Linear Algebra Quiz: Invertible Matrices
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Invertible MatricesQuestion 1 of 20

Let AA and BB be n×nn \times n matrices. If the product C=ABC = AB is an invertible matrix, what must be true about AA and BB?

Both AA and BB must be invertible.
At least one of AA or BB must be invertible.
Neither AA nor BB is necessarily invertible.
AA must be the inverse of BB.
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Linear Algebra Quiz

Linear Algebra Quiz: Invertible Matrices

Practice Invertible Matrices in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Invertible Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let AA and BB be n×nn \times n matrices. If the product C=ABC = AB is an invertible matrix, what must be true about AA and BB?

  1. Both AA and BB must be invertible. (correct answer)
  2. At least one of AA or BB must be invertible.
  3. Neither AA nor BB is necessarily invertible.
  4. AA must be the inverse of BB.
Explanation: A matrix is invertible if and only if its determinant is non-zero. The determinant of a product of matrices is the product of their determinants: det(AB)=det(A)det(B)\det(AB) = \det(A) \det(B). Since ABAB is invertible, we know det(AB)0\det(AB) \neq 0. Therefore, det(A)det(B)0\det(A) \det(B) \neq 0. This implies that both det(A)0\det(A) \neq 0 and det(B)0\det(B) \neq 0. Consequently, both matrix AA and matrix BB must be invertible.

Question 2

A system of three linear equations in three variables can be interpreted as the intersection of three planes in R3\mathbb{R}^3. If the determinant of the coefficient matrix is zero, what is a possible geometric interpretation of the system's solutions?

  1. The three planes must intersect at a single, unique point.
  2. The three planes may intersect along a common line or not at all. (correct answer)
  3. The three planes must be parallel to each other.
  4. The three planes must be coincident (i.e., they are all the same plane).
Explanation: A zero determinant for the coefficient matrix means the system does not have a unique solution. Geometrically, this means the three planes do not intersect at a single point. This can happen in two ways: there are infinitely many solutions (the planes intersect along a line or are all coincident) or there is no solution (the planes are parallel, two are parallel, or they intersect to form a triangular prism with no common intersection). The most general correct description is that they might intersect along a line (infinite solutions) or have no common intersection point (no solution).

Question 3

Let T:RnRnT: \mathbb{R}^n \to \mathbb{R}^n be a linear transformation with standard matrix AA. If the equation T(x)=0T(x)=0 has only the trivial solution, which statement about the equation T(x)=bT(x)=b must be true?

  1. It has a unique solution for every bRnb \in \mathbb{R}^n. (correct answer)
  2. It has a solution only when bb is the zero vector.
  3. It has infinitely many solutions for every non-zero bRnb \in \mathbb{R}^n.
  4. It may have no solution for some vectors bRnb \in \mathbb{R}^n.
Explanation: The condition that T(x)=0T(x)=0 (which is equivalent to Ax=0Ax=0) has only the trivial solution (x=0x=0) is a key part of the Invertible Matrix Theorem. It implies that the matrix AA is invertible. If AA is invertible, then for any vector bRnb \in \mathbb{R}^n, the equation Ax=bAx=b (or T(x)=bT(x)=b) has a unique solution given by x=A1bx = A^{-1}b.

Question 4

Consider the system of linear equations Ax=0Ax=0, where AA is the matrix A=(1k0241352)A = \begin{pmatrix} 1 & k & 0 \\ 2 & 4 & 1 \\ 3 & 5 & 2 \end{pmatrix} For which value of kk does this system have non-trivial solutions?

