For the matrix $A(t)=\left[\begin{array}{ccc}t& 1& 2\\ 3& t& 4\\ 5& 6& t\end{array}\right]$, find all real values of $t$ for which $A(t)$ is singular (i.e., $detA(t)=0$).

Step-by-step solution

1. Write the determinant as a polynomial in $t$ using cofactor expansion along the first row.
2. First term: $t\times \det \left[\begin{array}{cc}t& 4\\ 6& t\end{array}\right]=t(t^2-24)$.
3. Second term (with sign $-$): $-1\times \det \left[\begin{array}{cc}3& 4\\ 5& t\end{array}\right]=-(3t-20)=-3t+20$.
4. Third term: $+2\times \det \left[\begin{array}{cc}3& t\\ 5& 6\end{array}\right]=2(18-5t)=36-10t$.
5. Combine: $t(t^2-24)-3t+20+36-10t=t^3-37t+56$.
6. Set the polynomial equal to zero: $t^3-37t+56=0$.
7. Since the cubic has no rational roots, use numerical methods (e.g., Newton's method) to locate the real root. Evaluate the sign changes: $f(1)=20>0$, $f(2)=-10<0$, so a root lies between 1 and 2. Refine to $t\approx 1.6305$.
8. Hence the matrix is singular only for $t\approx 1.63$ (the other two roots are complex).

Concept tested: Determinant as a polynomial in a parameter

Published 17 September 2026 · How this solution was prepared and checked