If A= 2& &-3 0&2&5 1&1&3 , then A^ -1 exists if: 1. =2 2. 2 3. -2… - Vidyadip

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

$\text{None of these}$

Solution:

$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$

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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