  1. k=1k=1
  2. k=2k=2
  3. k=3k=3 (correct answer)
  4. Any value of kk except 33
Explanation: A homogeneous system Ax=0Ax=0 has non-trivial solutions if and only if the matrix AA is singular, which means its determinant is zero. We calculate the determinant of AA: det(A)=1det(4152)kdet(2132)+0det(2435)\det(A) = 1 \cdot \det\begin{pmatrix} 4 & 1 \\ 5 & 2 \end{pmatrix} - k \cdot \det\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} + 0 \cdot \det\begin{pmatrix} 2 & 4 \\ 3 & 5 \end{pmatrix} det(A)=1(4215)k(2213)+0\det(A) = 1(4\cdot2 - 1\cdot5) - k(2\cdot2 - 1\cdot3) + 0 det(A)=1(85)k(43)=3k\det(A) = 1(8 - 5) - k(4 - 3) = 3 - k For non-trivial solutions, we must have det(A)=0\det(A) = 0, so 3k=03 - k = 0, which gives k=3k=3.

Question 5

Consider the matrix A=(21342k639)A = \begin{pmatrix} 2 & -1 & 3 \\ -4 & 2 & k \\ 6 & -3 & 9 \end{pmatrix} . For which value of kk does the system Ax=bA\mathbf{x} = \mathbf{b} have a solution for every vector bR3\mathbf{b} \in \mathbb{R}^3?

  1. k=6k = -6 makes the system consistent for all b\mathbf{b}
  2. k=0k = 0 makes the matrix invertible and ensures solutions exist
  3. k=3k = 3 maximizes the rank and guarantees invertibility
  4. No value of kk makes the system solvable for every b\mathbf{b} (correct answer)
Explanation: The correct answer is D. Notice that row 3 is exactly -3 times row 1: (6,3,9)=3(2,1,3)(6, -3, 9) = -3(2, -1, 3). Also, row 2 is -2 times row 1 plus some adjustment in the third column: (4,2,k)=2(2,1,3)+(0,0,k+6)(-4, 2, k) = -2(2, -1, 3) + (0, 0, k+6). Since rows 1 and 3 are linearly dependent, the matrix can have rank at most 2, regardless of the value of kk. A 3×33 \times 3 matrix with rank less than 3 cannot be invertible, so the system cannot have a solution for every bR3\mathbf{b} \in \mathbb{R}^3. Choice A is wrong because even with k=6k = -6, the rank is still 2. Choice B is wrong because k=0k = 0 doesn't eliminate the linear dependence between rows 1 and 3. Choice C is wrong because no value of kk can make the rank equal to 3 when two rows are already linearly dependent.

Question 6

Matrix BB is 4×44 \times 4 and has the property that B2=IB^2 = I, where II is the 4×44 \times 4 identity matrix. If det(B)=1\det(B) = -1, which of the following statements about the invertibility of (B+I)(B + I) is correct?

  1. Matrix (B+I)(B + I) is always invertible since BB is invertible
  2. Matrix (B+I)(B + I) is not invertible because 1-1 is an eigenvalue of BB (correct answer)
  3. Matrix (B+I)(B + I) may or may not be invertible depending on other eigenvalues
  4. Matrix (B+I)(B + I) is invertible if and only if det(B+I)0\det(B + I) \neq 0
Explanation: The correct answer is B. Since B2=IB^2 = I, we have B2I=0B^2 - I = 0, which factors as (BI)(B+I)=0(B-I)(B+I) = 0. This means that every eigenvalue λ\lambda of BB satisfies λ2=1\lambda^2 = 1, so λ=1\lambda = 1 or λ=1\lambda = -1. Since det(B)=1\det(B) = -1 and the determinant is the product of eigenvalues, BB must have an odd number of eigenvalues equal to 1-1. If 1-1 is an eigenvalue of BB, then 0=1(1)=1+10 = -1 - (-1) = -1 + 1 is an eigenvalue of B+IB + I, making B+IB + I singular (not invertible). Choice A is wrong because invertibility of BB doesn't guarantee invertibility of B+IB + I. Choice C is wrong because we can definitively determine that B+IB + I is not invertible. Choice D is technically correct but doesn't provide the deeper insight that det(B+I)=0\det(B + I) = 0 necessarily.

Question 7

Let AA and CC be n×nn \times n matrices where AA is invertible. Consider the block matrix M=(AB0C)M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} where BB is n×nn \times n and 00 is the n×nn \times n zero matrix. Under what conditions is MM invertible?

