Linear algebra
worked examples

(a) Uniqueness via the determinant.

Write the system as \(A\mathbf{x}=\mathbf{b}\) with

$$A=\begin{pmatrix}2&0&1\\1&2&1\\2&\delta&1\end{pmatrix}, \qquad \mathbf{b}=\begin{pmatrix}1\\2\\1\end{pmatrix}.$$

A unique solution exists precisely when \(\lvert A\rvert\neq 0\). Expanding along the first row,

$$\lvert A\rvert = 2\begin{vmatrix}2&1\\\delta&1\end{vmatrix} - 0 + 1\begin{vmatrix}1&2\\2&\delta\end{vmatrix} = 2(2-\delta) + (\delta - 4) = 4 - 2\delta + \delta - 4 = -\delta.$$

So \(\lvert A\rvert = -\delta\), which is nonzero iff \(\delta\neq 0\). The system therefore fails to have a unique solution exactly when \(\delta = 0\); for every \(\delta\neq 0\) it has a unique solution.

(b) Solving at \(\delta = 2\) via the inverse.

At \(\delta=2\), \(\lvert A\rvert = -2\neq 0\), so \(A\) is invertible and \(\mathbf{x}=A^{-1}\mathbf{b}\). Computing the cofactors of \(A=\begin{pmatrix}2&0&1\\1&2&1\\2&2&1\end{pmatrix}\) and transposing gives the adjugate, and dividing by \(\lvert A\rvert=-2\),

$$A^{-1} = \begin{pmatrix}0&-1&1\\-\tfrac12&0&\tfrac12\\1&2&-2\end{pmatrix}.$$

(You can verify \(AA^{-1}=I\) directly.) Then

$$\mathbf{x} = A^{-1}\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}0(1)-1(2)+1(1)\\ -\tfrac12(1)+0(2)+\tfrac12(1)\\ 1(1)+2(2)-2(1)\end{pmatrix} = \begin{pmatrix}-1\\0\\3\end{pmatrix}.$$

So the unique solution is \(\mathbf{x} = (-1, 0, 3)\).

(c) Adding a fourth equation at \(\delta = 2\).

The fourth equation is \(\delta x_1 + x_2 = -2\), i.e. \(2x_1 + x_2 = -2\) when \(\delta=2\). The first three equations already pin the unique candidate \(\mathbf{x}=(-1,0,3)\); substituting into the fourth gives \(2(-1) + 0 = -2\), which matches the right-hand side. The added equation is therefore consistent, and the enlarged four-equation system still has the unique solution \((-1,0,3)\) at \(\delta = 2\).

Augment \(A\) with the identity and reduce the left block to \(I\); the right block becomes \(A^{-1}\). Begin with \(\left(\begin{array}{cc|cc}1&-3&1&0\\2&0&0&1\end{array}\right)\).

Step 1 — clear the entry below the pivot.

Add \((-2)\times\)row 1 to row 2:

$$\left(\begin{array}{cc|cc}1&-3&1&0\\0&6&-2&1\end{array}\right).$$

Step 2 — clear the entry above the second pivot.

Add \(\tfrac12\times\)row 2 to row 1:

$$\left(\begin{array}{cc|cc}1&0&0&\tfrac12\\0&6&-2&1\end{array}\right).$$

Step 3 — normalise the second pivot.

Multiply row 2 by \(\tfrac16\):

$$\left(\begin{array}{cc|cc}1&0&0&\tfrac12\\0&1&-\tfrac13&\tfrac16\end{array}\right).$$

The left block is now the identity, so

$$A^{-1} = \begin{pmatrix}0&\tfrac12\\-\tfrac13&\tfrac16\end{pmatrix}.$$

Cross-check with the \(2\times2\) formula: \(\lvert A\rvert = 1(0)-2(-3) = 6\), and \(\tfrac{1}{6}\begin{pmatrix}0&3\\-2&1\end{pmatrix} = \begin{pmatrix}0&\tfrac12\\-\tfrac13&\tfrac16\end{pmatrix}\). ✓

Step 1 — the characteristic equation.

Because the first column of \(A-\lambda I\) has a single nonzero entry, expand \(\lvert A-\lambda I\rvert\) along it:

$$\lvert A-\lambda I\rvert = (2-\lambda)\begin{vmatrix}4-\lambda&3\\1&2-\lambda\end{vmatrix} = (2-\lambda)\big[(4-\lambda)(2-\lambda) - 3\big].$$

Now \((4-\lambda)(2-\lambda) - 3 = \lambda^2 - 6\lambda + 8 - 3 = \lambda^2 - 6\lambda + 5 = (\lambda-1)(\lambda-5)\). Hence

$$\lvert A-\lambda I\rvert = (2-\lambda)(\lambda-1)(\lambda-5) = 0,$$

so the eigenvalues are \(\lambda = 1,\ 2,\ 5\). (Check against the trace: \(2+4+2 = 8 = 1+2+5\). ✓)

Step 2 — eigenvector for \(\lambda = 2\).

Solve \((A-2I)\mathbf{x}=\mathbf{0}\): \(\begin{pmatrix}0&-1&1\\0&2&3\\0&1&0\end{pmatrix}\mathbf{x}=\mathbf{0}\). The third row gives \(x_2=0\), then the first gives \(x_3=0\), while \(x_1\) is free. An eigenvector is \(\mathbf{x}=(1,0,0)^{\top}\).

Step 3 — eigenvector for \(\lambda = 1\).

Solve \((A-I)\mathbf{x}=\mathbf{0}\): \(\begin{pmatrix}1&-1&1\\0&3&3\\0&1&1\end{pmatrix}\mathbf{x}=\mathbf{0}\). Rows two and three give \(x_2 = -x_3\); the first gives \(x_1 = x_2 - x_3 = -2x_3\). Taking \(x_3=1\), an eigenvector is \(\mathbf{x}=(-2,-1,1)^{\top}\).

Step 4 — eigenvector for \(\lambda = 5\).

Solve \((A-5I)\mathbf{x}=\mathbf{0}\): \(\begin{pmatrix}-3&-1&1\\0&-1&3\\0&1&-3\end{pmatrix}\mathbf{x}=\mathbf{0}\). The second row gives \(x_2 = 3x_3\); the first gives \(-3x_1 - x_2 + x_3 = 0 \Rightarrow x_1 = -\tfrac{2}{3}x_3\). Taking \(x_3=3\), an eigenvector is \(\mathbf{x}=(-2,9,3)^{\top}\).

So the eigenpairs are \(\big(1,(-2,-1,1)\big)\), \(\big(2,(1,0,0)\big)\) and \(\big(5,(-2,9,3)\big)\), each determined up to a nonzero scalar multiple.