Fall 2026: Math 590 Quiz 7 Solutions

Problem 1

Solve the following system of first order linear differential equations with the given initial conditions.

\begin{align*} x_1'(t) &= 2x_1(t) + x_3(t)\\ x_2'(t) &= 2x_2(t) + x_3(t)\\ x_3'(t) &= 3x_3(t), \end{align*}

\(x_1(0) = 1\), \(x_2(0) = 0\), \(x_3(0) = -1\). You may use the fact that the coefficient matrix is diagonalizable with eigenvalues \(2, 2, 3\) and diagonalizing matrix \(P = \begin{pmatrix} 1 & 0 & 1\\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}\) and \(P^{-1} = \begin{pmatrix} 1 & 0 & -1\\0 & 1 & -1\\0 & 0 & 1\end{pmatrix}\). (5 points)

Solution. In matrix form the system is \(X'(t) = AX(t)\), where

A = \begin{pmatrix} 2 & 0 & 1\\ 0 & 2 & 1\\ 0 & 0 & 3\end{pmatrix}, \qquad X(t) = \begin{pmatrix} x_1(t)\\ x_2(t)\\ x_3(t)\end{pmatrix},

and \(P^{-1}AP = D = \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{pmatrix}\). The columns of \(P\),

v_1 = \begin{pmatrix} 1\\0\\0\end{pmatrix}, \qquad v_2 = \begin{pmatrix} 0\\1\\0\end{pmatrix}, \qquad v_3 = \begin{pmatrix} 1\\1\\1\end{pmatrix},

are eigenvectors of \(A\) for the eigenvalues \(2, 2, 3\) respectively.

Substitute \(X = PY\). Then \(PY' = APY\), so \(Y' = P^{-1}APY = DY\), which is the decoupled system

y_1' = 2y_1, \qquad y_2' = 2y_2, \qquad y_3' = 3y_3,

with solutions

y_1(t) = y_1(0)e^{2t}, \qquad y_2(t) = y_2(0)e^{2t}, \qquad y_3(t) = y_3(0)e^{3t}.

Initial conditions for \(Y\). Since \(X = PY\), we have \(Y = P^{-1}X\), and in particular \(Y(0) = P^{-1}X(0)\). With \(X(0) = \begin{pmatrix} 1\\ 0\\ -1\end{pmatrix}\),

Y(0) = P^{-1}X(0) = \begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 1\\ 0\\ -1\end{pmatrix} = \begin{pmatrix} 1\cdot 1 + 0\cdot 0 + (-1)(-1)\\ 0\cdot 1 + 1\cdot 0 + (-1)(-1)\\ 0\cdot 1 + 0\cdot 0 + 1\cdot(-1)\end{pmatrix} = \begin{pmatrix} 2\\ 1\\ -1\end{pmatrix}.

So \(y_1(0) = 2\), \(y_2(0) = 1\), \(y_3(0) = -1\), and therefore

Y(t) = \begin{pmatrix} 2e^{2t}\\ e^{2t}\\ -e^{3t}\end{pmatrix}.

Back to \(X\). Finally,

X(t) = PY(t) = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\end{pmatrix} \begin{pmatrix} 2e^{2t}\\ e^{2t}\\ -e^{3t}\end{pmatrix} = \begin{pmatrix} 2e^{2t} - e^{3t}\\ e^{2t} - e^{3t}\\ -e^{3t}\end{pmatrix}.

Therefore

\boxed{\,x_1(t) = 2e^{2t} - e^{3t}, \qquad x_2(t) = e^{2t} - e^{3t}, \qquad x_3(t) = -e^{3t}.\,}

Problem 2

Find an orthogonal matrix \(Q\) that diagonalizes \(A = \begin{pmatrix} 1 & 2\\2 & -2\end{pmatrix}\). Explain why \(Q\) is orthogonal. (5 points)

Solution. The characteristic polynomial is

p_A(x) = \det\begin{pmatrix} x-1 & -2\\ -2 & x+2\end{pmatrix} = (x-1)(x+2) - 4 = x^2 + x - 6 = (x-2)(x+3),

so the eigenvalues are \(\lambda = 2\) and \(\lambda = -3\).

For \(\lambda = 2\): \(A - 2I = \begin{pmatrix} -1 & 2\\ 2 & -4\end{pmatrix}\), so \(E_2\) is spanned by \(\begin{pmatrix} 2\\1\end{pmatrix}\).

For \(\lambda = -3\): \(A + 3I = \begin{pmatrix} 4 & 2\\ 2 & 1\end{pmatrix}\), so \(E_{-3}\) is spanned by \(\begin{pmatrix} 1\\-2\end{pmatrix}\).

These eigenvectors are orthogonal, since \(\langle (2,1), (1,-2)\rangle = 2 - 2 = 0\) (as must happen, because \(A\) is symmetric and the eigenvalues are distinct). Each has length \(\sqrt{5}\), so normalizing gives the orthonormal basis

u_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} 2\\1\end{pmatrix}, \qquad u_2 = \frac{1}{\sqrt{5}}\begin{pmatrix} 1\\-2\end{pmatrix}.

Let

Q = \frac{1}{\sqrt{5}}\begin{pmatrix} 2 & 1\\ 1 & -2\end{pmatrix}. \qquad \text{Then} \qquad Q^{-1}AQ = Q^TAQ = \begin{pmatrix} 2 & 0\\ 0 & -3\end{pmatrix}.

Why \(Q\) is orthogonal: The columns of \(Q\) are orthogonal and have length one.