  1. MM is invertible if and only if CC is invertible, regardless of BB (correct answer)
  2. MM is invertible if and only if both AA and CC are invertible and B=0B = 0
  3. MM is invertible if and only if det(A)+det(C)0\det(A) + \det(C) \neq 0
  4. MM is always invertible since AA is given to be invertible
Explanation: The correct answer is A. For a block upper triangular matrix M=(AB0C)M = \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} , we have det(M)=det(A)det(C)\det(M) = \det(A) \cdot \det(C). Since AA is given to be invertible, det(A)0\det(A) \neq 0. Therefore, det(M)0\det(M) \neq 0 if and only if det(C)0\det(C) \neq 0, which means MM is invertible if and only if CC is invertible. The matrix BB can be any n×nn \times n matrix without affecting invertibility. Choice B is wrong because BB doesn't need to be zero. Choice C is wrong because determinants multiply, not add, for block triangular matrices. Choice D is wrong because MM also requires CC to be invertible.

Question 8

Let PP be an invertible n×nn \times n matrix and let AA be any n×nn \times n matrix. Consider the relationship between the systems Ax=bA\mathbf{x} = \mathbf{b} and (PAP1)y=Pb(PAP^{-1})\mathbf{y} = P\mathbf{b}. How do the solution sets of these two systems relate?

  1. The systems have identical solution sets with x=y\mathbf{x} = \mathbf{y} for all solutions
  2. The systems are equivalent with solutions related by y=Px\mathbf{y} = P\mathbf{x}
  3. The second system always has more solutions than the first system
  4. The systems have the same number of solutions with x=Py\mathbf{x} = P\mathbf{y} relating them (correct answer)
Explanation: The correct answer is D. If x\mathbf{x} is a solution to Ax=bA\mathbf{x} = \mathbf{b}, then setting y=Px\mathbf{y} = P\mathbf{x} gives us (PAP1)y=(PAP1)(Px)=PAx=Pb(PAP^{-1})\mathbf{y} = (PAP^{-1})(P\mathbf{x}) = PA\mathbf{x} = P\mathbf{b}, so y\mathbf{y} solves the second system. Conversely, if y\mathbf{y} solves the second system, then x=P1y\mathbf{x} = P^{-1}\mathbf{y} solves the first. Since PP is invertible, this establishes a bijection between solution sets, so they have the same cardinality. The relationship is y=Px\mathbf{y} = P\mathbf{x} or equivalently x=P1y\mathbf{x} = P^{-1}\mathbf{y}. Choice A is wrong because xy\mathbf{x} \neq \mathbf{y} in general. Choice B states the relationship in one direction but is incomplete. Choice C is wrong because the systems have the same number of solutions (including the possibility of zero or infinitely many).

Question 9

Let AA be a 5×55 \times 5 matrix with nullity(A)=2\text{nullity}(A) = 2. Consider the matrix equation AX=BAX = B where XX and BB are 5×35 \times 3 matrices. What is the maximum possible number of linearly independent columns that matrix BB can have for this equation to be consistent?

