For any square matrices A,B of same order, det(I+AB) equals det(I+BA) when
A Only if invertible
B Only if symmetric
C Always true
D Only for 2×2
AB and BA have the same nonzero eigenvalues (with multiplicities). Hence characteristic polynomials match for (I+AB) and (I+BA), giving det(I+AB)=det(I+BA) for same order.
If A is 3×3 with A²=0 and A≠0, then rank(A) can be
A 3 only
B 1 only
C 2 only
D 0 only
A²=0 implies Im(A) ⊆ Ker(A). For 3×3, rank(A) ≤ nullity(A), and rank+nullity=3. This forces rank(A) ≤ 1; since A≠0, rank(A)=1.
For 3×3 matrix A, if A³=0 but A²≠0, rank(A) can be
A 2 only
B 1 only
C 0 only
D 3 only
If A³=0 and A²≠0 (nilpotent index 3), the Jordan form must contain a 3×3 nilpotent block, which has rank 2. So rank(A)=2 in this case.
If A is 2×2 and A²=0, then A must be
A Identity
B Zero matrix
C Orthogonal
D Nonzero possible
A nonzero 2×2 nilpotent matrix exists, e.g., [[0,1],[0,0]] where A²=0 but A≠0. Such matrices are singular and have determinant 0.
For invertible A and B, det(A⁻¹BA) equals
A det(A)det(B)
B det(B)
C det(B)/det(A)
D 1
det(A⁻¹BA)=det(A⁻¹)det(B)det(A)=(1/det(A))·det(B)·det(A)=det(B). This shows determinant is invariant under similarity transformations.
If A is 3×3 and det(A)=2, then det(A⁻¹+A⁻ᵀ) can be simplified using
A Trace rule
B Swap rows
C Factor A⁻¹
D Cramer’s rule
A⁻¹ + A⁻ᵀ = A⁻¹(I + (A⁻¹)ᵀA). Factoring helps use determinant properties det(PQ)=det(P)det(Q) to reduce computation, especially for expressions with inverses.
If A is invertible and Aᵀ = A⁻¹, then A is
A Nilpotent
B Orthogonal
C Singular
D Skew-symmetric
Aᵀ=A⁻¹ implies AᵀA=I. This is exactly the definition of an orthogonal matrix. Such matrices preserve lengths and have determinant ±1.
If A is 3×3, det(A)=1 and A is orthogonal, then det(A) must be
A −1
B 0
C 1
D Any real
Orthogonal matrices have det(A)=±1. Given det(A)=1 already, it must be +1. Such matrices represent proper rotations (no reflection).
If A is 3×3 and adj(A)=0, then rank(A) must be
A 3
B 2
C Exactly 2
D ≤1
For 3×3, entries of adj(A) are 2×2 cofactors. adj(A)=0 means all 2×2 minors are zero, which implies rank(A)≤1 (cannot reach rank 2).
If A is 3×3 and det(A)=0 but adj(A)≠0, then rank(A) is
A 3
B 2
C 1
D 0
For 3×3, det(A)=0 means rank<3. If rank≤1 then adj(A)=0. Since adj(A)≠0, rank cannot be ≤1, so rank(A)=2.
If A is 4×4 and rank(A)=3, then det(A) is
A Zero
B Nonzero
C Always 1
D Always −1
For an n×n matrix, full rank n is required for det≠0. If rank(A)=3<4, rows are dependent, so det(A)=0 and A is singular.
If A is 2×2 with A²=I and trace(A)=0, then det(A) is
A 1
B 0
C −1
D 2
A²=I implies eigenvalues are ±1. Trace is sum of eigenvalues; trace 0 means eigenvalues are 1 and −1. Determinant is product, giving det(A)=−1.
For 2×2 matrix A, if A²=A and A≠0,I, then det(A) equals
A 1
B 0
C −1
D 2
Idempotent matrices have eigenvalues 0 or 1. If A is not 0 or I, it must have both eigenvalues 0 and 1. Product is 0, so det(A)=0.
If A is 3×3 with det(A)=5, then det(adj(A)) equals
A 5
B 125
C 1/25
D 25
For n×n, det(adj(A))=(det(A))^(n−1). Here n=3, so exponent is 2. Thus det(adj(A))=5²=25.
For 4×4 matrix A with det(A)=2, det(adj(A)) equals
A 4
B 8
C 16
D 64
det(adj(A))=(det(A))^(n−1). For 4×4, exponent is 3. So det(adj(A))=2³=8. This holds for all square matrices.
If A is invertible, then adj(A⁻¹) equals
A adj(A)⁻¹
B (1/det(A))adj(A)
C det(A)A
D det(A)A⁻¹
Using adj(A)=det(A)A⁻¹ for invertible A, we can derive adj(A⁻¹)=det(A⁻¹)(A⁻¹)⁻¹=(1/det(A))A. Also equals (adj(A))⁻¹.
