Systems of Linear Equations Questions and Answers

Answer: A

The true statement is There exist at least \(m - n\) linearly independent solutions to this system.

The rank of A is at most n, its number of rows. By the rank-nullity theorem the solution space of \(Ax = 0\) has dimension \(m - \text{rank}(A) \ge m - n\), so there are at least \(m - n\) linearly independent solutions.

The second statement claims exactly \(m - n\) vectors span every solution, which fails when the rank is less than n; for A = 0 every vector in \(\mathbb{R}^{m}\) is a solution. The third fails in general: a non-zero solution is guaranteed only with at least \(m - n - 1\) zero variables. The fourth fails, for example, for A = [1 0 0; 0 1 0], whose solutions have only the third variable non-zero, fewer than n = 2.

Answer: C

Only III is true.

Statement I is false: a system with fewer equations than unknowns can still be inconsistent, for example \(x + y + z = 1\) and \(x + y + z = 2\) (\(m = 2\), \(n = 3\)).

Statement II is false: more equations than unknowns can still be consistent, for example \(x = 1\), \(2x = 2\), \(3x = 3\) (\(m = 3\), \(n = 1\)); any homogeneous system also has the solution 0.

Statement III is true: it only claims that some square system has a solution, and \(x = 1\) (or any system with an invertible coefficient matrix) is one.

So options including I or II are wrong, and “none of them” ignores III.

Answer: D

The values are \(L_{32} = -\frac{1}{2}\), \(U_{33} = -\frac{1}{2}\) and \(x_{1} = 0\).

Use Doolittle's form, with 1s on the diagonal of \(L\). The first row of \(U\) is the first equation, (1, 1, −2), and the multipliers are \(L_{21} = 1\), \(L_{31} = 2\).

\[ \begin{aligned} R_{2} - R_{1} &: (0,\ 2,\ 1) \\ R_{3} - 2R_{1} &: (0,\ -1,\ -1) \\ L_{32} &= \tfrac{-1}{2} = -\tfrac{1}{2} \\ U_{33} &= -1 - \left(-\tfrac{1}{2}\right)(1) = -\tfrac{1}{2} \end{aligned} \]

Forward substitution with \(Ly = b\) gives \(y = (4,\ 3,\ \tfrac{1}{2})\), and back substitution with \(Ux = y\) gives:

\[ \begin{aligned} x_{3} &= \tfrac{1/2}{-1/2} = -1 \\ 2x_{2} + x_{3} &= 3 \Rightarrow x_{2} = 2 \\ x_{1} + x_{2} - 2x_{3} &= 4 \Rightarrow x_{1} = 0 \end{aligned} \]

Check in the third equation: \(0 + 2 + 5 = 7\). The options with \(L_{32} = 2\) take the multiplier of row 1 instead of row 2, and \(U_{33} = 2\) or \(x_{1} = -1\) come from sign errors in the elimination.

Answer: C

The answer is P is TRUE and Q is FALSE.

P: rank(M) = n means the n columns of M are linearly independent. Mx is a combination of the columns with weights from x, so Mx = 0 forces every weight to be zero, that is x = 0. P is true; m \(\gt\) n only makes the matrix tall.

Q: from \(E^{3} = 0\) we get \(E^{4} = 0\), so

\[ (I + E^{2})(I - E^{2}) = I - E^{4} = I \]

Thus \(I + E^{2}\) has the inverse \(I - E^{2}\) and is non-singular, so Q is false. In general, I plus a nilpotent matrix is always invertible. That rules out every option calling Q true.

Answer: D

The other solution is \(\begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix}\).

\(A\) has 3 columns and rank 2, so its null space has dimension \(3 - 2 = 1\). The difference of the two given solutions lies in it:

\[ \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} \]

so the null space is spanned by \((1,1,1)^{T}\) and every solution has the form \((1,2,3)^{T} + s(1,1,1)^{T}\). Taking \(s = 6\) gives \((7,8,9)^{T}\).

\((3,3,3)^{T}\) solves \(Ax = 0\), not \(Ax = b\), since \(b \neq 0\). \((5,7,9)^{T}\) is the sum of the two solutions, so \(A\) sends it to \(2b \neq b\). \((5,6,9)^{T}\) differs from \((1,2,3)^{T}\) by \((4,4,6)^{T}\), which is not a multiple of \((1,1,1)^{T}\).