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Algebra & Matrices Mastery Hub: The Industry Foundation Prac

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Q1Domain Verified
Consider a square matrix $A$ such that $A^2 = A$. If the determinant of $A$ is $\det(
$A$ has eigenvalues of 0 and 1 only.
$A$ is the zero matrix.
$A$ is invertible.
= 1/2$, which of the following statements must be true? A) $A$ is the identity matrix.
Q2Domain Verified
asks which statement *must be true* given $\det(
$T$ has a non-trivial null space.
= 1/2$. A matrix is invertible if and only if its determinant is non-zero. Since $\det(A) = 1/2 \neq 0$, $A$ must be invertible. Option A is incorrect because if $A$ were the identity matrix ($I$), then $I^2 = I$ is true, but $\det(I) = 1$, not $1/2$. Option B is incorrect because if $A$ were the zero matrix ($O$), then $O^2 = O$ is true, but $\det(O) = 0$, not $1/2$. Option C is incorrect because, as shown, the condition $A^2=A$ implies eigenvalues are 0 or 1. If $\det(A)=1/2$, this would imply a product of eigenvalues that cannot be $1/2$ if they are restricted to 0 and 1. This highlights that the premise itself might be impossible for standard matrices, but the invertibility is a direct consequence of the non-zero determinant. Question: Let $V$ be a vector space and $T: V \to V$ be a linear transformation. If $T$ is surjective (onto), which of the following is necessarily true? A) $T$ is injective (one-to-one).
The kernel of $T$ is the zero vector space.
The image of $T$ is the zero vector space.
Q3Domain Verified
Consider the system of linear equations $Ax = b$. If the augmented matrix $[A | b]$ has rank $r$, and the coefficient matrix $A$ has rank $k$, what condition must hold for the system to have a unique solution?
$r = k$ and $k = n$, where $n$ is the number of variables.
$r < k$.
$r = k$ and $k < n$, where $n$ is the number of variables.
$r > k$.

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