Identity Matrices - Linear Algebra

Let $I$ be the $2 \times 2$ identity matrix and let $A = \begin{pmatrix} 7 & 3 \\ 2 & 6 \end{pmatrix}$ . For which value of the scalar $k$ is the determinant of the matrix $A - kI$ equal to 6?
Answer: 3 First, construct the matrix $A - kI$ . $A - kI = \begin{pmatrix} 7 & 3 \\ 2 & 6 \end{pmatrix} - k\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 2 & 6 \end{pmatrix} - \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = \begin{pmatrix} 7-k & 3 \\ 2 & 6-k \end{pmatrix}$ Next, find the determinant of this matrix. For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is $ad-bc$ . $\det(A - kI) = (7-k)(6-k) - (3)(2)$ We are given that this determinant is equal to 6. So, we set up the equation: $(7-k)(6-k) - 6 = 6$ Expand the product and simplify: $42 - 7k - 6k + k^2 - 6 = 6$ $k^2 - 13k + 36 = 6$ $k^2 - 13k + 30 = 0$ Factor the quadratic equation: $(k-3)(k-10) = 0$ The possible values for $k$ are $k=3$ and $k=10$ . Of the choices given, 3 is an option. Distractors B and C are the values of $k$ for which the determinant is 0 (the eigenvalues of A). Distractor D is the magnitude of the coefficient of the $k$ term in the characteristic polynomial.
Consider a $2 \times 2$ matrix $A$ such that $A^2 - 3A + 2I_2 = 0$ , where $I_2$ is the $2 \times 2$ identity matrix. If $\text{tr}(A) = 3$ (where $\text{tr}$ denotes trace), what is $\det(A - I_2)$ ?
Answer: 0 From the equation $A^2 - 3A + 2I_2 = 0$ , we can factor this as $(A - I_2)(A - 2I_2) = 0$ . This means that the eigenvalues of $A$ are 1 and 2. Since $\text{tr}(A) = 3$ , we confirm that the sum of eigenvalues is $1 + 2 = 3$ . The matrix $A - I_2$ has eigenvalues $1 - 1 = 0$ and $2 - 1 = 1$ . Therefore, $\det(A - I_2) = 0 \cdot 1 = 0$ . Alternatively, since $(A - I_2)(A - 2I_2) = 0$ and we can write this as $(A - I_2)A - 2(A - I_2) = 0$ , we see that $A - I_2$ must be singular (has determinant 0) because one of its eigenvalues is 0.
Let $A$ and $B$ be $n \times n$ matrices and let $I$ be the $n \times n$ identity matrix. Which of the following statements is not always true?
Answer: If $A^2=I$ , then $A=I$ or $A=-I$ . Let's analyze each statement: A) If $A^2=I$ , we can take the determinant of both sides: $\det(A^2) = \det(I)$ . This means $(\det(A))^2 = 1$ , so $\det(A)$ must be $1$ or $-1$ . In either case, $\det(A) \neq 0$ , so $A$ is invertible. This statement is always true. B) This is the distributive property of matrix multiplication over addition, combined with the property $AI=A$ . $A(B+I) = AB + AI = AB+A$ . This statement is always true. C) If $A$ and $B$ are invertible, then $\det(A) \neq 0$ and $\det(B) \neq 0$ . The determinant of the product is $\det(AB) = \det(A)\det(B)$ . Since neither determinant is zero, their product is also not zero. Therefore, $AB$ is invertible. This statement is always true. D) This statement is not always true. While $I^2 = I$ and $(-I)^2 = I$ , there can be other matrices whose square is the identity. For example, consider the $2 \times 2$ matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ . Then $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$ . However, $A$ is neither $I$ nor $-I$ . Therefore, this statement is not always true.
A system of linear equations is represented by the matrix equation $A\mathbf{x} = \mathbf{b}$ . It is known that the matrix $A$ satisfies the equation $A^2 - 3A + 2I = 0$ , where $0$ is the zero matrix and $I$ is the identity matrix. Which of the following expressions represents the solution vector $\mathbf{x}$ ?
