91Ó°ÊÓ

We want to check if there are non-zero constants \(c_A\) and \(c_B\), such that: \(c_A A + c_B B = \left(\begin{array_ll}0 & 0 \\ 0 & 0\end{array}\right)\)

Try to solve for non-zero constants \(c_A\) and \(c_B\): \( \begin{cases} c_A = 0 \\ c_A + c_B = 0 \end{cases} \) The first equation tells us that \(c_A=0\). Plugging this into the second equation, we get \(c_B=0\). Both constants are zero, meaning there is no non-trivial linear combination that can produce a zero vector, hence they are linearly independent. (b) Vectors: \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\; B=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)\) and \(C=\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right)\)

We want to check if there are non-zero constants \(c_A\), \(c_B\) and \(c_C\), such that: \(c_A A + c_B B + c_C C = \left(\begin{array_ll}0 & 0 \\ 0 & 0\end{array}\right)\)

Try to solve for non-zero constants \(c_A\), \(c_B\), and \(c_C\): \( \begin{cases} c_A = 0 \\ c_B = 0 \\ c_C = 0 \end{cases} \) All of the constants are zero, so there is no non-trivial linear combination that produces a zero vector, meaning these vectors are linearly independent. (c) Vectors: \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right),\; B=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)\) and \(D=\left(\begin{array}{ll}2 & 3 \\ 0 & 2\end{array}\right)\)

We want to check if there are non-zero constants \(c_A\), \(c_B\) and \(c_D\), such that: \(c_A A + c_B B + c_D D = \left(\begin{array_ll}0 & 0 \\ 0 & 0\end{array}\right)\)

Try to solve for non-zero constants \(c_A\), \(c_B\), and \(c_D\): \( \begin{cases} c_A + 2c_D = 0 \\ c_B + 3c_D = 0 \\ c_A + 2c_D = 0 \end{cases} \) Notice there is a redundancy among the equations, and we only need two (since these are 2x2 matrices). Using the first two equations, we find the following adjusted system of equations: \( \begin{cases} c_A + 2c_D = 0 \\ c_B + 3c_D = 0 \end{cases} \) Here we have 3 sources of dependency among two equations, which means we have a freedom to choose one parameter independently. If we take \(c_D = 1\), then the solution is: \(\begin{cases} c_A = -2 \\ c_B = -3 \end{cases} \) Since we could find non-zero constants for a linear combination that produces a zero vector, these vectors are linearly dependent.