  1. At most 5 linearly independent columns since BB is in R5×3\mathbb{R}^{5 \times 3}
  2. At most 2 linearly independent columns since nullity(A)=2\text{nullity}(A) = 2
  3. At most 3 linearly independent columns since rank(A)=3\text{rank}(A) = 3 (correct answer)
  4. Exactly 3 linearly independent columns for any consistent system
Explanation: When you encounter matrix equations like AX=BAX = B with given nullity information, you need to connect the rank-nullity theorem to the concept of column space and consistency. Since AA is 5×55 \times 5 with nullity(A)=2\text{nullity}(A) = 2, the rank-nullity theorem tells us that rank(A)=52=3\text{rank}(A) = 5 - 2 = 3. This means the column space of AA is a 3-dimensional subspace of R5\mathbb{R}^5. For the equation AX=BAX = B to be consistent, every column of BB must lie in the column space of AA. Since the column space of AA has dimension 3, any set of vectors within it can have at most 3 linearly independent vectors. Therefore, BB can have at most 3 linearly independent columns, making answer C correct. Let's examine why the other options fail. Option A incorrectly focuses on the ambient space R5×3\mathbb{R}^{5 \times 3} rather than the constraint that BB's columns must lie in AA's column space. Option B confuses nullity with the dimension of the column space - the nullity tells us about the kernel, not directly about linear independence in BB. Option D is too restrictive, claiming exactly 3 independent columns are required, when fewer might suffice depending on the specific BB. Study tip: Always connect rank and nullity to dimensional constraints on solution spaces. When you see "consistent system," immediately ask yourself what subspace the right-hand side must belong to, then use rank-nullity to find that subspace's dimension.

Question 10

Suppose AA is a 3×33 \times 3 matrix with rank(A)=2\text{rank}(A) = 2. Consider the augmented matrix [Ab][A | \mathbf{b}] for the system Ax=bA\mathbf{x} = \mathbf{b}. If this system has infinitely many solutions, what can be concluded about rank([Ab])\text{rank}([A | \mathbf{b}])?

  1. rank([Ab])=3\text{rank}([A | \mathbf{b}]) = 3 because the system is consistent
  2. rank([Ab])=2\text{rank}([A | \mathbf{b}]) = 2 since the system has infinitely many solutions (correct answer)
  3. rank([Ab])=1\text{rank}([A | \mathbf{b}]) = 1 because there are multiple free variables
  4. rank([Ab])\text{rank}([A | \mathbf{b}]) could be either 2 or 3 depending on b\mathbf{b}
Explanation: The correct answer is B. For a system Ax=bA\mathbf{x} = \mathbf{b} to have infinitely many solutions, it must be consistent (have at least one solution) and have free variables. The system is consistent if and only if rank(A)=rank([Ab])\text{rank}(A) = \text{rank}([A | \mathbf{b}]). Since rank(A)=2\text{rank}(A) = 2 and the system has infinitely many solutions, we must have rank([Ab])=2\text{rank}([A | \mathbf{b}]) = 2. If rank([Ab])=3\text{rank}([A | \mathbf{b}]) = 3, the system would be inconsistent. Choice A is wrong because rank 3 would make the system inconsistent. Choice C is wrong because the rank cannot decrease when adding a column. Choice D is wrong because once we know the system has infinitely many solutions, the rank relationship is determined.

Question 11

Let AA be a 3×33 \times 3 matrix such that the system Ax=bA\mathbf{x} = \mathbf{b} has a unique solution for some vector b\mathbf{b}, but the homogeneous system Ax=0A\mathbf{x} = \mathbf{0} has infinitely many solutions. Which of the following statements about AA is correct?

  1. Matrix AA is invertible and has rank 3
  2. Matrix AA is not invertible and has rank less than 3
  3. The given conditions are contradictory and cannot occur simultaneously (correct answer)
  4. Matrix AA must be the zero matrix with rank 0
Explanation: The correct answer is C. If the homogeneous system Ax=0A\mathbf{x} = \mathbf{0} has infinitely many solutions, then AA has a non-trivial null space, meaning AA is not invertible. However, if AA is not invertible, then for any vector b\mathbf{b} not in the column space of AA, the system Ax=bA\mathbf{x} = \mathbf{b} has no solution, and for any b\mathbf{b} in the column space, the system has infinitely many solutions (never a unique solution). The conditions are contradictory. Choice A is wrong because an invertible matrix cannot have infinitely many solutions to the homogeneous system. Choice B is wrong because while AA would indeed not be invertible, such a matrix cannot produce a unique solution to any non-homogeneous system. Choice D is wrong because the zero matrix cannot produce a unique solution to any non-zero system.

