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.

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.

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) = 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.

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.

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],[0,0,0]]. When we add A + O = [[5+0,0+0,-2+0],[1+0,3+0,4+0]] = [[5,0,-2],[1,3,4]], each entry of A is added to 0, giving result equals A, confirming that O is the additive identity. For multiplication, A·O would result in the zero matrix if dimensions allow, showing that the zero matrix 'absorbs' under multiplication. Choice B is correct because it correctly states zero matrix property under addition. 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.

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]]. Choice C is correct because it correctly states the identity property with 1s on the diagonal and 0s elsewhere. Choice B gives the zero matrix or wrong matrix as the identity matrix, but the identity must have 1s on the main diagonal and 0s elsewhere. 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.

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.

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.

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 [[(21 + -10), (20 + -11)], [(31 + 40), (30 + 41)]] = [[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.

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, $A \cdot I = I \cdot A = A$, just as multiplying a number by 1 leaves it unchanged. 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. For the 3×3 identity matrix $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, $det(I_3) = 1$. Choice D is correct because it correctly states the identity property. 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 $det(AB) = det(A) \cdot det(B)$ (determinants multiply, not add). 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 ($A \cdot O = O$).