Answer: $\mathbf{x} = \frac{1}{2}(3I - A)\mathbf{b}$ When you encounter a matrix equation like $A\mathbf{x} = \mathbf{b}$ along with a polynomial equation involving the matrix $A$ , you're being asked to use the polynomial to find the inverse of $A$ without direct computation. Starting with the given equation $A^2 - 3A + 2I = 0$ , you can rearrange this to isolate terms containing $A$ . First, subtract $2I$ from both sides: $A^2 - 3A = -2I$ . Now factor out $A$ on the left: $A(A - 3I) = -2I$ . Dividing both sides by $-2$ : $A \cdot \frac{1}{2}(3I - A) = I$ . This reveals that $A^{-1} = \frac{1}{2}(3I - A)$ . Since $A\mathbf{x} = \mathbf{b}$ , multiplying both sides by $A^{-1}$ gives $\mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{2}(3I - A)\mathbf{b}$ . This matches answer choice B. Looking at the incorrect options: Choice A has $\frac{1}{2}(A - 3I)$ instead of $\frac{1}{2}(3I - A)$ , which would give $-A^{-1}$ . Choice C omits the crucial factor of $\frac{1}{2}$ , giving you $-2A^{-1}$ instead of $A^{-1}$ . Choice D combines both errors: it has the wrong sign and uses 2 instead of $\frac{1}{2}$ , yielding $-\frac{1}{2}A^{-1}$ . Strategy tip: When given a polynomial equation in a matrix $A$ , always try to factor it to find $A^{-1}$ . Rearrange to get $A \times (\text{something}) = I$ , and that "something" is your inverse.
Consider the matrix equation $XI_n + I_nX = 2X$ , where $X$ is an $n \times n$ unknown matrix and $I_n$ is the $n \times n$ identity matrix. Which of the following describes the solution set?
Answer: $X$ can be any $n \times n$ matrix The equation $XI_n + I_nX = 2X$ simplifies to $X + X = 2X$ (since $XI_n = X$ and $I_nX = X$ for any matrix $X$ ). This gives us $2X = 2X$ , which is always true. Therefore, any $n \times n$ matrix $X$ satisfies this equation. The identity matrix property $AI_n = I_nA = A$ for any matrix $A$ makes this equation trivially satisfied for all matrices.
Let $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ . Which of the following matrices is equal to $(I+J)^2$ ?
Answer: $\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$ This question tests matrix addition and multiplication, fundamental operations you'll encounter throughout linear algebra. When you see matrices being added and then squared, work step-by-step rather than trying mental shortcuts. First, compute $I + J$ : $I + J = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ Next, square this result by multiplying it by itself: $(I + J)^2 = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ Using matrix multiplication rules: - Top-left: $(1)(1) + (1)(-1) = 0$ - Top-right: $(1)(1) + (1)(1) = 2$ - Bottom-left: $(-1)(1) + (1)(-1) = -2$ - Bottom-right: $(-1)(1) + (1)(1) = 0$ This gives us $\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$ , which is choice B. Choice A (zero matrix) might tempt you if you incorrectly think $I + J = 0$ , but these matrices don't cancel out. Choice C could result from adding corresponding entries incorrectly or confusing matrix operations. Choice D has the right structure but wrong signs—you might get this by making sign errors during multiplication. Study tip: Always compute matrix operations in stages rather than trying to do everything at once. Write out each step clearly, especially for matrix multiplication where sign errors are common. Double-check your arithmetic in each entry.
Let $A$ be a $3 \times 3$ matrix such that $A^2 = I_3$ , where $I_3$ is the $3 \times 3$ identity matrix. If $\det(A) = -1$ , which of the following must be true about the matrix $A + I_3$ ?
Answer: $\det(A + I_3) = 0$ Since $A^2 = I_3$ , we have $A^2 - I_3 = 0$ , which factors as $(A-I_3)(A+I_3) = 0$ . This means that for any vector $v$ in the nullspace of $A+I_3$ , we have $(A+I_3)v = 0$ , so $Av = -v$ . Since $A^2 = I_3$ , the eigenvalues of $A$ are $\pm 1$ . Given $\det(A) = -1$ , $A$ has an odd number of eigenvalues equal to $-1$ . Since $(A-I_3)(A+I_3) = 0$ , the matrix $A+I_3$ must be singular, so $\det(A+I_3) = 0$ .
Let $A = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix}$ and let $I$ be the $2 \times 2$ identity matrix. If $B$ is a matrix such that $A - 2B = I$ , what is the matrix $B$ ?
Answer: $\begin{pmatrix} -1.5 & 1.5 \\ 0.5 & -1 \end{pmatrix}$ The matrix equation is $A - 2B = I$ . To solve for $B$ , first isolate the term with $B$ : $A - I = 2B$ . Then, multiply by the scalar $\frac{1}{2}$ : $B = \frac{1}{2}(A - I)$ . First, calculate $A - I$ : $A - I = \begin{pmatrix} -2 & 3 \\ 1 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -2-1 & 3-0 \\ 1-0 & -1-1 \end{pmatrix} = \begin{pmatrix} -3 & 3 \\ 1 & -2 \end{pmatrix}$ Now, multiply by $\frac{1}{2}$ : $B = \frac{1}{2}\begin{pmatrix} -3 & 3 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} -1.5 & 1.5 \\ 0.5 & -1 \end{pmatrix}$ Distractor B results from solving $A+2B=I$ instead of $A-2B=I$ . Distractor C correctly calculates $A-I$ but omits the final step of multiplying by $\frac{1}{2}$ . Distractor D results from calculating $\frac{1}{2}(A+I)$ instead of $\frac{1}{2}(A-I)$ .
Let $A$ and $B$ be any $n \times n$ matrices and let $I$ be the $n \times n$ identity matrix. Which of the following expressions is always equivalent to $(A+I)(B+I) - AB$ ?
Answer: $A+B+I$ This question tests matrix multiplication and distribution properties. When you see matrix expressions that need to be expanded, treat them like algebraic expressions but remember that matrix multiplication is generally not commutative. Let's expand $(A+I)(B+I) - AB$ step by step. Using the distributive property of matrix multiplication: $(A+I)(B+I) = A(B+I) + I(B+I) = AB + AI + IB + I^2$ Since $I$ is the identity matrix, we know that $AI = A$ , $IB = B$ , and $I^2 = I$ . Therefore: $(A+I)(B+I) = AB + A + B + I$ Now substituting back into our original expression: $(A+I)(B+I) - AB = (AB + A + B + I) - AB = A + B + I$ This matches answer choice C. Let's examine why the other options are incorrect. Choice A ( $I$ ) would only be true if $A + B = 0$ , which isn't given. Choice B ( $A + B$ ) is missing the identity matrix term that appears in our expansion. Choice D ( $A + B + 2I$ ) suggests an extra $I$ term that doesn't exist in our calculation. The key study tip here is to always expand matrix expressions systematically using distributive properties, and remember the special properties of the identity matrix: $AI = IA = A$ and $I^2 = I$ . Don't rush through the algebra—matrix problems often contain more terms than you might initially expect.
Let $A$ be a $3 \times 3$ matrix with real entries. If $A^3 = -I$ , where $I$ is the $3 \times 3$ identity matrix, what is a possible value for the determinant of $A$ ?
Answer: -1 When you encounter a matrix equation like $A^3 = -I$ , you're dealing with a constraint that severely limits what values certain properties of $A$ can take. The key insight is to use the determinant function, which has the crucial property that $\det(AB) = \det(A)\det(B)$ . Taking the determinant of both sides of $A^3 = -I$ , we get: $\det(A^3) = \det(-I)$ Using the multiplicative property repeatedly: $\det(A^3) = \det(A \cdot A \cdot A) = \det(A)^3$ . For the right side, $\det(-I) = \det(-1 \cdot I) = (-1)^3 \det(I) = (-1)^3 \cdot 1 = -1$ . Therefore: $\det(A)^3 = -1$ , which means $\det(A) = -1$ . The answer is C) -1. Let's see why the other options fail: A) If $\det(A) = 0$ , then $A$ would be singular (non-invertible), making $A^3$ also singular. But $-I$ is invertible, so this is impossible. B) If $\det(A) = 1$ , then $\det(A)^3 = 1^3 = 1 \neq -1$ . D) If $\det(A) = -3$ , then $\det(A)^3 = (-3)^3 = -27 \neq -1$ . Study tip: Whenever you see matrix equations involving powers (like $A^n = B$ ), immediately think about taking determinants of both sides. The multiplicative property $\det(A^n) = \det(A)^n$ often transforms complex matrix equations into simple algebraic ones you can solve directly.
Let $I$ be the $n \times n$ identity matrix. If a matrix polynomial is defined as $p(A) = 2A^2 - 5A + 4I$ , what is the resulting matrix $p(I)$ ?
Answer: $I$ To evaluate the matrix polynomial $p(I)$ , we substitute $I$ for $A$ in the expression. The expression is $p(I) = 2I^2 - 5I + 4I$ . A key property of the identity matrix is that for any positive integer $k$ , $I^k = I$ . Therefore, $I^2 = I$ . Substituting this into the expression gives: $p(I) = 2I - 5I + 4I$ Combining the terms: $p(I) = (2 - 5 + 4)I = 1I = I$ Distractor B might be chosen if there is a calculation error leading to a coefficient of 0. Distractor C could result from adding the absolute values of the coefficients: $(2+5+4)I=11I$ . Distractor D could result from a sign error, such as $2I - 5I - 4I = -7I$ , though this specific error is less likely.
Suppose $A$ is an $n \times n$ matrix such that $A^k = I_n$ for some positive integer $k$ , where $I_n$ is the identity matrix. If $B = I_n + A + A^2 + \cdots + A^{k-1}$ , what can we conclude about the matrix $(A - I_n)B$ ?
Answer: $(A - I_n)B = 0$ (the zero matrix) When you encounter a matrix equation where $A^k = I_n$ , you're dealing with what's called a nilpotent-like relationship. The key insight is recognizing that this setup creates a telescoping pattern when you multiply out $(A - I_n)B$ . Let's work through the multiplication directly. We have $B = I_n + A + A^2 + \cdots + A^{k-1}$ , so: $(A - I_n)B = (A - I_n)(I_n + A + A^2 + \cdots + A^{k-1})$ Distributing gives us: $A(I_n + A + A^2 + \cdots + A^{k-1}) - I_n(I_n + A + A^2 + \cdots + A^{k-1})$ This simplifies to: $(A + A^2 + A^3 + \cdots + A^k) - (I_n + A + A^2 + \cdots + A^{k-1})$ Notice the telescoping: most terms cancel out, leaving only $A^k - I_n$ . Since we're given that $A^k = I_n$ , we get $I_n - I_n = 0$ . Answer A ( $A^k - I_n$ ) represents the intermediate step before applying the constraint $A^k = I_n$ . Answer B ( $A^{k-1} - I_n$ ) misses the telescoping pattern entirely. Answer C ( $kI_n$ ) confuses this with a trace-like calculation and doesn't follow from the algebra. Study tip: When you see $A^k = I_n$ combined with a geometric series like $I + A + A^2 + \cdots$ , immediately think "telescoping." The constraint $A^k = I_n$ is usually the final key that makes everything collapse to zero or the identity.
Let $A$ be a non-square $m \times n$ matrix and let $I$ be an identity matrix. If the matrix product $AI$ is well-defined, what can be concluded about the product $IA$ ?
Answer: $IA$ is not a well-defined product. The rules of matrix multiplication state that for a product $XY$ to be defined, the number of columns in $X$ must equal the number of rows in $Y$ . We are given that $A$ is an $m \times n$ matrix and that the product $AI$ is well-defined. Let the identity matrix $I$ be $p \times p$ . For $AI$ to be defined, the number of columns of $A$ (which is $n$ ) must equal the number of rows of $I$ (which is $p$ ). Thus, $I$ must be the $n \times n$ identity matrix, $I_n$ . Now consider the product $IA$ . This would be the product of $I_n$ (an $n \times n$ matrix) and $A$ (an $m \times n$ matrix). For this product to be defined, the number of columns of $I_n$ (which is $n$ ) must equal the number of rows of $A$ (which is $m$ ). However, the problem states that $A$ is a non-square matrix, meaning $m \neq n$ . Because the inner dimensions do not match, the product $IA$ is not well-defined. Distractor A is a common mistake assuming commutativity and that the product is always defined. Distractors B and C are incorrect because the product cannot be computed at all.
Let $A$ and $B$ be invertible $n \times n$ matrices, and let $I$ be the $n \times n$ identity matrix. If $X = (A^{-1}B)^{-1}$ , which expression is equivalent to $AXA^{-1}$ ?
Answer: $AB^{-1}$ This problem requires using the properties of matrix inverses. First, simplify the expression for $X$ . The inverse of a product of matrices is the product of their inverses in reverse order: $(CD)^{-1} = D^{-1}C^{-1}$ . Applying this to $X = (A^{-1}B)^{-1}$ : $X = B^{-1}(A^{-1})^{-1}$ The inverse of an inverse is the original matrix, so $(A^{-1})^{-1} = A$ . This simplifies $X$ to: $X = B^{-1}A$ Now substitute this simplified expression for $X$ into $AXA^{-1}$ : $AXA^{-1} = A(B^{-1}A)A^{-1}$ Since matrix multiplication is associative, we can regroup the terms: $A(B^{-1}A)A^{-1} = (AB^{-1})(AA^{-1})$ A matrix multiplied by its inverse is the identity matrix, $AA^{-1} = I$ . $(AB^{-1})(I) = AB^{-1}$ Finally, any matrix multiplied by the identity matrix is itself. Thus, the expression simplifies to $AB^{-1}$ . Distractor B arises from incorrectly assuming that $(A^{-1}B)^{-1} = AB^{-1}$ . Distractor C is a result of oversimplification, perhaps thinking all terms cancel. Distractor D could result from errors in applying the inverse-of-a-product rule.
