We are given two vectors that belong to the null space of a matrix \(A\) of rank 2. By the rank–nullity theorem, the dimension of the null space is:
\[ \text{nullity}(A) = 4 - 2 = 2 \]
Hence, the null space consists of exactly two linearly independent vectors. Any vector in the null space must be a linear combination of the given two vectors:
\[ \begin{bmatrix}2 \\ 1 \\ 0 \\ 3\end{bmatrix}, \quad \begin{bmatrix}1 \\ 0 \\ 1 \\ 2\end{bmatrix} \]
Check Option A:
Test whether the vector
\[ \begin{bmatrix}1 \\ 1 \\ -1 \\ 1\end{bmatrix} \]
can be expressed as a linear combination of the given vectors. Assume:
\[ c_1 \begin{bmatrix}2 \\ 1 \\ 0 \\ 3\end{bmatrix} + c_2 \begin{bmatrix}1 \\ 0 \\ 1 \\ 2\end{bmatrix} = \begin{bmatrix}1 \\ 1 \\ -1 \\ 1\end{bmatrix} \]
This leads to the system of equations:
\[ 2c_1 + c_2 = 1 \] \[ c_1 = 1 \] \[ c_2 = -1 \] \[ 3c_1 + 2c_2 = 1 \]
The values \(c_1 = 1\) and \(c_2 = -1\) satisfy all equations.
Therefore, the vector \( \begin{bmatrix}1 \\ 1 \\ -1 \\ 1\end{bmatrix} \) is a linear combination of the given null-space vectors and hence belongs to the null space.
Final Answer:
\(\boxed{A}\)
Cholesky decomposition is carried out on the following square matrix [A]. \[ [A] = \begin{bmatrix} 8 & -5 \\ -5 & a_{22} \end{bmatrix} \] Let \( l_{ij} \) and \( a_{ij} \) be the (i,j)\textsuperscript{th elements of matrices [L] and [A], respectively. If the element \( l_{22} \) of the decomposed lower triangular matrix [L] is 1.968, what is the value (rounded off to the nearest integer) of the element \( a_{22} \)?}
| Point | Staff Readings Back side | Staff Readings Fore side | Remarks |
|---|---|---|---|
| P | -2.050 | - | 200.000 |
| Q | 1.050 | 0.95 | Change Point |
| R | - | -1.655 | - |