Question

Let

$A = \begin{bmatrix} 1 & 2 & 3 \\ 10 & 20 & 31 \\ 11 & 22 & k \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, where $k$ is a constant and $x, y, z$ are variables.

Statements

(37) Regardless of the value of $k$, the matrix $A$ is not invertible, i.e., there is no 3 x 3 matrix $B$ such that $BA$ = the 3 x 3 identity matrix.

(38) There is a unique $k$ such that determinant of $A$ is 0.

(39) The set of solutions $(x, y, z)$ of the matrix equation $A\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is either a line or a plane containing the origin.

(40) If the equation $A\mathbf{v} = \begin{bmatrix} p \\ q \\ r \end{bmatrix}$ has a solution, then it must be true that $q = 10p$.

Answer 1

(37), (39)

Answer 2

Let's analyze each statement.

Statement (37): Regardless of the value of $k$, the matrix $A$ is not invertible, i.e., there is no 3 x 3 matrix $B$ such that $BA$ = the 3 x 3 identity matrix. A square matrix is invertible if and only if its determinant is non-zero. Let's calculate the determinant of $A$:

$A = \begin{bmatrix} 1 & 2 & 3 \\ 10 & 20 & 31 \\ 11 & 22 & k \end{bmatrix}$

$\det(A) = 1 \cdot (20k - 31 \cdot 22) - 2 \cdot (10k - 31 \cdot 11) + 3 \cdot (10 \cdot 22 - 20 \cdot 11)$

$\det(A) = (20k - 682) - 2 \cdot (10k - 341) + 3 \cdot (220 - 220)$

$\det(A) = 20k - 682 - 20k + 682 + 3 \cdot 0$

$\det(A) = 0$.

The determinant of $A$ is 0 for all values of $k$. Since $\det(A) = 0$, the matrix $A$ is singular and thus not invertible, regardless of the value of $k$. Statement (37) is true.

Alternatively, observe that the second column $C_2 = \begin{bmatrix} 2 \\ 20 \\ 22 \end{bmatrix}$ is twice the first column $C_1 = \begin{bmatrix} 1 \\ 10 \\ 11 \end{bmatrix}$, i.e., $C_2 = 2C_1$. Since the columns are linearly dependent ($C_2 - 2C_1 = \mathbf{0}$), the determinant of $A$ is 0.

Statement (38): There is a unique $k$ such that determinant of $A$ is 0. We found that $\det(A) = 0$ for all values of $k$. This means there are infinitely many values of $k$ for which the determinant is 0, not a unique value. Statement (38) is false.

Statement (39): The set of solutions $(x, y, z)$ of the matrix equation $A\mathbf{v} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ is either a line or a plane containing the origin. The equation $A\mathbf{v} = \mathbf{0}$ is a homogeneous system of linear equations. The set of solutions is the null space of $A$, which is a vector subspace of $\mathbb{R}^3$. A subspace always contains the origin $\mathbf{0}$. The dimension of the null space (nullity) is given by $\text{nullity}(A) = \text{number of columns} - \text{rank}(A)$. Here, the number of columns is 3. Since $\det(A) = 0$, the matrix $A$ is singular, so its rank is less than 3. $\text{rank}(A) < 3$. The first two rows $R_1 = [1 \ 2 \ 3]$ and $R_2 = [10 \ 20 \ 31]$ are linearly independent because $R_2$ is not a scalar multiple of $R_1$ ($10/1 = 10$, $20/2 = 10$, but $31/3 \neq 10$). Since there are at least two linearly independent rows, the rank of $A$ is at least 2. Since the rank is less than 3 and at least 2, the rank must be exactly 2. $\text{rank}(A) = 2$. Now, calculate the nullity: $\text{nullity}(A) = 3 - \text{rank}(A) = 3 - 2 = 1$. The dimension of the solution space is 1. A 1-dimensional subspace of $\mathbb{R}^3$ is a line passing through the origin. Therefore, the set of solutions is a line containing the origin. A line is one of the possibilities mentioned in the statement ("either a line or a plane"). Statement (39) is true.

Statement (40): If the equation $A\mathbf{v} = \begin{bmatrix} p \\ q \\ r \end{bmatrix}$ has a solution, then it must be true that $q = 10p$. The equation $A\mathbf{v} = \mathbf{b}$ has a solution if and only if the vector $\mathbf{b} = \begin{bmatrix} p \\ q \\ r \end{bmatrix}$ is in the column space of $A$. The column space of $A$ is the span of its columns $C_1, C_2, C_3$. $C_1 = \begin{bmatrix} 1 \\ 10 \\ 11 \end{bmatrix}$, $C_2 = \begin{bmatrix} 2 \\ 20 \\ 22 \end{bmatrix}$, $C_3 = \begin{bmatrix} 3 \\ 31 \\ k \end{bmatrix}$. We observed that $C_2 = 2C_1$. So, the column space is spanned by $C_1$ and $C_3$. The column space is the set of all linear combinations $c_1 C_1 + c_3 C_3$.

$\begin{bmatrix} p \\ q \\ r \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 10 \\ 11 \end{bmatrix} + c_3 \begin{bmatrix} 3 \\ 31 \\ k \end{bmatrix} = \begin{bmatrix} c_1 + 3c_3 \\ 10c_1 + 31c_3 \\ 11c_1 + kc_3 \end{bmatrix}$.

So, if a solution exists, we must have:

$p = c_1 + 3c_3$

$q = 10c_1 + 31c_3$

$r = 11c_1 + kc_3$

Let's check if $q=10p$ is necessarily true. From the first equation, $c_1 = p - 3c_3$. Substitute this into the second equation:

$q = 10(p - 3c_3) + 31c_3$

$q = 10p - 30c_3 + 31c_3$

$q = 10p + c_3$.

This shows that if a solution exists, $q$ is related to $p$ and $c_3$ (where $c_3$ is determined by the specific linear combination of columns that forms $\mathbf{b}$). The condition $q = 10p$ requires $c_3$ to be 0. However, the vector $\mathbf{b}$ can be any vector in the column space. For example, if we choose $\mathbf{b} = C_3 = \begin{bmatrix} 3 \\ 31 \\ k \end{bmatrix}$ (by taking $c_1=0, c_3=1$), then $p=3$ and $q=31$. In this case, $q=31$ and $10p=10(3)=30$. Since $31 \neq 30$, the condition $q=10p$ is not satisfied. Therefore, if the equation has a solution, it is not necessarily true that $q=10p$. Statement (40) is false.

Let's summarize the truth values of the statements:

(37) True

(38) False

(39) True

(40) False

The question asks which of the statements are correct. Statements (37) and (39) are correct. The options are not provided in the question, so we list the correct statements.