Consistent vs. Inconsistent Systems - Linear Algebra

Let $\mathbf{x}_p = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ be a solution to the linear system $A\mathbf{x} = \mathbf{b}$ . You are also told that $\mathbf{x}_h = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ is a solution to the corresponding homogeneous system $A\mathbf{x} = \mathbf{0}$ . Which of the following vectors is also a solution to $A\mathbf{x} = \mathbf{b}$ ?
Answer: $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ The general solution to a non-homogeneous system $A\mathbf{x} = \mathbf{b}$ is given by $\mathbf{x} = \mathbf{x}_p + c\mathbf{x}_h$ , where $\mathbf{x}_p$ is a particular solution, $\mathbf{x}_h$ is a non-trivial solution to the associated homogeneous system $A\mathbf{x} = \mathbf{0}$ , and $c$ is any scalar. This is because $A(\mathbf{x}_p + c\mathbf{x}_h) = A\mathbf{x}_p + c(A\mathbf{x}_h) = \mathbf{b} + c(\mathbf{0}) = \mathbf{b}$ . We are given $\mathbf{x}_p = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ and $\mathbf{x}_h = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ . Any other solution will have the form $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + c \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1-c \\ c \\ 2 \end{pmatrix}$ . We check the options to see if they can be written in this form for some value of $c$ . (A) For $\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ : The second component implies $c=-1$ . Let's check if this works for the first component: $1 - c = 1 - (-1) = 2$ . The third component is 2, which matches. This vector is a solution for $c=-1$ . (B) For $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ : The second component implies $c=1$ . The third component should be 2, but it is 0. Not a solution. (C) For $\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ : The second component implies $c=1$ . The first component should be $1-1=0$ , but it is 2. Not a solution. (D) $\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ is a solution to the homogeneous system, not the non-homogeneous one (unless $\mathbf{b}=\mathbf{0}$ ).
Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ . The system $A\mathbf{x} = \mathbf{b}$ is consistent if and only if $\mathbf{b}$ is in the column space of $A$ . For which of the following vectors $\mathbf{b}$ is the system consistent?
Answer: $\mathbf{b} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$ The column space of $A$ is the set of all possible linear combinations of its column vectors. The columns of $A$ are $\mathbf{c}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{c}_2 = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ . Notice that $\mathbf{c}_2 = 2\mathbf{c}_1$ . This means the two columns are linearly dependent, and the column space is simply the set of all scalar multiples of the first column vector. So, a vector $\mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ is in the column space of $A$ if it is of the form $c \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ for some scalar $c$ . This implies that the second component must be twice the first component, i.e., $b_2 = 2b_1$ . We check the answer choices: (A) $3 \neq 2(1)$ (B) $6 = 2(3)$ . This vector is in the column space. (C) $2 \neq 2(2)$ (D) $1 \neq 2(0)$ Therefore, the system is consistent only for $\mathbf{b} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$ .
For what value of $k$ does the following system of equations have no solution? $\begin{align*} x + y - z &= 1 \\ 2x + 3y + kz &= 3 \\ x + ky + 3z &= 2 \end{align*}$
Answer: $k = -3$ We can use Gaussian elimination on the augmented matrix of the system. $\left[ \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & k & 3 \\ 1 & k & 3 & 2 \end{array} \right]$ Performing row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ gives: $\left[ \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & k-1 & 4 & 1 \end{array} \right]$ Next, performing $R_3 \to R_3 - (k-1)R_2$ : $\left[ \begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & 0 & 4 - (k-1)(k+2) & 1 - (k-1) \end{array} \right]$ The system has no solution if the last row represents an equation of the form $0 = c$ where $c \neq 0$ . This requires the coefficient of the third variable to be zero and the constant on the right to be non-zero. $4 - (k-1)(k+2) = 4 - (k^2+k-2) = -k^2-k+6 = -(k+3)(k-2) = 0$ . This occurs when $k=-3$ or $k=2$ . The constant term is $1 - (k-1) = 2-k$ . This must be non-zero, so $k \neq 2$ . Combining both conditions, the system is inconsistent only when $k=-3$ .
A system of equations has the reduced row echelon form $\begin{bmatrix} 1 & 0 & 2 & 0 & | & 5 \\ 0 & 1 & -1 & 0 & | & 3 \\ 0 & 0 & 0 & 1 & | & -2 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$ . Which statement best describes this system?
Answer: The system is consistent with infinitely many solutions because there is a free variable The system is consistent because there are no rows of the form $[0 \, 0 \, 0 \, 0 \, | \, c]$ where $c ≠ 0$ . The variables corresponding to columns 1, 2, and 4 are basic variables (leading variables), while the variable corresponding to column 3 is a free variable. Since there is at least one free variable, the system has infinitely many solutions.