If A is invertible, then adj(A) equals
A A/det(A)
B det(A)A
C det(A)A⁻¹
D A⁻¹/det(A)
From A·adj(A)=det(A)I, multiply by A⁻¹ to get adj(A)=det(A)A⁻¹. This identity is valid only when A is invertible.
If A is 3×3 and det(A)=1, then det(adj(A)) is
A 0
B 1
C 3
D −1
det(adj(A))=(det(A))^(n−1). With n=3 and det(A)=1, we get det(adj(A))=1²=1. Adjoint keeps determinant 1 for this case.
If A is 2×2, det(A)=3 and det(B)=2, then det(A⁻¹BA) equals
A 3
B 6
C 1/6
D 2
A⁻¹BA is similar to B, so determinant is unchanged: det(A⁻¹BA)=det(B)=2. Similar matrices have the same determinant and trace.
If A is 3×3 and det(A)=0, then det(adj(A)) must be
A 1
B −1
C 0
D Cannot decide
det(adj(A))=(det(A))^(n−1). If det(A)=0, any positive power remains 0. So det(adj(A))=0 for all singular square matrices.
If A is 3×3 and det(A)=2, then det(2A⁻¹) equals
A 1
B 4
C 2
D 8
det(A⁻¹)=1/det(A)=1/2. For 3×3, det(2A⁻¹)=2³·(1/2)=8·(1/2)=4. Use scaling power and inverse rule correctly.
If A is 4×4 with det(A)=2, then det(2A⁻¹) equals
A 1
B 2
C 8
D 4
For 4×4, scaling by 2 multiplies determinant by 2⁴=16. Since det(A⁻¹)=1/2, det(2A⁻¹)=16×1/2=8.
If A is 2×2 and det(A)=2, then det(2A⁻¹) equals
A 1
B 2
C 4
D 1/2
For 2×2, det(2A⁻¹)=2²·det(A⁻¹)=4·(1/2)=2. Determinant scaling uses power equal to order, then multiply by inverse determinant.
If det(A)=d for 3×3 invertible A, then det(A⁻ᵀ) equals
A 1/d
B d
C d²
D −d
A⁻ᵀ means (Aᵀ)⁻¹. det(Aᵀ)=det(A)=d. So det(A⁻ᵀ)=1/det(Aᵀ)=1/d. Same as det(A⁻¹).
If A is 3×3 and det(A)=d, then det(AᵀA⁻¹) equals
A 0
B d
C 1
D d²
det(AᵀA⁻¹)=det(Aᵀ)det(A⁻¹)=d·(1/d)=1. This holds for every invertible square matrix, independent of size or sign.
If A is 2×2 and det(A)=d, then det(A + I) depends on
A det only
B trace and det
C trace only
D order only
For 2×2, characteristic polynomial is λ² − (trace A)λ + det(A). det(A+I) equals value at λ=−1, which involves both trace and determinant.
If A is 2×2 and A has eigenvalues 2 and 3, then det(A) is
A 5
B 1
C 0
D 6
Determinant equals product of eigenvalues for square matrices. With eigenvalues 2 and 3, det(A)=2×3=6. Trace would be their sum, 5.
If A is 3×3 with eigenvalues 1,1,2, then det(A) is
A 4
B 1
C 2
D 0
Determinant equals product of eigenvalues: 1×1×2=2. This gives a quick way to evaluate determinant when eigenvalues are known.
If A is 3×3 with eigenvalues 0,2,3, then A is
A Invertible
B Singular
C Orthogonal
D Idempotent
If any eigenvalue is 0, determinant becomes 0 because determinant is product of eigenvalues. So A is singular and has no inverse.
For invertible A,B, det(A⁻¹B⁻¹AB) equals
A det(A)
B det(B)
C det(A)det(B)
D 1
det(A⁻¹B⁻¹AB)=det(A⁻¹)det(B⁻¹)det(A)det(B)=(1/det(A))(1/det(B))det(A)det(B)=1. It is determinant of a commutator.
If AB=BA and both invertible, then (A+B)⁻¹ always exists when
A det(A+B)≠0
B det(A)+det(B)≠0
C trace(A+B)≠0
D rank(A+B)=1
Inverse exists precisely when determinant is nonzero. Even if A and B commute and are invertible, A+B can still be singular, so det(A+B) must be checked.
If A is 2×2 and A has trace 0 and det 1, then A satisfies
A A²=I
B A²=A
C A²=−I
D A³=0
Characteristic polynomial is λ² − (tr A)λ + det(A)=λ²+1. So A satisfies A²+I=0 by Cayley–Hamilton, giving A²=−I.
If A is 2×2 with trace 2 and det 1, then A satisfies
A A²+2A+I=0
B A²−2A+I=0
C A²−I=0
D A²=A
For 2×2, characteristic polynomial is λ² − (tr A)λ + det(A). With trace 2 and det 1, it is λ²−2λ+1=(λ−1)², so A satisfies A²−2A+I=0.
