Question

Let \(L = \begin{pmatrix} 3 & -1 & -1 & -1 \\ -1 & 2 & -1 & 0 \\ -1 & -1 & 3 & 0 \\ -1 & 0 & -1 & 1 \end{pmatrix}\). Which of the following are TRUE?

• A. The matrix \( L \) is row equivalent to \[ \begin{pmatrix} 0 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ -1 & -1 & 3 & 0 \\ -1 & 0 & -1 & 1 \end{pmatrix} \]
• B. The linear system \(Lx = b\) has a solution for all \(b\)
• C. For \( b \neq \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \), the system \( Lx = b \) has a solution.
• D. Rank of the matrix \(L\) is 3

Hint:

For square matrices in exam questions, always check for simple linear dependencies first, such as rows/columns summing to zero, or one row being a multiple of another. This can quickly determine if the matrix is singular (determinant=0, rank<full rank) and save a lot of time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question tests fundamental concepts of linear algebra, including row equivalence, rank of a matrix, and the conditions for the existence of solutions to a linear system \(Ax=b\). A key step is to determine the rank of the matrix L, as it governs the other properties.
Step 2: Key Formula or Approach:
1. Rank: The number of linearly independent rows (or columns) of a matrix. It can be found by reducing the matrix to row echelon form and counting the number of non-zero rows.
2. Row Equivalence: Two matrices are row equivalent if one can be obtained from the other by a sequence of elementary row operations.
3. Existence of Solutions: The system \(Lx=b\) has a solution for all vectors \(b\) in \(\mathbb{R}^4\) if and only if the rank of the 4x4 matrix L is 4 (i.e., L is invertible). If the rank is less than 4, solutions only exist if \(b\) is in the column space of L.
Step 3: Detailed Explanation or Calculation:
Let's analyze the matrix L. A common technique to find dependencies is to check for simple combinations of rows or columns. Let's sum the four rows of L: Row 1: (3, -1, -1, -1)
Row 2: (-1, 2, -1, 0)
Row 3: (-1, -1, 3, 0)
Row 4: (-1, 0, -1, 1)
Sum of Rows = (3-1-1-1, -1+2-1+0, -1-1+3-1, -1+0+0+1) = (0, 0, 0, 0)
Since the sum of the rows is the zero vector (\(R_1+R_2+R_3+R_4 = \vec{0}\)), the rows are linearly dependent. This immediately tells us that the determinant of L is 0, and the rank is less than 4.
Now let's evaluate the options based on this finding:
(A) The matrix L is row equivalent to the given matrix M.
Let \(M = \begin{pmatrix} 0 & 0 & 0 & 0
-1 & 2 & -1 & 0
-1 & -1 & 3 & 0
-1 & 0 & -1 & 1 \end{pmatrix}\). We can perform the elementary row operation \(R_1 ⇒ R_1 + R_2 + R_3 + R_4\) on matrix L. As we calculated, the new first row will be (0, 0, 0, 0). The other rows remain unchanged. This transformation results exactly in matrix M. Therefore, L is row equivalent to M. This statement is TRUE.
(D) Rank of the matrix L is 3.
Since L is row equivalent to M, they have the same rank. The rank of M is the rank of the submatrix formed by its three non-zero rows (rows 2, 3, and 4). Let's check if these three rows are linearly independent. Consider the submatrix: \[ S = \begin{pmatrix} -1 & 2 & -1 & 0
-1 & -1 & 3 & 0
-1 & 0 & -1 & 1 \end{pmatrix} \] We can check the rank by finding a non-zero 3x3 minor. Let's take the determinant of the first three columns: \[ \det \begin{pmatrix} -1 & 2 & -1
-1 & -1 & 3
-1 & 0 & -1 \end{pmatrix} = -1( (-1)(-1) - (3)(0) ) - 2( (-1)(-1) - (3)(-1) ) + (-1)( (-1)(0) - (-1)(-1) ) \] \[ = -1(1) - 2(1 + 3) - 1(-1) = -1 - 8 + 1 = -8 \] Since there is a 3x3 minor with a non-zero determinant, the rank of this submatrix is 3. Therefore, the rank of L is 3. This statement is TRUE.
(B) The linear system Lx = b has a solution for all b.
For a 4x4 system, a solution exists for all \(b\) if and only if the rank is 4. Since we found the rank is 3, this statement is FALSE.
(C) For \(b \neq (1,1,1,1)^T\), the system has a solution.
A solution exists if and only if \(b\) lies in the column space of L. Since rank(L) = 3, the column space is a 3-dimensional subspace of \(\mathbb{R}^4\). There are infinitely many vectors \(b\) for which no solution exists. A solution exists if \(b\) satisfies the same dependency as the rows, which is \(b_1+b_2+b_3+b_4=0\). For \(b=(1,0,0,0)^T\), the sum is 1, so no solution exists. Thus, the statement is FALSE.
Step 4: Final Answer:
The true statements are (A) and (D).
Step 5: Why This is Correct:
Direct calculation shows that the rows of L are linearly dependent, with their sum being the zero vector. This proves that L is row-equivalent to a matrix with a zero row (A) and that its rank is less than 4. Calculation of a 3x3 minor confirms the rank is exactly 3 (D). A rank of 3 for a 4x4 matrix implies that solutions to \(Lx=b\) do not exist for all \(b\), making (B) and (C) false.