Consider the system of equations represented by the matrix equation $A\mathbf{x} = \mathbf{b}$ , where $A$ is a $4 \times 5$ matrix with rank 3. Under what condition is this system guaranteed to be consistent?
Answer: When the rank of the augmented matrix $[A|\mathbf{b}]$ equals 3 A system $A\mathbf{x} = \mathbf{b}$ is consistent if and only if $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$ . Since $A$ has rank 3, the system is consistent when the augmented matrix also has rank 3, meaning $\mathbf{b}$ lies in the column space of $A$ . Option A would mean $\mathbf{b}$ is linearly independent from all columns of $A$ , making the system inconsistent. Option C describes when $A\mathbf{b} = \mathbf{0}$ , not $A\mathbf{x} = \mathbf{b}$ . Option D is irrelevant to consistency.
Suppose the system $A\mathbf{x} = \mathbf{b}$ is consistent and has more than one solution. Which of the following statements must be true?
Answer: The homogeneous system $A\mathbf{x} = \mathbf{0}$ has infinitely many solutions. If a non-homogeneous system $A\mathbf{x} = \mathbf{b}$ has more than one solution, it must have infinitely many. The general solution can be written as $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$ , where $\mathbf{x}_p$ is a particular solution to $A\mathbf{x} = \mathbf{b}$ and $\mathbf{x}_h$ is any solution to the corresponding homogeneous system $A\mathbf{x} = \mathbf{0}$ . For there to be more than one solution for $\mathbf{x}$ , there must be non-zero solutions for $\mathbf{x}_h$ . The existence of a non-trivial solution to the homogeneous system implies the existence of infinitely many solutions (any scalar multiple of $\mathbf{x}_h$ is also a solution). Therefore, the homogeneous system $A\mathbf{x} = \mathbf{0}$ must have infinitely many solutions.
Consider the system of equations: $\begin{align*} x + 2y - z &= 1 \\ 2x + 5y - z &= 3 \\ x + 3y &= k \end{align*}$ For which value of $k$ is the system consistent?
Answer: $k=2$ To determine when the system is consistent, we perform Gaussian elimination on the augmented matrix. $\left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 2 & 5 & -1 & 3 \\ 1 & 3 & 0 & k \end{array} \right]$ Apply row operations $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ : $\left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & k-1 \end{array} \right]$ Now apply the operation $R_3 \to R_3 - R_2$ : $\left[ \begin{array}{ccc|c} 1 & 2 & -1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & k-2 \end{array} \right]$ For the system to be consistent, the last row must not represent a contradiction. The equation from the last row is $0x + 0y + 0z = k-2$ . This equation is only true if $k-2=0$ , which means $k=2$ . If $k \neq 2$ , the last row becomes $0 = c$ for some non-zero $c$ , and the system is inconsistent.
Matrix $A$ is $4 \times 6$ with rank 4, and matrix $B$ is $4 \times 1$ . The system $A\mathbf{x} = B$ is known to be consistent. A second system $A\mathbf{x} = 2B$ is formed. Which statement about the relationship between these two systems is correct?
Answer: The second system is guaranteed to be consistent and has the same solution structure as the first When you encounter problems about linear systems with scalar multiples of the right-hand side vector, think about how consistency and solution structure relate to the column space and null space of the coefficient matrix. Since $A\mathbf{x} = B$ is consistent, vector $B$ lies in the column space of $A$ . The key insight is that column spaces are closed under scalar multiplication—if $B$ is in the column space, then any scalar multiple like $2B$ is also in the column space. This guarantees that $A\mathbf{x} = 2B$ is consistent. Moreover, both systems have identical solution structures. Since $A$ is $4 \times 6$ with rank 4, the null space has dimension $6 - 4 = 2$ . Any consistent system $A\mathbf{x} = \mathbf{c}$ will have infinitely many solutions (a particular solution plus the 2-dimensional null space). If $\mathbf{x}_0$ solves $A\mathbf{x} = B$ , then $2\mathbf{x}_0$ solves $A\mathbf{x} = 2B$ , and both have the same infinite solution set structure. Option A is wrong because scalar multiples preserve consistency—there's no mechanism for the second system to become inconsistent. Option B correctly notes both have the same number of solutions but gives an incomplete reason. Option C incorrectly suggests the second system might be inconsistent, ignoring that column spaces are closed under scalar operations. Study tip: Remember that if a vector is in a subspace (like the column space), then all its scalar multiples are too. This principle frequently appears in consistency problems involving scaled right-hand sides.
Consider a homogeneous system of linear equations $A\mathbf{x} = \mathbf{0}$ , where $A$ is an $m \times n$ matrix. Which statement about the consistency of this system is always true?
