Mathematics
Linear Algebra
510 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
-
$\begin{bmatrix}
\ \frac{1}{4} & 0 & 0 & 0 \\
\ 0 & \frac{1}{4} & 0 & 0 \\
\ 0 & 0 & \frac{1}{2} & 0 \\
\ 0 & 0 & 0 & \frac{1}{2} \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ \frac{1}{2} & 0 & 0 & 0 \\
\ 0 & \frac{1}{2} & 0 & 0 \\
\ 0 & 0 & \frac{1}{2} & 0 \\
\ 0 & 0 & 0 & \frac{1}{2} \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 1 & 0 & 0 & 0 \\
\ 0 & 1 & 0 & 0 \\
\ 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ \frac{1}{4} & 0 & 0 & 0 \\
\ 0 & \frac{1}{4} & 0 & 0 \\
\ 0 & 0 & \frac{1}{4} & 0 \\
\ 0 & 0 & 0 & \frac{1}{4} \\
\end{bmatrix}$
C
Correct answer
Explanation
$\text{From orthogonal matrix}
\\
\begin{bmatrix}
\ AA^T \
\end{bmatrix} = I
\\
\text{Since the inverse of I is I, thus}
\\
\begin{bmatrix}
\ AA^T \
\end{bmatrix} = I^{-1} = I
$
-
$\begin{bmatrix}
\ 3 \\\\
\ 2 \\\\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 4 \\\\
\ 3 \\\\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 2 \\\\
\ -1 \\\\
\end{bmatrix}$
-
$\begin{bmatrix}
\ -1 \\\\
\ 2 \\\\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 6 & 2 \\
\ 2 & 6 \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 4 & 6 \\
\ 6 & 4 \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 2 & 4 \\
\ 4 & 2\\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 4 & 8 \\
\ 8 & 4\\
\end{bmatrix}$
A
Correct answer
Explanation
Sum of the Eigen values must be equal to the sum of element of principal diagonal of matrix.
Only matrix $\begin{bmatrix}
\ 6 & 2 \
\ 2 & 6 \
\end{bmatrix}$ saisfy this condition.
C
Correct answer
Explanation
$\text{We have}
\\
A = \begin{bmatrix} \ 1 & 1 & 1 \ \ 1 & -1 & 0\ 1 & 1 &1 \ \end{bmatrix} \sim \begin{bmatrix} \ 1 & 1 & 1 \ \ 1 & -1 & 0\ 0 & 0 &0 \ \end{bmatrix} \hspace{2cm} R_3 - R_1
\\
\text{Since one full row is zero, $\rho(A) < 3$}
\\
\text{Now} \begin{vmatrix} \ 1 & 1 \ \ 1 & -1 \ 0 & 0 \ \end{vmatrix} = 2 \neq 0, thus \hspace{0.2cm}\rho(A) = 2$
-
p11p12 - p12p21 = 1
-
p11p22 - p12p21 = -1
-
p11p22 - p12p21 = 0
-
p11p22 + p12p21 = 0
C
Correct answer
Explanation
The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero.
Thus p11 p22 - p12 p21 = 0
-
$\begin{bmatrix}
\ 2e^{-2}-3e^{-1} & e^{-1}-e^{-2} \\
\ 2e^{-2}-2e^{-1} & 5e^{-2}-e^-{1} \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ e^{-1}+e^{-1} & 2e^{-2}-e^{-1} \\
\ 2e^{-1}-4e^{2} & 3e^{-1}+2e^{-2} \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 5e^{-2}-e^{-1} & 3e^{-1}-e^{-2} \\
\ 2e^{-2}-6e^{-1} & 4e^{-2}-6^{-1} \\
\end{bmatrix}$
-
$\begin{bmatrix}
\ 2e^{-1}-e^{-2} & e^{-1}-e^{-2} \\
\ -2e^{-1}+2e^{-2} & -e^{-1}+2e^{-2} \\
\end{bmatrix}$
-
3, 3 + 5j, 6 - j
-
-
3 + j, 3 - j, 5 + j
-
3, - 1 + 3j, - 1 - 3j
D
Correct answer
Explanation
Sum of the principL diagonal el;ement of matrix is equal to the sum of Eigen values. Sum of the diagonal element is -1 - 1 + 3 = 1. In only option (D), the sum of Eigen values is 1.
-
(MT)T = M
-
(cM)T = c(M)T
-
(M+ N)T = MT + NT
-
MN = NM
D
Correct answer
Explanation
$= \begin{bmatrix}
\ n_1m_1 + n_2m_3 & n_1m_2 + n_2m_4 \
\ n_3m_1 + n_4m_3 & n_3m_2 + n_4m_4 \
\end{bmatrix}
\\
MN \neq NM
\\
\text{i.e. property does not hold always.}$
-
always zero
-
always purely imaginary
-
either zero or purely imaginary
-
always real
C
Correct answer
Explanation
Eigen values of a skew-symmetric matrix are either zero or pure imaginary occuring in conjugate pairs.
-
5 + j
-
5 – j
-
1 – 5j
-
1 + 5j
-
$
\left[
\begin{array}
\ 0 & e^{-t} \\\\
e^{-t} & 0
\end{array}
\right]
$
-
$
\left[
\begin{array}
\ e^{-t} & 0 \\\\
0 & e^t &
\end{array}
\right]
$
-
$
\left[
\begin{array}
\ e^{-t} & 0 \\\\
0 & e^{-t} &
\end{array}
\right]
$
-
$
\left[
\begin{array}
\ 0 & e^{t} \\\\
e^{t} & 0
\end{array}
\right]
$
-
Diagonal matrix
-
Single matrix
-
Special matrix
-
Equal matrix
-
Matrices with a relatively high proportion of zero entries
-
Matrices with a relatively high proportion of non-zero entries
-
Matrices with a relatively high proportion of 1's
-
Matrices with a relatively high proportion of negative entries
-
Matrices with all zero entries
A
Correct answer
Explanation
Matrices with a relatively high proportion of zero entries are called sparse matrices.