answer:The null-space of matrix A, denoted as N(A), consists of vectors x for which Ax = 0. In the case of A = zz^T, a vector x is in N(A) if and only if zz^Tx = 0. This condition is satisfied precisely when the dot product (or inner product) of z and x is zero, which implies that x is orthogonal to z^T, or equivalently, to z. To understand the rank of A, we can employ the Rank-Nullity Theorem, which states that for any linear transformation, the sum of the rank (the dimension of the image) and the nullity (the dimension of the null space) is equal to the dimension of the domain. Here, the domain is R^n (n-dimensional real space) and the null space N(A) has a dimension of n-1, as it is the orthogonal complement of the one-dimensional subspace spanned by z. This means that any vector in R^n can be written as the sum of a vector in the null space and a scalar multiple of z. Since N(A) has dimension n-1, and R^n is an n-dimensional space, the Rank-Nullity Theorem implies that rank(A) = n - (n-1) = 1. This is because the non-zero vector z spans the image of A, and all other vectors are in the null space. Thus, A has rank 1 because it can map at most one vector (z) to a non-zero vector, while all others are mapped to the zero vector.

answer:To determine the stability of the equilibrium solutions, we need to linearize the system about each equilibrium point and examine the eigenvalues of the resulting Jacobian matrix. For the equilibrium point (0, 0), the Jacobian matrix is: {eq}begin{bmatrix} 10 & 0 0 & -5 end{bmatrix} {/eq} The eigenvalues are 10 and -5, which are both real and have opposite signs. Therefore, the equilibrium point (0, 0) is a saddle point and is unstable. For the equilibrium point (10, 0), the Jacobian matrix is: {eq}begin{bmatrix} -10 & -200 0 & -5 end{bmatrix} {/eq} The eigenvalues are -10 and -5, which are both real and negative. Therefore, the equilibrium point (10, 0) is a stable node. For the equilibrium point (100, -9/2), the Jacobian matrix is: {eq}begin{bmatrix} -900 & -900 -9/20 & -5 end{bmatrix} {/eq} The eigenvalues are approximately -900.05 and -4.95, which are both real and negative. Therefore, the equilibrium point (100, -9/2) is also a stable node.

answer:To find a basis for the null space of the given matrix M, we solve the equation Mv = 0 for vector v = [x_1, x_2]^T. The augmented matrix for this equation is: left[ begin{array}{cc|c} 0 & -10 & 0 -7 & -4 & 0 end{array} right] Reducing this matrix to row echelon form: 1. Swap row 1 with row 2: left[ begin{array}{cc|c} -7 & -4 & 0 0 & -10 & 0 end{array} right] 2. Divide row 2 by -10: left[ begin{array}{cc|c} -7 & -4 & 0 0 & 1 & 0 end{array} right] 3. Add 4 times row 2 to row 1: left[ begin{array}{cc|c} -7 & 0 & 0 0 & 1 & 0 end{array} right] 4. Divide row 1 by -7: left[ begin{array}{cc|c} 1 & 0 & 0 0 & 1 & 0 end{array} right] The reduced row echelon form shows that the matrix is already in its identity form, which indicates that there are no free variables in the system. The null space of the matrix is therefore the set containing only the zero vector, (0, 0). Hence, a basis for the null space of the matrix is: {, (0, 0), }

answer:Resistance (R) in a liquid flow system, when compared to electrical circuits through hydraulic analogies, can be defined using the relationship between the change in level difference (or head, which is analogous to electrical potential or voltage) and the change in flow rate (which corresponds to electric current). The correct analogy is represented by answer B: ( R = frac{Delta H}{Q} ), where ΔH is the change in level difference and Q is the change in flow rate.