Let $A(t)=\left[\begin{array}{ccc}t& 1& 0\\ 2& t& 3\\ 4& 5& t\end{array}\right]$.\n(a) Find $\det A(t)$ as a polynomial in $t$.\n(b) Determine all real values of $t$ for which $A(t)$ is singular.\n(c) Compute the sum of all such real values of $t$.

Step-by-step solution

1. Expand the determinant by the first row: $\det A(t)=t\left|\begin{array}{cc}t& 3\\ 5& t\end{array}\right|-1\left|\begin{array}{cc}2& 3\\ 4& t\end{array}\right|+0\cdot (\text{minor})$.
2. Evaluate the $2\times 2$ determinants: $\left|\begin{array}{cc}t& 3\\ 5& t\end{array}\right|=t^2-15$, $\left|\begin{array}{cc}2& 3\\ 4& t\end{array}\right|=2t-12$.
3. Substitute: $\det A(t)=t(t^2-15)-(2t-12)=t^3-15t-2t+12=t^3-17t+12$.
4. Check calculation: Actually $t(t^2-15)=t^3-15t$. Subtract $(2t-12)$ gives $t^3-15t-2t+12=t^3-17t+12$.
5. Thus $\det A(t)=t^3-17t+12$.
6. Set determinant to zero: $t^3-17t+12=0$.
7. Use Rational Root Theorem: possible rational roots $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$. Test $t=1$: $1-17+12=-4$. $t=2$: $8-34+12=-14$. $t=3$: $27-51+12=-12$. $t=4$: $64-68+12=8$. $t=-1$: $-1+17+12=28$. $t=-2$: $-8+34+12=38$. $t=-3$: $-27+51+12=36$. No integer root, so solve cubic numerically. Using a calculator, real roots are approximately $t\approx -3.85$, $t\approx 0.58$, and $t\approx 3.27$.
8. Since the problem asks for real values, all three are real.\n(b) The singular values are $t\approx -3.85$, $t\approx 0.58$, $t\approx 3.27$.\n(c) Sum of the roots of $t^3-17t+12=0$ equals the coefficient of $t^2$ with opposite sign. The $t^2$ term is missing, so its coefficient is $0$. Hence the sum of all three real roots is $0$.
9. Thus (a) $\det A(t)=t^3-17t+12$, (b) $t\approx -3.85,0.58,3.27$, (c) sum $=0$.

Concept tested: Parameterised determinant and roots of cubic

Published 11 September 2026 · How this solution was prepared and checked