Zero and Identity Matrices and Determinants - Precalculus

Let $A=\begin{bmatrix}3 & 5\\ 2 & 4\end{bmatrix}$ . Based on the determinant, does matrix $A$ have an inverse? Why or why not?
Answer: Yes; $\det(A)=2\neq 0$ , so $A$ has an inverse. 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.
Let $I_3=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ . What is $\det(I_3)$ ?
Answer: $1$ 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.
Let $M=\begin{bmatrix}2 & 4\\ 1 & 2\end{bmatrix}$ . Does matrix $M$ have an inverse? Why or why not? (Use the determinant to decide.)
Answer: No; $M$ has no inverse because $\det(M)=0$ . 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) = ad - bc = (2)(2) - (4)(1) = 4 - 4 = 0. Since det(M) = 0, and this is equal to zero, matrix M is singular and has no inverse. Choice C is correct because it correctly applies invertibility criterion. Choice D incorrectly claims the matrix has no inverse because det(M) ≠ 0, but the criterion is that det(M) = 0 means singular (no inverse) and det(M) ≠ 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.
Let $A = \begin{pmatrix} 2 & -1 \\ 4 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . If $C = A \cdot B - B \cdot A$ , what is the determinant of matrix $C$ ?
Answer: $0$ Since $B$ is the identity matrix, $A \cdot B = A$ and $B \cdot A = A$ . Therefore, $C = A - A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ , which is the zero matrix. The determinant of the zero matrix is $0$ . Choice B is the determinant of $A$ ( $2 \cdot 3 - (-1) \cdot 4 = 10$ ). Choice C reflects a sign error in computing the determinant of $A$ . Choice D assumes $C$ equals the identity matrix.
Let $A$ and $B$ be $2\times2$ matrices with $\det(A)=3$ and $\det(B)=-2$ . What is $\det(AB)$ ?
Answer: $-6$ 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 B is correct because it properly applies the multiplicative property of determinants: det(AB) = det(A)·det(B) = 3·(-2) = -6. Choice A would result from adding the determinants (3 + (-2) = 1) instead of multiplying them, a common error. 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.
Let $O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ be the $2\times 2$ zero matrix. What is $\det(O_2)$ ?
Answer: $0$ This question tests understanding of zero matrices and their properties. 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 the zero matrix O₂ = [0 0; 0 0], we compute det(O₂) = (0)(0) - (0)(0) = 0 - 0 = 0. Choice B is correct because det(O) = 0 for any zero matrix—all entries are zero, so any product of entries is zero. Choice A incorrectly claims det(O₂) = 1, confusing the zero matrix with the identity matrix, but only the identity matrix has determinant 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). The zero matrix having determinant 0 confirms it's singular (non-invertible)—you cannot find a matrix that, when multiplied by the zero matrix, gives the identity.
Let $A$ be a $2 \times 2$ matrix with $\det(A) = -3$ . Based on the determinant, is $A$ invertible?
Answer: Yes, because $\det(A) \neq 0$ . This question tests understanding of determinants and invertibility. The determinant of a $2 \times 2$ 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 $\det(A) \neq 0$ , the matrix is invertible (non-singular), but if $\det(A) = 0$ , the matrix is singular (no inverse exists). Since $\det(A) = -3$ , 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 $\det(A) = -3$ , but the criterion is that $\det(A) = 0$ means singular (no inverse) and $\det(A) \neq 0$ means invertible. For invertibility, just compute the determinant: if $\det(A) \neq 0$ , 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 $2 \times 2$ 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 $\det \neq 0$ means it's invertible.
Let $A$ be any $2\times2$ matrix and let $O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . What property does the zero matrix have under multiplication (with compatible dimensions), i.e., what is $A\cdot O_2$ ?
Answer: $A\cdot O_2=O_2$ 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. For multiplication, A·O = [[show result is O]], showing that the zero matrix 'absorbs' under multiplication. Choice C is correct because it correctly states the zero matrix property under multiplication. Choice A claims A·O = A, but the zero matrix absorbs under multiplication, so A·O = O. 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.
Given the $2\times2$ matrix $M=\begin{bmatrix}2&4\\1&2\end{bmatrix}$ , does matrix $M$ have an inverse? Why or why not?
Answer: No; $\det(M)=0$ , so $M$ is singular. 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) = ad - bc = (2)(2) - (4)(1) = 4 - 4 = 0. Since det(M) = 0, matrix M is singular and has no inverse. Choice B is correct because it correctly identifies that det(M) = 0, making M singular (non-invertible). Choice A incorrectly calculates the determinant as 4, while choice C makes the logical error of claiming that det = 0 means invertible (the opposite is true). 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.
Given $N=\begin{bmatrix}1 & 2\\ 2 & 4\end{bmatrix}$ , why is matrix $N$ singular (has no inverse)?
Answer: Because $\det(N)=1\cdot4-2\cdot2=0$ . 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) = 1*4 - 2*2 = 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 1*4 + 2*2 = 8 instead of 1*4 - 2*2. 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.
Let $A=\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix}$ and let $O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ be the $2\times2$ zero matrix. For matrix $A$ , what is $A+O_2$ ?
Answer: $\begin{bmatrix}5 & 0\\ -2 & 1\end{bmatrix}$ 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.
Let $A=\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ and let $I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ be the $2\times 2$ identity matrix. For matrix $A$ , what is $A\cdot I_2$ (matrix multiplication)?
