Jul 23, 2017 · First think about what it means to be linearly independent. Linearly independent means that every row/column cannot be represented by the other rows/columns. Hence it is …

The matrix you wrote down $$\begin{pmatrix} 1 & a_1 & a_1^2 \\ 1 & a_2 & a_2^2 \\ 1 & a_3 & a_3^2 \end{pmatrix}$$ is exactly the matrix you need to check for nondegeneracy. In fact, this …

Jan 25, 2023 · Say we suspect the columns of a matrix are independent and want to verify that fact quickly by hand. What is the best way to do it? I'm currently studying MITx 18.033 where …

$\mbox{row}_1$ and $\mbox{row}_2$ are linearly independent; $\mbox{row}_2$ and $\mbox{row}_3$ are linearly independent; $\mbox{row}_1$ and $\mbox{row}_3$ are linearly …

Jan 24, 2016 · $\begingroup$ For your explanation on how you find the coefficients of the dependent columns, $(2, -1)$ and $(3, -1)$, these are the coefficient of the first and second …

Apr 3, 2018 · Hence the relationship between the invertibility of a matrix and the determinant (for a $ 2 \times 2 $ matrix). Invertibility is equivalent to the matrix having linearly independent …

Can the determinant (assuming it's non-zero) be used to determine that the vectors given are linearly independent, span the subspace and are a basis of that subspace? (In other words …

Ultimately, a linear dependency is where vectors satisfy some relationship with each other. Conversely, a set of vectors is linearly independent if they satisfy no linearity equation other …

With regard to the non-homogenous linear system Ax = b while A is an augmented matrix including b (in contrast with the above-mentioned coefficient matrix), it has the standard …

My question is is it also linear indepedent because the vectors are not multiples of the first vector. No. Just because the second and third columns are not multiples of the first, it does not mean …