Question 12

Consider the 3×33 \times 3 matrix A=(12304500k)A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & k \end{pmatrix} where kk is a parameter. For the matrix (A2I)(A - 2I), determine the condition on kk that makes this matrix invertible.

  1. k2k \neq 2 is the only condition needed for invertibility (correct answer)
  2. k0k \neq 0 and k4k \neq 4 are required for invertibility
  3. k2k \neq 2 and k4k \neq 4 ensure that (A2I)(A - 2I) is invertible
  4. k1k \neq -1, k2k \neq 2, and k3k \neq 3 are needed for invertibility
Explanation: The correct answer is A. We have A2I=(1223042500k2)=(12302500k2)A - 2I = \begin{pmatrix} 1-2 & 2 & 3 \\ 0 & 4-2 & 5 \\ 0 & 0 & k-2 \end{pmatrix} = \begin{pmatrix} -1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & k-2 \end{pmatrix} . Since this is an upper triangular matrix, its determinant is the product of its diagonal entries: det(A2I)=(1)(2)(k2)=2(k2)\det(A - 2I) = (-1)(2)(k-2) = -2(k-2). The matrix is invertible if and only if this determinant is nonzero, which occurs when k20k - 2 \neq 0, or k2k \neq 2. Choice B incorrectly focuses on making AA invertible rather than A2IA - 2I. Choice C incorrectly includes the condition k4k \neq 4, which isn't necessary. Choice D lists irrelevant conditions that don't affect the determinant of A2IA - 2I.

Question 13

If a 3×33 \times 3 matrix AA is singular, what can be concluded about the solution set of the system Ax=bAx=b?

  1. The system must have a unique solution.
  2. The system must have infinitely many solutions.
  3. The system has either no solution or infinitely many solutions. (correct answer)
  4. The system must have no solution.
Explanation: A singular matrix is a square matrix that is not invertible. For a system Ax=bAx=b where AA is singular, the system is not guaranteed to have a solution. If the vector bb is not in the column space of AA, the system is inconsistent (no solution). If bb is in the column space of AA, the system is consistent and will have infinitely many solutions because there will be at least one free variable. Therefore, the system has either no solution or infinitely many solutions, depending on bb.

Question 14

Let AA be an n×nn \times n matrix. If the columns of AA do not span Rn\mathbb{R}^n, which statement regarding the equation Ax=bAx=b is correct?

  1. The equation Ax=bAx=b has a unique solution for any bRnb \in \mathbb{R}^n.
  2. The equation Ax=bAx=b must be inconsistent for any non-zero bRnb \in \mathbb{R}^n.
  3. The equation Ax=bAx=b has either no solution or infinitely many solutions. (correct answer)
  4. The equation Ax=0Ax=0 has only the trivial solution.
Explanation: The statement that the columns of AA do not span Rn\mathbb{R}^n is one of the conditions for a square matrix to be non-invertible (singular) according to the Invertible Matrix Theorem. For a singular matrix AA, the system Ax=bAx=b cannot have a unique solution. Depending on whether bb lies in the column space of AA, the system will either have infinitely many solutions (if consistent) or no solution (if inconsistent).

Question 15

Consider the system of equations:

x+yz=2x + y - z = 2

x+2y+z=3x + 2y + z = 3

x+y+(c25)z=cx + y + (c^2 - 5)z = c

For which value of the constant cc will the system have no unique solution?

