System of Linear Equations Solvability for Matrix Subspace

To determine which of the given systems of linear equations has at least one solution for every $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \in S$, we first find the condition on $b_1, b_2, b_3$ for the set $S$.

Step 1: Characterize the set $S$

The system of equations defining $S$ is given by:

$$\begin{aligned} -x + 2y + 5z &= b_1 \\ 2x - 4y + 3z &= b_2 \\ x - 2y + 2z &= b_3 \end{aligned}$$

In matrix form, $AX = B$, where

$$A = \begin{bmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$$

Let's perform row operations on the augmented matrix $[A \mid B]$:

$$[A \mid B] = \begin{bmatrix} -1 & 2 & 5 & \mid & b_1 \\ 2 & -4 & 3 & \mid & b_2 \\ 1 & -2 & 2 & \mid & b_3 \end{bmatrix}$$

Applying $R_2 \to R_2 + 2R_1$ and $R_3 \to R_3 + R_1$:

$$\begin{bmatrix} -1 & 2 & 5 & \mid & b_1 \\ 0 & 0 & 13 & \mid & b_2 + 2b_1 \\ 0 & 0 & 7 & \mid & b_3 + b_1 \end{bmatrix}$$

Applying $R_3 \to 13R_3 - 7R_2$:

$$\begin{bmatrix} -1 & 2 & 5 & \mid & b_1 \\ 0 & 0 & 13 & \mid & 2b_1 + b_2 \\ 0 & 0 & 0 & \mid & 13(b_1 + b_3) - 7(2b_1 + b_2) \end{bmatrix}$$

For this system to have at least one solution, we must have:

$$13(b_1 + b_3) - 7(2b_1 + b_2) = 0 \implies -b_1 - 7b_2 + 13b_3 = 0 \implies b_1 + 7b_2 - 13b_3 = 0$$

Thus, the set $S$ is given by:

$$S = \left\{ \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \in \mathbb{R}^3 : b_1 + 7b_2 - 13b_3 = 0 \right\}$$

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Step 2: Evaluate the Options

A system of linear equations $MX = B$ has a unique solution for every $B \in \mathbb{R}^3$ if $\det(M) \neq 0$. Consequently, if $\det(M) \neq 0$, it will also have at least one solution for every $B \in S$.

Analysis of Option A:

The coefficient matrix is:

$$M_A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{bmatrix}$$

Calculating the determinant:

$$\det(M_A) = 1(24 - 10) - 2(0 - 5) + 3(0 - 4) = 14 + 10 - 12 = 12 \neq 0$$

Since $\det(M_A) \neq 0$, the system has a unique solution for every vector $B \in \mathbb{R}^3$, and hence for every $B \in S$. Option A is correct.

Analysis of Option B:

The coefficient matrix is:

$$M_B = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}$$

Calculating the determinant:

$$\det(M_B) = 1(-6 + 6) - 1(-15 + 12) + 3(-5 + 4) = 0 + 3 - 3 = 0$$

Row operations on $[M_B \mid B]$ yield:

• $R_1 + R_3 \implies -x = b_1 + b_3 \implies x = -b_1 - b_3$
• $R_1 \implies y + 3z = b_1 - x = 2b_1 + b_3$
• $R_2 \implies 5(-b_1 - b_3) + 2(2b_1 + b_3) = b_2 \implies b_1 + b_2 + 3b_3 = 0$

Thus, the system in Option B is consistent if and only if $b_1 + b_2 + 3b_3 = 0$. Since this plane is not identical to $b_1 + 7b_2 - 13b_3 = 0$, not every element in $S$ satisfies this condition. Option B is incorrect.

Analysis of Option C:

The coefficient matrix is:

$$M_C = \begin{bmatrix} -1 & 2 & -5 \\ 2 & -4 & 10 \\ 1 & -2 & 5 \end{bmatrix}$$

Notice that Row 2 = $-2 \times \text{Row 1}$ and Row 3 = $-1 \times \text{Row 1}$. Thus, for a solution to exist, we must have $b_2 = -2b_1$ and $b_3 = -b_1$. This defines a line in $\mathbb{R}^3$, whereas $S$ is a 2-dimensional plane. Hence, not every element in $S$ satisfies this condition. Option C is incorrect.

Analysis of Option D:

The coefficient matrix is:

$$M_D = \begin{bmatrix} 1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5 \end{bmatrix}$$

Calculating the determinant:

$$\det(M_D) = 1(0 - 12) - 2(-10 - 3) + 5(8 - 0) = -12 + 26 + 40 = 54 \neq 0$$

Since $\det(M_D) \neq 0$, the system has a unique solution for every vector $B \in \mathbb{R}^3$, and hence for every $B \in S$. Option D is correct.

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Conclusion

The correct options are A and D.