Answer: $\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ 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 = [[2, -1], [3, 4]] and I2 = [[1, 0], [0, 1]], when we compute A·I2, the result is [[(2*1 + -1*0), (2*0 + -1*1)], [(3*1 + 4*0), (3*0 + 4*1)]] = [[2, -1], [3, 4]], confirming the identity property. Choice C is correct because it correctly states the identity property by showing A·I2 = A. Choice A states that A·I2 = I2, but the identity property is A·I2 = 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.
Let $O_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . What is $\det(O_2)$ ?
Answer: $0$ 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( $O_2$ ) = 0*0 - 0*0 = 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.
Let $A=\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ be a $2\times 2$ matrix and let $I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ be the $2\times 2$ identity matrix. For matrix $A$ , what is $A\cdot I_2$ ?
Answer: $\begin{bmatrix}2 & -1\\ 3 & 4\end{bmatrix}$ 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 = [[2,-1],[3,4]] · [[1,0],[0,1]] = [[2*1 + (-1)*0, 2*0 + (-1)*1], [3*1 + 4*0, 3*0 + 4*1]] = [[2,-1],[3,4]], we get back A, confirming the identity property. Choice C is correct because it correctly states the identity property by showing A·I2 equals A. 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.
Let $A$ be any $2\times2$ matrix and let $O$ be the $2\times2$ zero matrix $O=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ . What property does the zero matrix have under multiplication, i.e., what is $A\cdot O$ ?
Answer: $A\cdot O=O$ This question tests understanding of zero matrices and their properties under multiplication. 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. For multiplication, when any matrix A is multiplied by the zero matrix O, the result is always the zero matrix: A·O = O. This is because each entry in the product is computed as a sum of products, and since all entries of O are zero, every sum equals zero. Choice A is correct because A·O = O for any matrix A—the zero matrix 'absorbs' under multiplication. Choice B incorrectly states A·O = A, confusing the multiplication property with the addition property (A + O = A). Remember that while the zero matrix is the additive identity (A + O = A), it acts as an absorbing element under multiplication (A·O = O·A = O), similar to how multiplying any number by 0 gives 0.
Suppose $A$ and $B$ are $2 \times 2$ matrices with $\det(A)=3$ and $\det(B)=-2$ . What is $\det(AB)$ ?
Answer: $-6$ This question tests understanding of determinant properties. The determinant has the property that $\det(AB) = \det(A) \cdot \det(B)$ 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 $\det(AB) = \det(A) \cdot \det(B) = 3 \cdot (-2) = -6$ . Choice B is correct because it correctly applies the determinant property $\det(AB) = \det(A) \cdot \det(B)$ . Choice D uses $\det(AB) = \det(A) + \det(B) = 3 + (-2) = 1$ , but the correct property is $\det(AB) = \det(A) \cdot \det(B)$ (multiplication). Determinant properties: $\det(I) = 1$ (identity always invertible), $\det(O) = 0$ (zero matrix never invertible), and $\det(AB) = \det(A) \cdot \det(B)$ (determinants multiply, not add). Key to determinants: for a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , 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 $\det \neq 0$ means it's invertible.
Let $A=\begin{bmatrix}5&-2\\0&3\end{bmatrix}$ and let the $2\times2$ identity matrix be $I_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ . For matrix $A$ , what is $A\cdot I_2$ ?
Answer: $\begin{bmatrix}5&-2\\0&3\end{bmatrix}$ 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 I₂ = [1 0; 0 1]. When we compute A·I₂ = [5 -2; 0 3]·[1 0; 0 1], each row of A multiplied by each column of I₂ gives us back the original matrix A = [5 -2; 0 3], confirming the identity property. Choice C is correct because it correctly states that A·I₂ = A, showing that multiplying by the identity matrix leaves A unchanged. Choice A gives the identity matrix itself, confusing the result with the multiplier, while choice B gives the zero matrix, confusing additive and multiplicative identities. 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).
If $M$ is a $2 \times 2$ matrix such that $M + 3I_2$ has determinant $8$ and $M$ has determinant $2$ , what is the trace of matrix $M$ ?
Answer: $-1$ Let $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . Then $M + 3I_2 = \begin{pmatrix} a+3 & b \\ c & d+3 \end{pmatrix}$ . We have $\det(M + 3I_2) = (a+3)(d+3) - bc = ad + 3a + 3d + 9 - bc = \det(M) + 3(a+d) + 9 = 8$ . Since $\det(M) = 2$ , we get $2 + 3(a+d) + 9 = 8$ , so $3(a+d) = -3$ , giving $a+d = -1$ . The trace is $a+d = -1$ . Other choices result from computational errors in expanding the determinant or applying the given conditions.
Given the $2\times 2$ matrix $A=\begin{bmatrix}3 & 1\\ 2 & 4\end{bmatrix}$ , what is the determinant $\det(A)$ ? (Use $\det\left(\begin{bmatrix}a & b\\ c & d\end{bmatrix}\right)=ad-bc$ .)
Answer: $10$ 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.
Let $A=\begin{bmatrix}1 & -2\\ 3 & 5\end{bmatrix}$ and let $I_2=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ . For matrix $A$ , what is $I_2\cdot A$ (matrix multiplication)?
Answer: $\begin{bmatrix}1 & -2\\ 3 & 5\end{bmatrix}$ 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 [[(1*1 + 0*3), (1*-2 + 0*5)], [(0*1 + 1*3), (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.

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