First slide

Introduction to Determinants

Question

If $B = [\begin{matrix} 5 & 2 α & 1 \\ 0 & 2 & 1 \\ α & 3 & - 1 \end{matrix}]$ is the inverse of a $3 \times 3$ matrix A, then the sum of all values of $α$ for which det(A)+1=0, is

Moderate

A

0
B

-1
C

1
D

2

Solution

$Given matrix B is the inverse matrix of 3 \times 3 matrix A,$
$Where B = [\begin{array}{ccc} 5 & 2 α & 1 \\ 0 & 2 & 1 \\ α & 3 & - 1 \end{array}]$
We know that ,
$\det (A) = \frac{1}{d e t (B)} [∵ \det (A^{- 1}) = \frac{1}{\det (A)}]$
Since, det(A) +1=0
$\begin{array}{l} \frac{1}{\det (B)} + 1 = 0 \\ \Rightarrow \det (B) = - 1 \\ \Rightarrow 5 (- 2 - 3) - 2 α (0 - α) + 1 (0 - 2 α) = - 1 \\ \Rightarrow - 25 + 2 α^{2} - 2 α = - 1 \\ \Rightarrow 2 α^{2} - 2 α - 24 = 0 \\ \Rightarrow α^{2} - α - 12 = 0 \\ \Rightarrow (α - 4) (α + 3) = 0 \end{array}$
$\Rightarrow α = - 3, 4$
$So, required sum of all values of α is 4 - 3 = 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App