If A is 3×3 and A is orthogonal with det −1, it represents
A Proper rotation
B Nilpotent map
C Projection map
D Reflection-type
Orthogonal matrices preserve lengths. Determinant +1 indicates proper rotation, while determinant −1 indicates orientation reversal, typically a reflection or rotation combined with reflection.
For invertible A, det(AᵀA) is always
A Negative
B Positive
C Zero
D Equal to trace
det(AᵀA)=det(A)² which is always ≥0, and for invertible A, det(A)≠0 so det(A)²>0. Hence determinant is strictly positive.
If A is 3×3 and det(A)=−2, then det(AᵀA) equals
A −4
B −2
C 4
D 2
det(AᵀA)=det(A)². With det(A)=−2, square gives 4. This shows AᵀA has positive determinant, consistent with being symmetric positive definite when invertible.
For 3×3, det(adj(M))=(det(M))². Take M=A⁻¹, so det(M)=1/det(A). Hence det(adj(A⁻¹))=(1/det(A))².
If A is 2×2 invertible, adj(adj(A)) equals
A det(A)A
B (det A)⁰A
C det(A)I
D A
For 2×2, adj(A)=det(A)A⁻¹. Then adj(adj(A))=det(adj(A))·(adj(A))⁻¹. Here det(adj(A))=det(A), and (adj(A))⁻¹=(1/det(A))A, giving A.
If A is 3×3 invertible, adj(adj(A)) equals
A A
B (det A)²A
C det(A)A
D (det A)I
For 3×3, formula is adj(adj(A))=(det(A))^{(3−2)}A=(det(A))¹A=det(A)A. This is a standard adjoint-of-adjoint identity.
For n×n invertible A, adj(adj(A)) equals
A A
B (det A)^{n−2}A
C (det A)A
D (det A)^{n}A
Using adj(A)=det(A)A⁻¹ and applying adj again gives adj(adj(A))=(det(A))^{n−2}A for invertible A. It generalizes 2×2 and 3×3 cases.
If A is 2×2 with trace 0 and det 0, then A must be
A Orthogonal
B Identity
C Involutory
D Nilpotent
Characteristic polynomial is λ² − (tr A)λ + det(A)=λ². By Cayley–Hamilton, A²=0, so A is nilpotent. It may be zero or nonzero.
If A is 3×3 with Aᵀ = A and A² = I, then A represents
A Projection
B Rotation only
C Reflection
D Nilpotent
Symmetric and involutory (A²=I) implies A is orthogonally diagonalizable with eigenvalues ±1. Such matrices act like reflections across subspaces, common in geometry.
If A is 2×2 with Aᵀ=A and A is orthogonal, then A must satisfy
A A²=I
B A=I only
C A²=0
D det=0
Symmetric orthogonal implies Aᵀ=A and AᵀA=I, so A²=I. Hence such matrices are involutory, typically representing reflections (eigenvalues ±1).
If A is 3×3 and A is orthogonal, then Aᵀ = A⁻¹ implies
A A is singular
B columns orthonormal
C det(A)=0
D trace(A)=0
AᵀA=I means columns (and rows) form an orthonormal set: each column has length 1 and different columns are perpendicular. This property defines orthogonality.
If A is 3×3 and A has eigenvalue 0, then det(A) is
A 1
B −1
C Cannot decide
D 0
Determinant equals product of eigenvalues. If any eigenvalue is 0, the product becomes 0. Therefore det(A)=0 and A is singular, so inverse does not exist.
If A is 2×2 with eigenvalues 1 and 1, then A must be
A Identity only
B Singular always
C Not necessarily identity
D Nilpotent always
Having eigenvalues 1,1 implies trace 2 and det 1, but A could be non-diagonalizable, e.g., [[1,1],[0,1]]. So A need not be identity.
If A is 2×2 with eigenvalues 1 and −1, then A² equals
A 0 always
B I always
C −I always
D Depends on A
Eigenvalues of A² are squares of eigenvalues: 1² and (−1)² both become 1. For 2×2 with distinct eigenvalues 1 and −1, A is diagonalizable, so A²=I.
If A is 2×2 with trace 0 and det −1, then eigenvalues are
A 1, −1
B i, −i
C 0, 0
D 2, −2
Eigenvalues satisfy sum = trace = 0 and product = det = −1. The pair (1, −1) matches. So A is invertible and typically involutory-type after similarity.
If a 3×3 system has det(A)=0 and rank(A)=rank([A|B])=3, then it has
A Unique solution
B Impossible case
C No solution
D Infinite solutions
For 3 variables, rank cannot exceed 3. If rank(A)=3, then A has full rank, which forces det(A)≠0. So det(A)=0 with rank 3 cannot happen.