  1. c=2c=2 only
  2. c=2c=-2 only
  3. c=5c=\sqrt{5} or c=5c=-\sqrt{5}
  4. c=2c=2 or c=2c=-2 (correct answer)
Explanation: When you encounter a system of linear equations with a parameter and need to find when there's "no unique solution," you're looking for values that make the system either inconsistent (no solutions) or dependent (infinitely many solutions). Both cases occur when the coefficient matrix becomes singular. To solve this, set up the augmented matrix and find when the determinant of the coefficient matrix equals zero: [11112111c25]\begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ 1 & 1 & c^2-5 \end{bmatrix} Computing the determinant: det=1(2(c25)1)1(1(c25)1)+(1)(12)\det = 1(2(c^2-5) - 1) - 1(1(c^2-5) - 1) + (-1)(1 - 2) Simplifying: det=2c2101c2+5+1=c24\det = 2c^2 - 10 - 1 - c^2 + 5 + 1 = c^2 - 4 The system has no unique solution when c24=0c^2 - 4 = 0, which gives us c2=4c^2 = 4, so c=2c = 2 or c=2c = -2. Looking at the wrong answers: Choice A suggests only c=2c = 2 works, missing the negative solution. Choice B suggests only c=2c = -2, missing the positive solution. Choice C gives c=±5c = \pm\sqrt{5}, which comes from incorrectly solving c25=0c^2 - 5 = 0 instead of setting the determinant to zero—this is a common trap where students confuse the parameter in the matrix with the condition for singularity. Study tip: For parametric systems, always find when the determinant equals zero to identify non-unique solutions. Don't forget that quadratic equations in the parameter typically yield two critical values, and both positive and negative roots matter.

Question 16

Given an invertible n×nn \times n matrix AA and a vector bRnb \in \mathbb{R}^n, the solution to Ax=bAx=b is given by x=A1bx = A^{-1}b. Which of the following is a common error in expressing this solution?

  1. x=bAx = \frac{b}{A}
  2. x=bA1x = bA^{-1} (correct answer)
  3. x=(AI)bx = (A-I)b
  4. x=bAx = b-A
Explanation: Matrix multiplication is not commutative, meaning the order of multiplication matters. The correct derivation is to left-multiply both sides of Ax=bAx=b by A1A^{-1}: A1(Ax)=A1bA^{-1}(Ax) = A^{-1}b, which simplifies to Ix=A1bIx = A^{-1}b, or x=A1bx = A^{-1}b. The expression x=bA1x = bA^{-1} represents right-multiplication. For this to even be a valid operation, bb would have to be a 1×n1 \times n row vector, and the result would be a row vector, which is inconsistent with xx being an n×1n \times 1 column vector. This highlights the non-commutative nature of matrix multiplication.

Question 17

Let AA be a 4×44 \times 4 matrix such that its reduced row echelon form has a row of zeros. Which of the following statements must be true?

  1. The system Ax=bAx=b has a unique solution for every bR4b \in \mathbb{R}^4.
  2. The determinant of AA is non-zero.
  3. The columns of AA span R4\mathbb{R}^4.
  4. The system Ax=0Ax=0 has infinitely many solutions. (correct answer)
Explanation: When you encounter questions about reduced row echelon form and rows of zeros, you're dealing with fundamental concepts about matrix rank and solution spaces. A row of zeros in the reduced row echelon form directly tells you about the matrix's rank and the behavior of associated linear systems. If matrix AA has a row of zeros in its reduced row echelon form, then rank(A)<4\text{rank}(A) < 4. This means AA cannot have full rank, which has immediate consequences for the homogeneous system Ax=0Ax = 0. Since the rank is less than the number of variables (4), the null space has dimension greater than zero, meaning there are infinitely many solutions to Ax=0Ax = 0. The system will have free variables that can take any values, generating infinitely many solution vectors. Choice A is incorrect because when rank(A)<4\text{rank}(A) < 4, the system Ax=bAx = b either has no solution or infinitely many solutions for any given bb, never a unique solution. Choice B is wrong because matrices with rank less than their size always have determinant zero - the row of zeros makes this clear. Choice C fails because if rank(A)<4\text{rank}(A) < 4, the columns cannot span all of R4\mathbb{R}^4; they span a subspace of smaller dimension. Remember this key connection: a row of zeros in reduced row echelon form means the matrix lacks full rank, which always implies the homogeneous system has infinitely many solutions. This relationship between rank deficiency and non-trivial null spaces appears frequently in linear algebra problems.

Question 18

If an n×nn \times n matrix AA has rank nn, which of the following is a necessary consequence?