Answer: The system is always consistent. A homogeneous system of linear equations is a system of the form $A\mathbf{x} = \mathbf{0}$ . By definition, a system is consistent if it has at least one solution. The vector $\mathbf{x} = \mathbf{0}$ (the zero vector) is always a solution to any homogeneous system, because $A\mathbf{0} = \mathbf{0}$ . This solution is called the trivial solution. Since there is always at least one solution, a homogeneous system is always consistent, regardless of the dimensions or rank of the matrix $A$ . The other choices describe conditions for a unique solution or are common misconceptions about systems in general.
The augmented matrix of a system of linear equations has been reduced to the following form: $\left[ \begin{array}{ccc|c} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & a^2-9 & a-3 \end{array} \right]$ For what value(s) of $a$ is the system inconsistent?
Answer: $a = -3$ A system is inconsistent if its row-echelon form contains a row of the form $[0, 0, \dots, 0 | c]$ where $c$ is a non-zero constant. In this matrix, the last row corresponds to the equation $(a^2-9)x_3 = a-3$ . For this equation to represent a contradiction ( $0 = c$ with $c \neq 0$ ), we need the coefficient of $x_3$ to be zero and the constant on the right-hand side to be non-zero. 1. Set the coefficient to zero: $a^2 - 9 = 0 \implies (a-3)(a+3) = 0$ . This gives $a=3$ or $a=-3$ . 2. Set the constant term to be non-zero: $a-3 \neq 0 \implies a \neq 3$ . Both conditions must be met simultaneously. The value from the first condition that also satisfies the second condition is $a=-3$ . If $a=3$ , the last row becomes $[0, 0, 0 | 0]$ , which indicates infinitely many solutions, not an inconsistent system.
A manufacturer produces three products: P1, P2, and P3. Each unit of P1 requires 1 hour of labor and 2 lbs of material. Each unit of P2 requires 2 hours of labor and 4 lbs of material. Each unit of P3 requires 3 hours of labor and 6 lbs of material. For a specific production run, the manufacturer has exactly 40 hours of labor available. The total material required for this run is $k$ lbs. For which value of $k$ is it impossible to find a combination of products that meets these exact constraints?
Answer: $k = 100$ Let $x_1, x_2, x_3$ be the number of units produced for products P1, P2, and P3, respectively. The constraints can be modeled by a system of linear equations: Labor: $1x_1 + 2x_2 + 3x_3 = 40$ Material: $2x_1 + 4x_2 + 6x_3 = k$ Notice that the coefficients of the variables in the material equation are exactly twice the coefficients in the labor equation. If we multiply the entire first equation by 2, we get $2(1x_1 + 2x_2 + 3x_3) = 2(40)$ , which simplifies to $2x_1 + 4x_2 + 6x_3 = 80$ . For the system to be consistent (i.e., for a solution to exist), the second equation must be equivalent to this. This means we must have $k = 80$ . If $k=80$ , there are infinitely many production combinations possible. If $k$ is any value other than 80, the two equations contradict each other, and the system is inconsistent, meaning no combination of products is possible. Out of the given choices, $k=100$ makes the system inconsistent.
A homogeneous system $A\mathbf{x} = \mathbf{0}$ where $A$ is a $3 \times 5$ matrix has a non-trivial solution. If we modify this to the non-homogeneous system $A\mathbf{x} = \mathbf{b}$ where $\mathbf{b} \neq \mathbf{0}$ , which statement is necessarily true?
Answer: The non-homogeneous system has infinitely many solutions if it is consistent When you encounter questions about homogeneous versus non-homogeneous linear systems, focus on how the null space structure affects solution behavior. Since $A$ is $3 \times 5$ , there are more variables than equations. The homogeneous system $A\mathbf{x} = \mathbf{0}$ having a non-trivial solution tells us that $\text{rank}(A) < 5$ , meaning the null space of $A$ is non-trivial with dimension at least $5 - \text{rank}(A) \geq 1$ . This null space structure is crucial: if the non-homogeneous system $A\mathbf{x} = \mathbf{b}$ has any solution $\mathbf{x}_p$ , then its general solution is $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$ , where $\mathbf{x}_h$ is any solution to the homogeneous system. Since we know the homogeneous system has infinitely many non-trivial solutions, the non-homogeneous system must have infinitely many solutions whenever it's consistent. This confirms answer D. A is wrong because consistency depends on whether $\mathbf{b}$ lies in the column space of $A$ —this isn't automatically impossible just because $\mathbf{b} \neq \mathbf{0}$ . B is wrong because the non-trivial null space guarantees infinitely many solutions, never a unique solution. C is wrong because consistency depends on both $\text{rank}(A)$ and $\text{rank}([A|\mathbf{b}])$ —specifically, whether $\mathbf{b}$ is in the column space of $A$ . Key takeaway: When a homogeneous system has non-trivial solutions, any consistent non-homogeneous system with the same coefficient matrix will have infinitely many solutions due to the inherited null space structure.
