Mathematics

Linear Algebra

449 Questions

Linear algebra involves the study of matrices, vectors, and linear transformations. Common topics include finding the rank of a matrix, calculating eigenvalues and eigenvectors, and performing LU decomposition. These advanced mathematical concepts are frequently tested in engineering, statistics, and civil service examinations.

Matrix rank calculationLU decompositionEigenvalues and eigenvectorsMatrix invertibilityDeterminant properties

Linear Algebra Questions

Multiple choice

The eigen values of matrix $\begin{bmatrix} \ 9 & 5 \ \ 5 & 8 \ \end{bmatrix}$ are
    • 2.42 and 6.86

  2. 3.48 and 13.53

  3. 4.70 and 6.86

  4. 6.86 and 9.50
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The eigenvalues are found by solving det(A - lambda*I) = 0. For the matrix [[9, 5], [5, 8]], the characteristic equation is (9-lambda)(8-lambda) - 25 = 0, which simplifies to lambda^2 - 17*lambda + 47 = 0. The roots are approximately 3.48 and 13.52.

Multiple choice

[A] is a square matrix which is neither symmetric nor skew-symmetric and [A]T is its transpose. The sum and difference of these matrices are defined as [S] = [A] + [A]T and [D] = [A] – [A]T respectively. Which of the following statements is TRUE?

  1. Both [S] and [D] are symmetric.

  2. Both [S] and [D] are skew-symmetric.

  3. [S] is skew-symmetric and [D] is symmetric.

  4. [S] is symmetric and [D] is skew-symmetric.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For any square matrix A, S = A + A^T is always symmetric because S^T = (A + A^T)^T = A^T + A = S. Conversely, D = A - A^T is always skew-symmetric because D^T = (A - A^T)^T = A^T - A = -(A - A^T) = -D.

Multiple choice

If any two columns of a determinant P = $\begin{vmatrix} \ 4 & 7 & 8 \ \ 3 & 1 & 5 \ \ 9 & 6 & 2 \ \end{vmatrix}$are interchanged, which of the following statements regarding the value of the determinant is correct?

  1. The absolute value remains unchanged, but the sign will change.

  2. Both the absolute value and the sign will change.

  3. The absolute value will change, but the sign will not change.

  4. Both the absolute value and the sign will remain unchanged.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

It is the fundamental property of determinant that when any two columns or rows of a determinant are interchanged, then the absolute value remains unchanged, but the sign will change.

Multiple choice

The product of matrices (PQ)–1 and P is

  1. P–1

  2. Q–1

  3. P–1 Q–1

  4. PQP–1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The expression (PQ)^-1 * P is equal to (Q^-1 * P^-1) * P. Since P^-1 * P = I (the identity matrix), the expression simplifies to Q^-1 * I = Q^-1.

Multiple choice

Real matrices [A]3×1, [b]3×3, [C]3×5, [D]5×3, [E]5×5, [F]5×1 are given. Matrices [B] and [E] symmetric. Following statements are made with respect to these matrices. (I) Matrix product [F]T [C]T [B] [C] [F] is a scalar. (II) Matrix product [D]T [F] [D] is always symmetric. With reference to above statements, which of the following applies?

  1. Statement I is true but II is false.

  2. Statement I is false but II is true.

  3. Both the statements are true.

  4. Both the statements are false.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Statement I: F(T) is 1x5, C(T) is 5x3, B is 3x3, C is 3x5, F is 5x1. The product (1x5)(5x3)(3x3)(3x5)(5x1) results in a 1x1 matrix (scalar). Statement II: D(T) is 3x5, F is 5x1, D is 5x3. The product (3x5)(5x1)(5x3) is a 3x3 matrix, but it is not necessarily symmetric because F is not a symmetric matrix.

Multiple choice

A square matrix B is skew symmetric if

  1. BT = - B

  2. BT = B

  3. B-1 = B

  4. B-1 = BT
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A matrix is skew-symmetric by definition if its transpose is equal to the negative of the matrix itself, expressed as B^T = -B.

Multiple choice

For the matrix A = $\begin{bmatrix} \ 5 & 3 \ \ 1 & 3 \ \end{bmatrix}$, ONE of the normalized eigen vectors is given as

  1. $\begin{pmatrix} \ \frac{1}{2} \\ \ \frac{\sqrt{3}}{2} \\ \end{pmatrix}$
  2. $\begin{pmatrix} \ \frac{1}{\sqrt{2}} \\ \ \frac{-1}{\sqrt{2}} \\ \end{pmatrix}$
  3. $\begin{pmatrix} \ \frac{3}{\sqrt{10}} \\ \ \frac{-1}{\sqrt{10}} \\ \end{pmatrix}$
  4. $\begin{pmatrix} \ \frac{1}{\sqrt{5}} \\ \ \frac{2}{\sqrt{5}} \\ \end{pmatrix}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation