Precalculus Quiz: Zero And Identity Matrices And Determinants
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Zero And Identity Matrices And DeterminantsQuestion 1 of 20

Given the 2×22 \times 2 matrix A=[3124]A=\begin{bmatrix}3 & 1\\2 & 4\end{bmatrix}, what is the determinant det⁡(A)\det(A)? (Use det⁡ ⁣([abcd])=ad−bc\det\!\left(\begin{bmatrix}a & b\\c & d\end{bmatrix}\right)=ad-bc.)

1010
1111
−10-10
1414
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Precalculus Quiz

Precalculus Quiz: Zero And Identity Matrices And Determinants

Practice Zero And Identity Matrices And Determinants in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Zero And Identity Matrices And Determinants, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

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Question 1

Given the 2×22 \times 2 matrix A=[3124]A=\begin{bmatrix}3 & 1\\2 & 4\end{bmatrix}, what is the determinant det⁡(A)\det(A)? (Use det⁡ ⁣([abcd])=ad−bc\det\!\left(\begin{bmatrix}a & b\\c & d\end{bmatrix}\right)=ad-bc.)

  1. 1010 (correct answer)
  2. 1111
  3. −10-10
  4. 1414
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×22 \times 2 matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} is calculated as det⁡(A)=ad−bc\det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det⁡(A)≠0\det(A) \neq 0, the matrix is invertible (non-singular), but if det⁡(A)=0\det(A) = 0, the matrix is singular (no inverse exists). For matrix A=[3124]A = \begin{bmatrix} 3 & 1 \\ 2 & 4 \end{bmatrix}, we compute det⁡(A)=ad−bc=(3)(4)−(1)(2)=12−2=10\det(A) = ad - bc = (3)(4) - (1)(2) = 12 - 2 = 10. Choice A is correct because it properly calculates det⁡(A)=ad−bc\det(A) = ad - bc. Choice C reverses the subtraction in the determinant formula, computing bc−adbc - ad instead of ad−bcad - bc, which gives the wrong sign. Key to determinants: for a 2×22 \times 2 matrix [ab;cd][a b; c d], always use det⁡=ad−bc\det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det⁡=0\det = 0 means the matrix has no inverse (singular) while det⁡≠0\det \neq 0 means it's invertible. For invertibility, just compute the determinant: if det⁡(A)≠0\det(A) \neq 0, the matrix is invertible; if det⁡(A)=0\det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 2

Given A=[4121]A=\begin{bmatrix}4&1\\2&1\end{bmatrix}, does matrix AA have an inverse? Why or why not?

  1. Yes; det⁡(A)=2≠0\det(A)=2\neq 0, so AA is non-singular. (correct answer)
  2. No; det⁡(A)=2≠0\det(A)=2\neq 0, so AA is singular.
  3. Yes; det⁡(A)=0\det(A)=0, so AA is non-singular.
  4. No; det⁡(A)=0\det(A)=0, so AA is singular.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [4 1; 2 1], we compute det(A) = ad - bc = (4)(1) - (1)(2) = 4 - 2 = 2. Since det(A) = 2 ≠ 0, matrix A is non-singular and invertible. Choice A is correct because it properly calculates det(A) = 2 ≠ 0 and correctly applies the invertibility criterion. Choice D incorrectly claims det(A) = 0, making an arithmetic error in calculating (4)(1) - (1)(2), and wrongly concludes the matrix is singular. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test. Any non-zero determinant, whether positive or negative, indicates the matrix is invertible.

Question 3

Let A=[2−134]A=\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix} and let I2=[1001]I_2=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} be the 2×22\times2 identity matrix. For matrix AA, what is A⋅I2A\cdot I_2?

  1. [1001]\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}
  2. [0000]\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}
  3. [2−134]\begin{bmatrix}2 & -1\\3 & 4\end{bmatrix} (correct answer)
  4. [2134]\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}
Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. The identity matrix for 2×2 is I = [[1,0],[0,1]]. When we compute A·I, where A = [[2,-1],[3,4]], the result is [[21 + (-1)0, 20 + (-1)1], [31 + 40, 30 + 41]] = [[2,-1],[3,4]], which equals A, confirming the identity property. Choice C is correct because it properly applies the identity property, yielding A itself. Choice A states that A·I = I, but the identity property is A·I = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 4

Let A=[3524]A=\begin{bmatrix}3 & 5\\ 2 & 4\end{bmatrix}. Based on the determinant, does matrix AA have an inverse? Why or why not?

  1. No; det⁡(A)=0\det(A)=0, so AA is singular.
  2. Yes; det⁡(A)=2≠0\det(A)=2\neq 0, so AA is non-singular.
  3. Yes; det⁡(A)=2≠0\det(A)=2\neq 0, so AA has an inverse. (correct answer)
  4. No; det⁡(A)=2≠0\det(A)=2\neq 0, so AA has no inverse.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [[3,5],[2,4]], we compute det(A) = ad - bc = (3)(4) - (5)(2) = 12 - 10 = 2. Since det(A) = 2, and this is not equal to zero, matrix A is non-singular and invertible. Choice C is correct because it correctly applies invertibility criterion. Choice D incorrectly claims the matrix has no inverse when det(A)=2≠0, but the criterion is that det(A)=0 means singular (no inverse) and det(A)≠0 means invertible. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 5

Let A=[1−235]A=\begin{bmatrix}1 & -2\\ 3 & 5\end{bmatrix} and let I2=[1001]I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}. For matrix AA, what is I2⋅AI_2\cdot A (matrix multiplication)?

  1. [1001]\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}
  2. [0000]\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}
  3. [1−235]\begin{bmatrix}1 & -2\\ 3 & 5\end{bmatrix} (correct answer)
  4. [−12−3−5]\begin{bmatrix}-1 & 2\\ -3 & -5\end{bmatrix}
Explanation: This question tests understanding of identity matrices and their properties. The identity matrix I is a square matrix with 1s on the main diagonal and 0s elsewhere, serving as the multiplicative identity: for any matrix A of compatible dimensions, A·I = I·A = A, just as multiplying a number by 1 leaves it unchanged. For matrix A = [[1, -2], [3, 5]] and I2 = [[1, 0], [0, 1]], when we compute I2·A, the result is [[(11 + 03), (1*-2 + 05)], [(01 + 13), (0-2 + 1*5)]] = [[1, -2], [3, 5]], confirming the identity property. Choice C is correct because it correctly states the identity property by showing I2·A = A. Choice A states that I2·A = I2, but the identity property is I2·A = A—multiplying by the identity matrix leaves A unchanged, not replaces it with I2. Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 6

Given N=[1224]N=\begin{bmatrix}1 & 2\\ 2 & 4\end{bmatrix}, why is matrix NN singular (has no inverse)?

  1. Because det⁡(N)=1⋅4−2⋅2=0\det(N)=1\cdot4-2\cdot2=0. (correct answer)
  2. Because det⁡(N)=1⋅4+2⋅2=8\det(N)=1\cdot4+2\cdot2=8.
  3. Because det⁡(N)=2⋅2−1⋅4=4\det(N)=2\cdot2-1\cdot4=4.
  4. Because det⁡(N)≠0\det(N)\neq 0.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix N = [[1,2],[2,4]], we compute det(N) = 14 - 22 = 4 - 4 = 0. Since det(N) = 0, and this is equal to zero, matrix N is singular and has no inverse. Choice A is correct because it properly calculates det(N) = ad - bc = 0. Choice B reverses the subtraction in the determinant formula, computing 14 + 22 = 8 instead of 14 - 22. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible.

Question 7

Let A=[50−21]A=\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix} and let O2=[0000]O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix} be the 2×22\times2 zero matrix. For matrix AA, what is A+O2A+O_2?

  1. [0000]\begin{bmatrix}0&0\\0&0\end{bmatrix}
  2. [50−21]\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix} (correct answer)
  3. [502−1]\begin{bmatrix}5 & 0\\ 2 & -1\end{bmatrix}
  4. [1001]\begin{bmatrix}1&0\\0&1\end{bmatrix}
Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. The zero matrix O = [[0,0],[0,0]]. When we add A + O = [[5,0],[-2,1]] + [[0,0],[0,0]] = [[5+0,0+0],[-2+0,1+0]] = [[5,0],[-2,1]], each entry of A is added to 0, giving result equals A, confirming that O is the additive identity. Choice B is correct because it correctly states the zero matrix property that A + O = A. Choice A claims A + O = O, but the zero matrix is the additive identity, so A + O = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 8

Let I3=[100010001]I_3=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}. What is det⁡(I3)\det(I_3)?

  1. 00
  2. 33
  3. 11 (correct answer)
  4. −1-1
Explanation: This question tests understanding of determinant properties. The determinant has the property that det(AB) = det(A)·det(B) for square matrices A and B, det(I) = 1 for the identity matrix, and det(O) = 0 for the zero matrix, with the non-zero determinant criterion determining invertibility. The identity matrix for 3×3 is I = [[1,0,0],[0,1,0],[0,0,1]]. For det(I3), it equals 1 because the identity matrix is invertible, and its determinant is always 1. Choice C is correct because it correctly states det(I) = 1. Choice A claims det(I) = 0, but the determinant of the identity matrix is always 1. Determinant properties: det(I) = 1 (identity always invertible), det(O) = 0 (zero matrix never invertible), and det(AB) = det(A)·det(B) (determinants multiply, not add). For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 9

Let A=[50−21]A=\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix} and let O2=[0000]O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix} be the 2×22\times2 zero matrix. For matrix AA, what is A+O2A+O_2?​

  1. [0000]\begin{bmatrix}0&0\\0&0\end{bmatrix}
  2. [50−21]\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix} (correct answer)
  3. [502−1]\begin{bmatrix}5 & 0\\ 2 & -1\end{bmatrix}
  4. [1001]\begin{bmatrix}1&0\\0&1\end{bmatrix}
Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. The zero matrix O = [[0,0],[0,0]]. When we add A + O = [[5,0],[-2,1]] + [[0,0],[0,0]] = [[5+0,0+0],[-2+0,1+0]] = [[5,0],[-2,1]], each entry of A is added to 0, giving result equals A, confirming that O is the additive identity. Choice B is correct because it correctly states the zero matrix property that A + O = A. Choice A claims A + O = O, but the zero matrix is the additive identity, so A + O = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 10

Let AA be a 2×22 \times 2 matrix with det⁡(A)=−3\det(A) = -3. Based on the determinant, is AA invertible?

  1. Yes, because det⁡(A)≠0\det(A) \neq 0. (correct answer)
  2. No, because det⁡(A)<0\det(A) < 0.
  3. No, because det⁡(A)≠1\det(A) \neq 1.
  4. Yes, because det⁡(A)=0\det(A) = 0.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×22 \times 2 matrix A=[ab;cd]A = [a b; c d] is calculated as det⁡(A)=ad−bc\det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det⁡(A)≠0\det(A) \neq 0, the matrix is invertible (non-singular), but if det⁡(A)=0\det(A) = 0, the matrix is singular (no inverse exists). Since det⁡(A)=−3\det(A) = -3, and this is not equal to zero, matrix A is non-singular and invertible. Choice A is correct because it correctly applies the invertibility criterion. Choice B incorrectly claims the matrix is not invertible when det⁡(A)=−3\det(A) = -3, but the criterion is that det⁡(A)=0\det(A) = 0 means singular (no inverse) and det⁡(A)≠0\det(A) \neq 0 means invertible. For invertibility, just compute the determinant: if det⁡(A)≠0\det(A) \neq 0, the matrix is invertible; if det⁡(A)=0\det(A) = 0, the matrix is singular and has no inverse—this is the key test. Key to determinants: for a 2×22 \times 2 matrix [ab;cd][a b; c d], always use det⁡=ad−bc\det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det⁡=0\det = 0 means the matrix has no inverse (singular) while det⁡≠0\det \neq 0 means it's invertible.

Question 11

Let O2=[0000]O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}. What is det⁡(O2)\det(O_2)?

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. 22
Explanation: This question tests understanding of determinants and zero matrices. The determinant has the property that det(AB) = det(A)·det(B) for square matrices A and B, det(I) = 1 for the identity matrix, and det(O) = 0 for the zero matrix, with the non-zero determinant criterion determining invertibility. For the 2×2 zero matrix O_2 = [[0,0],[0,0]], det(O2O_2) = 00 - 00 = 0. Choice B is correct because it correctly states that det(O) = 0. Choice A claims det(O) = 1, but the determinant of the zero matrix is always 0. Determinant properties: det(I) = 1 (identity always invertible), det(O) = 0 (zero matrix never invertible), and det(AB) = det(A)·det(B) (determinants multiply, not add). For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 12

Let A=[52−13]A=\begin{bmatrix}5&2\\-1&3\end{bmatrix} and let OO be the 2×22\times 2 zero matrix O=[0000]O=\begin{bmatrix}0&0\\0&0\end{bmatrix}. For matrix AA, what is A+OA+O?

  1. [0000]\begin{bmatrix}0&0\\0&0\end{bmatrix}
  2. [52−13]\begin{bmatrix}5&2\\-1&3\end{bmatrix} (correct answer)
  3. [5003]\begin{bmatrix}5&0\\0&3\end{bmatrix}
  4. [5213]\begin{bmatrix}5&2\\1&3\end{bmatrix}
Explanation: This question tests understanding of zero matrices and their properties. The zero matrix O has all entries equal to zero and serves as the additive identity for matrices: A + O = O + A = A for any matrix A of the same dimensions, analogous to adding 0 to a number. When we add A + O = [5 2; -1 3] + [0 0; 0 0], we add corresponding entries: (5+0)=5, (2+0)=2, (-1+0)=-1, and (3+0)=3, giving us [5 2; -1 3], which equals A. Choice B is correct because it properly shows that A + O = A = [5 2; -1 3], confirming that O is the additive identity. Choice A incorrectly gives the zero matrix as the answer, claiming A + O = O, but the zero matrix is the additive identity, so A + O = A (adding zero doesn't change A). Remember the roles of special matrices: the identity matrix I is the multiplicative identity (AI = IA = A), while the zero matrix O is the additive identity (A + O = A) and absorbs under multiplication (A·O = O). Don't confuse the additive identity (zero matrix O where A + O = A) with the multiplicative identity (identity matrix I where A·I = A)—they play analogous roles to 0 and 1 for real numbers.

Question 13

Given M=[2412]M=\begin{bmatrix}2 & 4\\ 1 & 2\end{bmatrix}, does matrix MM have an inverse? Why or why not? (Use det⁡([abcd])=ad−bc\det\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=ad-bc.)

  1. Yes, because det⁡(M)=2⋅2+4⋅1=8≠0\det(M)=2\cdot2+4\cdot1=8\neq0.
  2. No, because det⁡(M)=2⋅2−4⋅1=0\det(M)=2\cdot2-4\cdot1=0. (correct answer)
  3. Yes, because det⁡(M)=4⋅1−2⋅2=0\det(M)=4\cdot1-2\cdot2=0.
  4. No, because det⁡(M)=2⋅2−4⋅1=4≠0\det(M)=2\cdot2-4\cdot1=4\neq0.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix M = [[2,4],[1,2]], we compute det(M) = 22 - 41 = 4 - 4 = 0. Since det(M) = 0, and this is equal to zero, matrix M is singular and has no inverse. Choice B is correct because it correctly applies the invertibility criterion. Choice A reverses the subtraction in the determinant formula, computing 22 + 41 instead of 22 - 41, which gives the wrong value. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible.

Question 14

Suppose AA and BB are 2×22\times 2 matrices with det⁡(A)=3\det(A)=3 and det⁡(B)=−2\det(B)=-2. What is det⁡(AB)\det(AB)?​​

  1. 11
  2. −5-5
  3. −6-6 (correct answer)
  4. 66
Explanation: This question tests understanding of determinant properties. The determinant has the property that det(AB) = det(A)·det(B) for square matrices A and B, det(I) = 1 for the identity matrix, and det(O) = 0 for the zero matrix, with the non-zero determinant criterion determining invertibility. Given det(A) = 3 and det(B) = -2, we use the property det(AB) = det(A)·det(B) = (3)(-2) = -6. Choice C is correct because it properly applies the multiplicative property of determinants: det(AB) = det(A)·det(B) = 3·(-2) = -6. Choice A uses det(AB) = det(A) + det(B) = 3 + (-2) = 1 (addition), when the correct property is det(AB) = det(A)·det(B) (multiplication). Determinant properties: det(I) = 1 (identity always invertible), det(O) = 0 (zero matrix never invertible), and det(AB) = det(A)·det(B) (determinants multiply, not add). This multiplicative property is crucial for understanding how determinants behave under matrix multiplication—they multiply together, preserving information about invertibility.

Question 15

For the 2×22\times2 matrix A=[412−1]A=\begin{bmatrix}4&1\\2&-1\end{bmatrix}, does matrix AA have an inverse? Why or why not?

  1. No; det⁡(A)=0\det(A)=0, so AA is singular.
  2. Yes; det⁡(A)=0\det(A)=0, so AA is invertible.
  3. Yes; det⁡(A)=−6≠0\det(A)=-6\neq 0, so AA is invertible. (correct answer)
  4. No; det⁡(A)=−6≠0\det(A)=-6\neq 0, so AA is singular.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [4 1; 2 -1], we compute det(A) = ad - bc = (4)(-1) - (1)(2) = -4 - 2 = -6. Since det(A) = -6 ≠ 0, matrix A is non-singular and invertible. Choice C is correct because it properly calculates det(A) = -6 and correctly concludes that since det(A) ≠ 0, the matrix is invertible. Choice A incorrectly calculates the determinant as 0, while choice D makes the logical error of claiming that a non-zero determinant means singular. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 16

Let A=(2−143)A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix} and B=(1001)B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} . If C=A⋅B−B⋅AC = A \cdot B - B \cdot A, what is the determinant of matrix CC?

  1. 00 (correct answer)
  2. 1010
  3. −10-10
  4. 11
Explanation: Since BB is the identity matrix, A⋅B=AA \cdot B = A and B⋅A=AB \cdot A = A. Therefore, C=A−A=(0000)C = A - A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} , which is the zero matrix. The determinant of the zero matrix is 00. Choice B is the determinant of AA (2⋅3−(−1)⋅4=102 \cdot 3 - (-1) \cdot 4 = 10). Choice C reflects a sign error in computing the determinant of AA. Choice D assumes CC equals the identity matrix.

Question 17

Given A=[4121]A=\begin{bmatrix}4 & 1\\ 2 & 1\end{bmatrix}, does matrix AA have an inverse? Why or why not?

