Question

[tex] Check,Whether \frac{1}{11}\left[\begin{array}{ccc}-1&8&α\\1&-19&14\\2&6&-5\end{array}\right] is and inverse of A=\left[\begin{array}{ccc}1&2&5\\3&1&1\\4&2&1\end{array}\right],if so, then α=.....

(a) -3

(b) 2

(c) -5

(d) not exists. [/tex]

Answer 1

Answer:

In order to the first matrix be the inverse of the second, we need $\alpha = -3$.

Step-by-step explanation:

$\frac{1}{11}\left[\begin{array}{ccc}-1&8&\alpha\\1&-19&14\\2&6&-5\end{array}\right] \times \left[\begin{array}{ccc}1&2&5\\3&1&1\\4&2&1\end{array}\right]  = \left[\begin{array}{ccc}\frac{1}{11}(4\alpha+23)&\frac{1}{11}(2\alpha+6)&\frac{1}{11}(\alpha+3)\\0&1&0\\0&0&1\end{array}\right]$

Then we need

$\frac{1}{11}(4\alpha+23) = 1 \Rightarrow \alpha = -3$

In this case, we have

$\frac{1}{11}(2\alpha+6) = \frac{1}{11}(\alpha+3) = 0$

As wished. So the correct option is (a)