The system $\begin{cases} x + y + z = 6 \\ 2x + 2y + 2z = 12 \\ 3x + 3y + 3z = k \end{cases}$ has infinitely many solutions. A student claims that any value of $k$ will work because "all equations are multiples of each other." What is wrong with this reasoning?
Answer: Only $k = 18$ gives infinitely many solutions; other values make the system inconsistent The first equation gives $x + y + z = 6$ . The second equation $2x + 2y + 2z = 12$ is indeed $2$ times the first equation, so it's redundant. For the third equation to be consistent, we need $3x + 3y + 3z = 3(x + y + z) = 3(6) = 18$ . Therefore, $k = 18$ . If $k ≠ 18$ , the third equation contradicts the first equation (scaled appropriately), making the system inconsistent. The student's error is assuming that visual similarity (coefficients being multiples) automatically ensures consistency without checking that the constants are also proportional.
A system of linear equations consists of three distinct lines in the xy-plane: $x+y=3$ , $2x+2y=k$ , and $x-y=1$ . For which value of $k$ is the system consistent?
Answer: $k=6$ The first two equations are $x+y=3$ and $2x+2y=k$ . The lines they represent are parallel because the coefficients of $x$ and $y$ in the second equation are twice those in the first. For a system containing these two equations to be consistent, the lines must be identical (coincident). This occurs if the constant term is also scaled by the same factor. Thus, we must have $k = 2 \times 3 = 6$ . If $k=6$ , the first two equations are redundant, and the system effectively becomes: $\begin{cases} x+y=3 \\ x-y=1 \end{cases}$ This is a system of two distinct, non-parallel lines, which has a unique solution. Adding the two equations gives $2x=4 \implies x=2$ , and substituting back gives $2+y=3 \implies y=1$ . The unique solution is $(2,1)$ . If $k \neq 6$ , the first two lines are parallel and distinct, meaning there is no point $(x,y)$ that satisfies both equations. In that case, the entire system is inconsistent.
A system of $m$ linear equations in $n$ variables is represented by the augmented matrix $[A|\mathbf{b}]$ . If the system is inconsistent, which of the following statements must be true?
Answer: $\mathrm{rank}(A) < \mathrm{rank}([A|\mathbf{b}])$ The Rouché–Capelli theorem states that a system of linear equations is consistent if and only if the rank of the coefficient matrix $A$ is equal to the rank of the augmented matrix $[A|\mathbf{b}]$ . An inconsistent system, by definition, has no solution. This occurs precisely when the rank of the coefficient matrix is less than the rank of the augmented matrix. This situation corresponds to a row in the row-echelon form of the augmented matrix that is of the form $[0, 0, \dots, 0 | c]$ where $c$ is a non-zero constant, which represents the impossible equation $0=c$ .
During Gaussian elimination of an augmented matrix, a student obtains the row $[0 \, 0 \, 0 \, 0 \, | \, 5]$ in the fourth position. However, they continue elimination and eventually this row becomes $[0 \, 0 \, 0 \, 0 \, | \, 0]$ . What can be concluded about the original system?
Answer: The original system was inconsistent, and the student made an error in their calculations Elementary row operations preserve the solution set of a system. Once a row $[0 \, 0 \, 0 \, 0 \, | \, 5]$ appears, it indicates the equation $0 = 5$ , which means the system is inconsistent. This row cannot be transformed to $[0 \, 0 \, 0 \, 0 \, | \, 0]$ through valid elementary row operations. The only way this could happen is if the student made a computational error. Elementary row operations can only: (1) multiply a row by a non-zero constant, (2) swap rows, or (3) add/subtract multiples of rows. None of these can change 5 to 0 in the augmented column when all other entries in that row are zero.
Consider the system of linear equations: $\begin{cases} 3x + 2y = 7 \\ 6x + ky = 10 \end{cases}$ For which value of $k$ is the system inconsistent?
Answer: $k = 4$ For a 2x2 system of linear equations to be inconsistent, the lines they represent must be parallel and have different y-intercepts. Two lines are parallel if their slopes are equal. The slope of the first line, $y = -\frac{3}{2}x + \frac{7}{2}$ , is $-\frac{3}{2}$ . The slope of the second line, $y = -\frac{6}{k}x + \frac{10}{k}$ , is $-\frac{6}{k}$ . Setting the slopes equal gives: $-\frac{3}{2} = -\frac{6}{k}$ This simplifies to $3k = 12$ , so $k=4$ . When $k=4$ , the second equation becomes $6x + 4y = 10$ , which simplifies to $3x + 2y = 5$ . The system is now $3x + 2y = 7$ and $3x + 2y = 5$ , which is a contradiction ( $7 \neq 5$ ). Therefore, the system is inconsistent when $k=4$ .

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