a) 120
b) -80
c) -175
d) 0
View Answer
Explanation: det\(\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{bmatrix}\)
= a11(a22a33-a32a23)-a12(a21a33-a31a23)+a13(a21a32-a31a22)
=3×(5×5−0×(−8))−2×(4×5−10×(−8))+1×(4×0−10×5)
=3×(25−0)−2×(20+80)+1×(0−50)
=3×25−2×100+1×(−50)
=75−200−50
=-175.
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