Let $E$ be an elementary matrix obtained by adding 3 times row 2 to row 1 of the $3 \times 3$ identity matrix $I_3$ . If $F$ is the elementary matrix that undoes this operation, what is the product $EFE$ ?
Answer: $E$ (the original elementary matrix) When you encounter elementary matrices and their operations, remember that these matrices represent fundamental row operations, and understanding their algebraic properties is key to matrix manipulation. Let's work through this step-by-step. The elementary matrix $E$ adds 3 times row 2 to row 1, so: $E = \begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ The matrix $F$ undoes this operation by subtracting 3 times row 2 from row 1: $F = \begin{pmatrix} 1 & -3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ Notice that $F = E^{-1}$ , so $EF = FE = I_3$ . Now for $EFE$ : since $EF = I_3$ , we have $EFE = I_3 \cdot E = E$ . The answer is D. Let's examine why the other options fail: - A ( $I_3$ ): This would be true for $EFE$ only if $E$ were symmetric, which it's not. - B ( $E^2$ ): This incorrectly assumes $F = E$ , but $F$ has the opposite effect of $E$ . - C ( $F$ ): This would require $E^2 = I_3$ , meaning $E$ is its own inverse, which isn't true here. Study tip: For elementary matrix problems, always remember that an elementary matrix and its inverse are designed to cancel each other out completely. When you see a product like $EFE$ where $F = E^{-1}$ , the middle terms will always eliminate to give you the identity, leaving just the outer matrix.
Consider the matrix equation $AX = B$ , where $A$ is an $n \times n$ invertible matrix, $X$ is an $n \times n$ unknown matrix, and $B$ is an $n \times n$ given matrix. If we know that $X = I_n$ is a solution, what can we conclude about the relationship between $A$ and $B$ ?
Answer: $B = A$ and the equation has a unique solution If $X = I_n$ is a solution to $AX = B$ , then substituting gives us $AI_n = B$ . Since $AI_n = A$ (multiplying any matrix by the identity matrix gives the original matrix), we have $B = A$ . Since $A$ is invertible, the matrix equation $AX = B = A$ has the unique solution $X = A^{-1}A = I_n$ .
In $\mathbb{R}^2$ , a linear transformation $T$ is represented by a matrix $M$ . If $T(\mathbf{v}) = \mathbf{v}$ for all vectors $\mathbf{v} \in \mathbb{R}^2$ , what is the trace of the matrix $M$ ?
Answer: 2 The condition $T(\mathbf{v}) = \mathbf{v}$ for all vectors $\mathbf{v}$ describes the identity transformation, which leaves every vector unchanged. The matrix representation of a linear transformation $T$ is a matrix $M$ such that $T(\mathbf{v}) = M\mathbf{v}$ . Therefore, the given condition is $M\mathbf{v} = \mathbf{v}$ for all $\mathbf{v} \in \mathbb{R}^2$ . The only matrix for which this is true is the identity matrix. In $\mathbb{R}^2$ , this is the $2 \times 2$ identity matrix: $M = I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ The trace of a square matrix is the sum of the elements on its main diagonal. For $M=I_2$ , the trace is: $\mathrm{tr}(M) = 1 + 1 = 2$ Distractor A (trace=0) and B (trace=1) are incorrect values. Distractor D is incorrect because the condition holds for all vectors, which uniquely determines that the matrix must be the identity matrix, so its trace is a fixed value.
Let $M$ be a $2 \times 2$ matrix with the property that $AM = A$ for any arbitrary $2 \times 2$ matrix $A$ . Which matrix is $M$ ?
Answer: $M = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ The statement $AM=A$ must hold for any $2 \times 2$ matrix $A$ . This allows us to choose a specific, convenient matrix for $A$ to determine $M$ . A powerful choice is an invertible matrix, such as the identity matrix itself, $A=I$ . Substituting $A=I$ into the property gives: $IM = I$ Since any matrix multiplied by the identity matrix is the matrix itself, $IM=M$ . Therefore, we have $M=I$ . So, $M$ must be the $2 \times 2$ identity matrix, $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . Distractor A does not satisfy the property for all $A$ . For example, if $A=I$ , $IA = I \neq A$ . Distractor C would imply $A=0$ for any $A$ , which is false. Distractor D is incorrect because the condition that the property holds for any matrix $A$ is precisely what provides enough information to uniquely determine $M$ .

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