91Ó°ÊÓ

We'll begin by setting up the equation \(k_1A + k_2B + k_3C = 0\): \[k_1\left[\begin{array}{lll}1 & 2 & 3 \\\ 4 & 0 & 5\end{array}\right] + k_2\left[\begin{array}{rrr}2 & 4 & 7 \\\ 10 & 1 & 13\end{array}\right] + k_3\left[\begin{array}{rrr}1 & 2 & 5 \\\ 8 & 2 & 11\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\\ 0 & 0 & 0\end{array}\right].\]

Now, we'll create the system of linear equations by setting the corresponding entries equal to each other: \[\left\{\begin{aligned} k_1 + 2k_2 + k_3 &= 0, \\ 2k_1 + 4k_2 + 2k_3 &= 0, \\ 3k_1 + 7k_2 + 5k_3 &= 0, \\ 4k_1 + 10k_2 + 8k_3 &= 0, \\ 5k_1 + 13k_2 + 11k_3 &= 0. \end{aligned}\right.\]

We can eliminate the first variable k_1 in the last two equations by multiplying the first equation by 2 and subtracting it from the second equation, and by multiplying the first equation by 4 and subtracting it from the fourth equation: \[\left\{\begin{aligned} 2k_1 + 4k_2 + 2k_3 &= 0, \\ k_3 &= -k_2, \\ 4k_1 + 10k_2 + 8k_3 &= 0, \\ 5k_1 + 13k_2 + 11k_3 &= 0. \end{aligned}\right.\] Now, we can use the second equation \(k_3 = -k_2\) to eliminate k_3 from the other equations by replacing k_3 with \(-k_2\): \[\left\{\begin{aligned} 2k_1 + 6k_2 &= 0, \\ 4k_1 + 18k_2 &= 0, \\ 5k_1 + 24k_2 &= 0. \end{aligned}\right.\]

Now, we'll solve the new system of linear equations: \[\left\{\begin{aligned} 2k_1 + 6k_2 &= 0, \\ 4k_1 + 18k_2 &= 0, \\ 5k_1 + 24k_2 &= 0. \end{aligned}\right.\] Dividing the first equation by 2, we get \(k_1 + 3k_2 = 0\). Now we can easily see that all three equations are the same, which means they are dependent. We can rewrite the equation as \(k_1 = -3k_2\).

Since the solution of the linear system is non-trivial (\(k_1=-3k_2\), \(k_2=k_2\), \(k_3=-k_2\), we can choose \(k_2 = 1\), then \(k_1 = -3\) and \(k_3 = -1\) so the linear dependence takes place), matrices A, B, and C are linearly dependent. Now, we can express B as a linear combination of A and C: \[B = k_1A + k_3C = -3\left[\begin{array}{lll}1 & 2 & 3 \\\ 4 & 0 & 5\end{array}\right] + (-1)\left[\begin{array}{rrr}1 & 2 & 5 \\\ 8 & 2 & 11\end{array}\right] = \left[\begin{array}{rrr}2 & 4 & 7 \\\ 10 & 1 & 13\end{array}\right].\] This means that B can be written as a linear combination of A and C, so the basis of the subspace W spanned by A, B, and C is just \(\{A, C\}\), i.e.: \[W = \operatorname{span} \left\{\left[\begin{array}{lll}1 & 2 & 3 \\\ 4 & 0 & 5\end{array}\right], \left[\begin{array}{rrr}1 & 2 & 5 \\\ 8 & 2 & 11\end{array}\right]\right\}.\] Since our basis has 2 matrices, the dimension of the subspace W is 2. Thus, we have found the dimension and basis for the subspace W.