Linear Algebra Practice Exam — Free

Eigenvalues are the roots of the characteristic polynomial $\det(A - \lambda I) = 0$:

So the eigenvalues are $\lambda = 2$ and $\lambda = 5$ (their sum is the trace 7 and their product is the determinant 10, a quick sanity check). The diagonal entries 4 and 3 are not eigenvalues (reading eigenvalues off the diagonal only works for triangular matrices, and $A$ is not triangular); that rules out option A.

For each eigenvalue, solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$.

$\lambda = 5$: $A - 5I = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}$, whose rows both say $-x + y = 0$, so $y = x$ and an eigenvector is $(1, 1)$.

$\lambda = 2$: $A - 2I = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}$, giving $2x + y = 0$, so $y = -2x$ and an eigenvector is $(1, -2)$.

Direct verification: $A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, confirming option C.

Options B and D each pair a genuine eigenvector with the wrong eigenvalue, a common slip when the eigenvalues are found correctly but the null spaces are mixed up: $A(1,1)^T = (5,5)^T$ is 5 times the vector, not 2 times, and $A(1,-2)^T = (2,-4)^T = 2(1,-2)^T$, not 5 times.

$\lambda = 5$ with eigenvector $(1, 1)$ (option C).

Expand along the first row, which contains a zero and so saves work (cofactor expansion along any row or column gives the same value):

Compute the $2\times 2$ determinants: $1\cdot 4 - 2\cdot(-1) = 6$ and $3\cdot(-1) - 1\cdot 1 = -4$. Hence

Since $\det(A) \ne 0$, $A$ is invertible and $A^{-1}$ exists.

For the second part use two determinant rules. First, $\det(A^{-1}) = \dfrac{1}{\det(A)}$, because $\det(A)\det(A^{-1}) = \det(I) = 1$. Second, multiplying an $n \times n$ matrix by a scalar $c$ multiplies every one of its $n$ rows by $c$, so $\det(cA) = c^n \det(A)$; here $n = 3$. Therefore

The common mistake is to pull out a single factor of 3 and get $\dfrac{3}{8}$; the scalar must be raised to the power of the matrix size.

$\det(A) = 8$ and $\det\left(3A^{-1}\right) = \dfrac{27}{8}$

(a) Each data point gives one equation $c_0 + c_1 x_i = y_i$:

Four equations in two unknowns have no exact solution, so we minimize $\lVert \mathbf{b} - A\mathbf{c}\rVert$; the minimizer satisfies the normal equations, because the residual must be orthogonal to the column space of $A$. Compute

(b) Solve $4c_0 + 6c_1 = 12$ and $6c_0 + 14c_1 = 25$. From the first, $c_0 = 3 - \tfrac32 c_1$. Substituting into the second: $6\left(3 - \tfrac32 c_1\right) + 14c_1 = 25$, so $5c_1 = 7$, $c_1 = 1.4$, and $c_0 = 3 - 2.1 = 0.9$. The least-squares line is $y = 0.9 + 1.4x$.

(c) The fitted values at $x = 0, 1, 2, 3$ are 0.9, 2.3, 3.7 and 5.1, so the residual vector is

Its entries sum to zero, as they must (the residual is orthogonal to the all-ones column of $A$). The error is the length of this vector:

$y = 0.9 + 1.4x$, with least-squares error $\sqrt{0.2} \approx 0.447$.