Review : Matrices & Vectors — Question 3

Question 3 – Solution

Strategy. Subtracting the first row exposes the parameter that controls both inversion and the exceptional case.

Step 1: Compute the determinant. Replace rows two and three by themselves minus row one. These operations do not change the determinant and produce $$\begin{pmatrix}1&1&1\\0&\tau-1&0\\0&0&\tau-1\end{pmatrix}.$$ Therefore $\boxed{\det A_\tau=(\tau-1)^2}$, and inversion is possible exactly when $\tau\ne 1$.

Step 2: Solve for a general right-hand side. Put $d=\tau-1\ne 0$. The subtracted equations give $x_2=(c_2-c_1)/d$, $x_3=(c_3-c_1)/d$, and $x_1=c_1-x_2-x_3$. Thus $$\boxed{A_\tau^{-1}=\begin{pmatrix} 1+2/d&-1/d&-1/d\\-1/d&1/d&0\\-1/d&0&1/d \end{pmatrix}.}$$ Substitution in the three original equations returns $c_1,c_2,c_3$ for every $c$, so multiplication $A_\tau A_\tau^{-1}=I_3$ is verified. The displayed matrices are symmetric, and transposing that equality also verifies multiplication in the opposite order.

Step 3: Handle the prescribed and singular systems. For $\tau\ne 1$, $$\boxed{x=\left(1-\frac 3d,\frac 1d,\frac 2d\right)^T.}$$ At $\tau=1$, all three left sides are $x_1+x_2+x_3$, while the entries of $b$ differ, so there is no solution. For a general $c$, consistency at $1$ is equivalent to $c_1=c_2=c_3$. If they coincide, the whole plane $x_1+x_2+x_3=c_1$ is the solution set; otherwise it is empty.

Step 4: Distinguish invertibility from bounded recovery. For $b=(1,2,3)^T$, the coordinate $x_2=1/(\tau-1)$ becomes unbounded near $1$. For $c=(1,1,1)^T$, the unique nearby solution is always $(1,0,0)^T$. Thus some data remain harmless while other fixed data yield arbitrarily large solutions. The factors $1/d$ in the inverse prevent a uniform bound near the singular parameter, even though every matrix with $d\ne 0$ is invertible.