Consider the linear system Ax = b. Given is that $A=\begin{bmatrix}t & 2 \\ -2 & t+3 \end{bmatrix}$ and $\textbf{b}=\begin{bmatrix}1 \\ -4 \end{bmatrix}$. For which value of t has the system a solution with $x_1 = 0$?

\begin{equation} x_1 = \frac{det(A_1(\textbf{b}))}{det(A)}=\frac{\begin{vmatrix}1 & 2\\ -4 & t+3\end{vmatrix}}{\begin{vmatrix}t & 2\\ -2 & t+3\end{vmatrix}}=\frac{t+11}{t^2+3t+4} \end{equation} To get $x_1 = 0$, we need $t = -11$.

Interview Questions Linear Algebra Using Cramer's Rule

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