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Mathematics Question on Determinants

If A=[1−11 21−3 111 ],10B[422 −50α 1−23 ]A =\begin{bmatrix} {1}&{-1} &{1}\\\ {2}&{1}& {-3} \\\ {1}&{1}&{1}\\\ \end{bmatrix} ,10B \begin{bmatrix} {4}&{2} &{2}\\\ {-5}&{0}& {\alpha} \\\ {1}&{-2}&{3}\\\ \end{bmatrix} and BB is the inverse of AA, then the value of α\alpha is

A

0

B

2

C

4

D

5

Answer

5

Explanation

Solution

Given, A=∣1−11 21−3 111∣A = \begin{vmatrix}1&-1&1\\\ 2&1&-3\\\ 1&1&1\end{vmatrix} and
10B=∣422 −50α 1−23∣10 B = \begin{vmatrix}4&2&2\\\ -5&0&\alpha\\\ 1&-2&3\end{vmatrix}
Since, B is the inverse of A
ie, B=A−1 B = A^{-1}
So, 10A−1=∣422 −50α 1−23∣10 A^{-1}= \begin{vmatrix}4&2&2\\\ -5&0&\alpha\\\ 1&-2&3\end{vmatrix}
⇒10A−1A=∣422 −50α 1−23∣A\Rightarrow 10 A^{-1} A = \begin{vmatrix}4&2&2\\\ -5&0&\alpha\\\ 1&-2&3\end{vmatrix} A
⇒10I=∣422 −50α 1−23∣∣1−11 21−3 111∣\Rightarrow 10I = \begin{vmatrix}4&2&2\\\ -5&0&\alpha\\\ 1&-2&3\end{vmatrix} \begin{vmatrix}1&-1&1\\\ 2&1&-3\\\ 1&1&1\end{vmatrix}
⇒∣1000 0100 0010∣=∣1000 −5+α5+α−5+α 0010∣\Rightarrow \begin{vmatrix}10&0&0\\\ 0&10&0\\\ 0&0&10\end{vmatrix} = \begin{vmatrix}10&0&0\\\ -5+\alpha& 5+\alpha &-5 +\alpha\\\ 0&0&10\end{vmatrix}
⇒−5+α=0\Rightarrow -5 +\alpha = 0
⇒α=5\Rightarrow \alpha = 5
Alternative : A−1=1∣A∣adj(A)A^{-1} = \frac{1}{\left|A\right|} \text{adj} \left(A\right)
adj(A)=∣422 −505 1−23∣\text{adj}\left(A\right) = \begin{vmatrix}4&2&2\\\ -5&0&5\\\ 1&-2&3\end{vmatrix}
∣A∣=1[1+3]+1[2+3]+1[2−1]=10\left|A\right| = 1\left[1+3\right] + 1 \left[2+3 \right]+1 \left[2-1\right] =10
∴A−1=110∣422 −505 1−23∣\therefore A^{-1} = \frac{1}{10} \begin{vmatrix}4&2&2\\\ -5&0&5\\\ 1&-2&3\end{vmatrix}
10B=∣422 −50α 1−23∣10B = \begin{vmatrix}4&2&2\\\ -5&0 &\alpha\\\ 1&-2&3\end{vmatrix}
=∣422 −505 1−23∣= \begin{vmatrix}4&2&2\\\ -5&0&5\\\ 1&-2&3\end{vmatrix}
⇒α=5\Rightarrow \alpha = 5