  1. Yes; AA has an inverse because det⁡(A)=2≠0\det(A)=2\neq 0. (correct answer)
  2. No; AA has no inverse because det⁡(A)=2≠0\det(A)=2\neq 0.
  3. Yes; AA has an inverse because det⁡(A)=0\det(A)=0.
  4. No; AA has no inverse because det⁡(A)=0\det(A)=0.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [[4, 1], [2, 1]], we compute det(A) = ad - bc = (4)(1) - (1)(2) = 4 - 2 = 2. Since det(A) = 2, and this is not equal to zero, matrix A is non-singular and invertible. Choice A is correct because it correctly applies invertibility criterion. Choice B incorrectly claims the matrix has no inverse because det(A) ≠ 0, but the criterion is that det(A) = 0 means singular (no inverse) and det(A) ≠ 0 means invertible. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 18

Given the 2×22\times2 matrix A=[1236]A=\begin{bmatrix}1 & 2\\3 & 6\end{bmatrix}, does matrix AA have an inverse? Why or why not?

  1. Yes; det⁡(A)=12≠0\det(A)=12\ne0, so AA is invertible.
  2. No; det⁡(A)=0\det(A)=0, so AA is singular. (correct answer)
  3. Yes; det⁡(A)=0\det(A)=0, so AA is invertible.
  4. No; det⁡(A)=12≠0\det(A)=12\ne0, so AA is singular.
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [[1,2],[3,6]], we compute det(A) = ad - bc = (1)(6) - (2)(3) = 6 - 6 = 0. Since det(A) = 0, and this is equal to zero, matrix A is singular and has no inverse. Choice B is correct because it correctly applies the invertibility criterion. Choice A incorrectly claims the matrix is invertible when det(A) = 12, but the criterion is that det(A) = 0 means singular (no inverse) and det(A) ≠ 0 means invertible. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 19

Given the 2×22\times 2 matrix A=[3124]A=\begin{bmatrix}3 & 1\\ 2 & 4\end{bmatrix}, what is the determinant det⁡(A)\det(A)? (Use det⁡([abcd])=ad−bc\det\left(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\right)=ad-bc.)

  1. 1010 (correct answer)
  2. −10-10
  3. 1414
  4. 55
Explanation: This question tests understanding of determinants and invertibility. The determinant of a 2×2 matrix A = [a b; c d] is calculated as det(A) = ad - bc, giving a scalar value that determines whether the matrix has an inverse: if det(A) ≠ 0, the matrix is invertible (non-singular), but if det(A) = 0, the matrix is singular (no inverse exists). For matrix A = [[3, 1], [2, 4]], we compute det(A) = ad - bc = (3)(4) - (1)(2) = 12 - 2 = 10. Choice A is correct because it properly calculates det(A) = ad - bc. Choice B reverses the subtraction in the determinant formula, computing bc - ad instead of ad - bc, which gives the wrong sign. Key to determinants: for a 2×2 matrix [a b; c d], always use det = ad - bc (products along main diagonal minus products along anti-diagonal), and remember that det = 0 means the matrix has no inverse (singular) while det ≠ 0 means it's invertible. For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.

Question 20

Let O2=[0000]O_2=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} be the 2×22\times 2 zero matrix. What is det⁡(O2)\det(O_2)?

  1. 11
  2. 00 (correct answer)
  3. −1-1
  4. 22
Explanation: This question tests understanding of determinant properties. The determinant has the property that det(AB) = det(A)·det(B) for square matrices A and B, det(I) = 1 for the identity matrix, and det(O) = 0 for the zero matrix, with the non-zero determinant criterion determining invertibility. The zero matrix O = [[0,0],[0,0]]. For det(O2), it equals 0 because the zero matrix is singular and never invertible. Choice B is correct because it correctly states det(O) = 0. Choice A claims det(O) = 1, but the determinant of the zero matrix is always 0. Determinant properties: det(I) = 1 (identity always invertible), det(O) = 0 (zero matrix never invertible), and det(AB) = det(A)·det(B) (determinants multiply, not add). For invertibility, just compute the determinant: if det(A) ≠ 0, the matrix is invertible; if det(A) = 0, the matrix is singular and has no inverse—this is the key test.