91Ó°ÊÓ

We have 3 matrices \(A\), \(B\), and \(C\). Let's see if we can find scalars \(x\), \(y\), and \(z\), so that their linear combination equals to the zero matrix: \[x A + y B + z C = 0\]

Expanding the matrix equation, we get: \[\left[\begin{array}{lll} x(1) + y(2) + z(1) & x(2) + y(4) + z(2) & x(1) + y(3) + z(3) \\ x(3) + y(7) + z(5) & x(1) + y(5) + z(7) & x(2) + y(6) + z(6) \end{array}\right] = \left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\] Now, set up a homogeneous system of linear equations: \[\begin{cases} x + 2y + z = 0 \\ 3x + 7y + 5z = 0 \\ 2x + 4y + 2z = 0 \\ x + 5y + 7z = 0 \\ 2x + 6y + 6z = 0 \end{cases}\]

Notice that the third equation is just double of the first equation. Similarly, the fifth equation is also double of the second equation. So, we can remove these redundant equations from the system: \[\begin{cases} x + 2y + z = 0 \\ 3x + 7y + 5z = 0 \\ x + 5y + 7z = 0 \end{cases}\]

Now, let's solve the system of linear equations: \[\begin{cases} x + 2y + z = 0 \\ 3x + 7y + 5z = 0 \\ x + 5y + 7z = 0 \end{cases}\] Subtract the first equation from the third equation to eliminate x: \[\begin{cases} x + 2y + z = 0 \\ 3x + 7y + 5z = 0 \\ 3y + 6z = 0 \end{cases}\] Then divide the third equation by 3: \[\begin{cases} x + 2y + z = 0 \\ 3x + 7y + 5z = 0 \\ y + 2z = 0 \end{cases}\] Now, subtract three times the second equation from the first equation to eliminate \(y\): \[\begin{cases} x + 3z = 0 \\ 3x + 7y + 5z = 0 \\ y + 2z = 0 \end{cases}\] From the first equation, we get \(x = -3z\). Then, substituting the value of \(x\) in the second equation, we get: \[3(-3z) + 7y + 5z = 0\] \[-9z + 7y + 5z = 0\] Solving the equation, we get \(y = 2z\).

Since both \(x\) and \(y\) are expressed in terms of the single parameter \(z\), the dimension of the subspace is 1. The basis of the subspace can be found by substituting the expressions of \(x\) and \(y\) in terms of \(z\) into the matrices \(A\), \(B\), and \(C\): \[x = -3z\] \[y = 2z\] So, the linear combination of the matrices can be represented as: \[(-3z)A + (2z)B + zC\] Now, substitute the values of \(A\), \(B\), and \(C\): \[(-3z)\left[\begin{array}{lll} 1 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right] + (2z)\left[\begin{array}{lll} 2 & 4 & 3 \\ 7 & 5 & 6 \end{array}\right] + z\left[\begin{array}{lll} 1 & 2 & 3 \\ 5 & 7 & 6 \end{array}\right]\] Adding the three matrices, we get: \[\left[\begin{array}{lll} -3z & -6z & -3z \\ -9z & -3z & -6z \end{array}\right] + \left[\begin{array}{lll} 4z & 8z & 6z \\ 14z & 10z & 12z \end{array}\right] + \left[\begin{array}{lll} z & 2z & 3z \\ 5z & 7z & 6z \end{array}\right] = \left[\begin{array}{lll} 2z & 4z & 6z \\ 10z & 14z & 12z \end{array}\right]\] So the basis of the subspace \(W\) is: \[\left\{\left[\begin{array}{lll} 2z & 4z & 6z \\ 10z & 14z & 12z \end{array}\right]\right\}\] where \(z\) is any scalar. The dimension of the subspace \(W\) is \(1\), and a basis for the subspace is the set with one single matrix \(\left\{\left[\begin{array}{lll} 2 & 4 & 6 \\ 10 & 14 & 12 \end{array}\right]\right\}\) (choosing \(z = 1\)).