The unique least-squares estimator exists if and only if the design matrix, which is formed by stacking the vectors \mathbf x_i as rows, has full rank.
Let's form the design matrix for each choice:
Choice 1:
\mathbf X = \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 1 \end{pmatrix}
Choice 2:
\mathbf X = \begin{pmatrix} 0 & 1 \\ 0 & 3 \end{pmatrix}
Choice 3:
\mathbf X = \begin{pmatrix} 0 & 1 \\ 0 & 1 \\ 0 & 1 \\ 3 & 0 \end{pmatrix}
Choice 4:
\mathbf X = \begin{pmatrix} 0 & 1 \\ 0 & 3 \\ 0 & -1 \end{pmatrix}
To check if the design matrix has full rank, we need to check if its determinant is non-zero.
Choice 1:
\det(\mathbf X) = 0*1 - 0*1 = 0
The design matrix does not have full rank.
Choice 2:
\det(\mathbf X) = 0*3 - 0*1 = 0
The design matrix does not have full rank.
Choice 3:
\det(\mathbf X) = (0*(-1) - 0*0)*1 - (0*0 - 3*0)*1 + (0*0 - 3*1)*1 = -3
The design matrix has full rank.
Choice 4:
\det(\mathbf X) = (0*(-1) - 0*0)*(-1) - (0*0 - 3*0)*1 + (0*0 - 3*(-1))*14 = 42
The design matrix has full rank.
Therefore, only choice 4 admits a unique least-squares estimator.
n n\times n matrix \mathbf{M} has “full rank"i.e. \, \text {rank}(\mathbf{M})=n\, if and only if its determinant is non-zero.
Each choice below specifies a collection of \mathbf x's in \mathbb {R}^2 and y's in \mathbb {R}, where each \mathbf x can be written as \mathbf x= (x^{(1)},x^{(2)}). Which one admits a unique least-squares estimator \hat{{\boldsymbol \beta }} = (\hat{\beta }_0, \hat{\beta }_1, \hat{\beta }_2) for the linear model y = \beta _0 + \beta _1 x^{(1)} + \beta _2 x^{(2)}?
(x^{(1)}_1,x^{(2)}_1,y_1) = (0,1,10), (x^{(1)}_2,x^{(2)}_2,y_2) = (0,1,9), (x^{(1)}_3,x^{(2)}_3,y_3) = (0,1,-3)
(x^{(1)}_1,x^{(2)}_1,y_1) = (0,1,10), (x^{(1)}_2,x^{(2)}_2,y_2) = (0,3,9)
(x^{(1)}_1,x^{(2)}_1,y_1) = (0,1,10), (x^{(1)}_2,x^{(2)}_2,y_2) = (0,1,9), (x^{(1)}_3,x^{(2)}_3,y_3) = (0,0,7), (x^{(1)}_4,x^{(2)}_4,y_4) = (3,0,-1)
(x^{(1)}_1,x^{(2)}_1,y_1) = (0,1,10), (x^{(1)}_2,x^{(2)}_2,y_2) = (0,3,9), (x^{(1)}_3,x^{(2)}_3,y_3) = (0,-1,14)
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