Question
If $\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix},$ then $\text{A}^{-1}$ exists if:
- $\lambda=2$
- $\lambda\neq2$
- $\lambda\neq-2$
- $\text{None of these}$
✓
Answer
- $\text{None of these}$
$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$
The inverse of a matrix exists if its determinant is not equal to 0.
Consider,
$|\text{A}|=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}\neq0$
$\Rightarrow|\text{A}| = 2 (6 – 5) – \lambda (0 – 5) + (-3) (0 – 2)\neq0$
$\Rightarrow2 + 5\lambda + 6 \neq 0$
$\Rightarrow5\lambda + 8 \neq 0$
$\Rightarrow5\lambda \neq -8$
$\Rightarrow\lambda\neq\frac{-8}{5}$
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