  1. The equation Ax=bAx=b has a unique solution for any bRnb \in \mathbb{R}^n. (correct answer)
  2. The equation Ax=bAx=b has infinitely many solutions for any bRnb \in \mathbb{R}^n.
  3. The equation Ax=0Ax=0 has a non-trivial solution.
  4. The determinant of AA is 0.
Explanation: For an n×nn \times n matrix, having rank nn is one of the equivalent conditions for the matrix to be invertible (by the Invertible Matrix Theorem). If a matrix AA is invertible, then the system of equations Ax=bAx=b has a unique solution, x=A1bx = A^{-1}b, for every vector bb in Rn\mathbb{R}^n. The other choices describe properties of non-invertible matrices.

Question 19

Which of the following conditions is sufficient to guarantee that a system of nn linear equations in nn unknowns, represented by Ax=bAx=b, has a unique solution?

  1. The system is consistent.
  2. The number of equations equals the number of unknowns.
  3. The vector bb is not the zero vector.
  4. The columns of the coefficient matrix AA are linearly independent. (correct answer)
Explanation: When you encounter questions about unique solutions to linear systems, you're dealing with fundamental concepts of linear independence, matrix invertibility, and solution existence. The key is understanding what mathematical conditions guarantee exactly one solution. For a system Ax=bAx = b to have a unique solution, the coefficient matrix AA must be invertible (non-singular). This happens precisely when the columns of AA are linearly independent. When columns are linearly independent in an n×nn \times n matrix, the matrix has full rank (rank = nn), making it invertible. This means you can solve for xx uniquely as x=A1bx = A^{-1}b. Therefore, answer D is correct. Let's examine why the other options fail. A is insufficient because consistency only means a solution exists—it could be one solution or infinitely many. For example, if AA has dependent columns, consistent systems have infinitely many solutions. B is a common trap: having equal numbers of equations and unknowns is necessary but not sufficient. You could have nn equations in nn unknowns where some equations are redundant (linearly dependent), leading to infinitely many solutions or no solution. C is irrelevant—the nature of vector bb doesn't determine uniqueness of solutions. Even when b=0b = 0, you can have unique solutions (just x=0x = 0) if AA is invertible. Study tip: Remember that for square systems, linear independence of columns is equivalent to the matrix being invertible, having non-zero determinant, and guaranteeing unique solutions. These concepts are all interconnected in linear algebra.

Question 20

Consider a 4×44 \times 4 matrix AA such that A3=AA^3 = A. If AA has exactly two distinct eigenvalues, which of the following statements about the invertibility of AA is most accurate?

  1. Matrix AA is always invertible because it satisfies a polynomial equation
  2. Matrix AA is invertible if and only if 00 is not an eigenvalue (correct answer)
  3. Matrix AA cannot be invertible since A3=AA^3 = A implies det(A)=0\det(A) = 0
  4. The invertibility depends on whether the eigenvalues are real or complex
Explanation: The correct answer is B. Since A3=AA^3 = A, we have A3A=0A^3 - A = 0, which factors as A(A2I)=A(AI)(A+I)=0A(A^2 - I) = A(A-I)(A+I) = 0. This means every eigenvalue λ\lambda satisfies λ(λ21)=λ(λ1)(λ+1)=0\lambda(\lambda^2 - 1) = \lambda(\lambda - 1)(\lambda + 1) = 0, so the possible eigenvalues are 0,1,10, 1, -1. Since AA has exactly two distinct eigenvalues, the possibilities are {0,1}\{0,1\}, {0,1}\{0,-1\}, or {1,1}\{1,-1\}. Matrix AA is invertible if and only if 00 is not an eigenvalue, which occurs only when the eigenvalues are {1,1}\{1,-1\}. Choice A is wrong because having 00 as an eigenvalue makes AA singular. Choice C is wrong because AA can be invertible if its eigenvalues are {1,1}\{1,-1\}. Choice D is wrong because all eigenvalues here are real, and invertibility depends on whether 00 is an eigenvalue, not on real vs. complex.