Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible
Answer
Let
\[A=\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}.\]
Then $\frac{1}{a}R_1+R_3 \rightarrow R_3$ gives
\[\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & x & x^2+\frac{c}{a} \end{pmatrix},\]
and $-\frac{x}{a}R_2+R_3 \rightarrow R_3$ gives
\[\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ 0 & 0 & \frac{b}{a}x+x^2+\frac{c}{a} \end{pmatrix}.\]
Hence
\[\text{det}(A)=a(\frac{b}{a}x+x^2+\frac{c}{a})=0\]
\[\Rightarrow ax^2+b x+c=0\]
\[\Rightarrow x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]
Hence the matrix $A$ is invertible if and only if
\[x\neq\frac{-b \pm \sqrt{b^2-4ac}}{2a}.\]

574
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 2488 views
- $7.00
Related Questions
- Representation theory quick question
- Find $x$ so that $\begin{bmatrix} 2 & 0 & 10 \\ 0 & x+7 & -3 \\ 0 & 4 & x \end{bmatrix} $ is invertible
- Is the $\mathbb{C}$-algebra $Fun(X,\mathbb{C})$ semi-simple?
- Solving for two unknown angles, from two equations.
- Use Rouche’s Theorem to show that all roots of $z ^6 + (1 + i)z + 1 = 0$ lines inside the annulus $ \frac{1}{2} \leq |z| \leq \frac{5}{4}$
- Character of 2-dimensional irreducible representation of $S_4$
- Clock Problem
- Prove that $tan x +cot x=sec x csc x$