First slide

Applications of determinant

Question

$If S = [\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 \end{array}], A = \frac{1}{2} [\begin{array}{ccc} b + c & c - a & b - a \\ c - b & c + a & a - b \\ b - c & a - c & a + b \end{array}], then S A S^{- 1} =$

Moderate

A

$[\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}]$
B

$\frac{1}{2} [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}]$
C

$2 [\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}]$
D

none

Solution

$\begin{array}{l} S = [\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}] \\ \Rightarrow \det S = 0 (0 - 1) - 1 (0 - 1) + 1 (1 - 0) \\ = 0 + 1 + 1 \\ = 2 \end{array}$
The cofactors of elements of A are
$\begin{array}{l} A_{11} = + (0 - 1) = - 1, A_{12} = - (0 - 1) = 1, A_{13} = + (1 - 0) = 1 \\ A_{21} = - 1 (0 - 1) = 1, A_{22} = + (0 - 1) = - 1, A_{23} = - (0 - 1) = 1 \\ A_{31} = + (1 - 0) = 1, A_{32} = - (0 - 1) = 1, A_{33} = + (0 - 1) = - 1 \\ Adj A = [\begin{array}{ccc} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] = [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] ∴ A^{- 1} = \frac{A d j A}{\det A} = \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] \end{array}$
$\begin{array}{l} S A S^{- 1} = [\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}] \frac{1}{2} [\begin{array}{ccc} b + c & c - a & b - a \\ c - b & c + a & a - b \\ b - c & a - c & a + b \end{array}] \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] \\ = \frac{1}{4} [\begin{array}{ccc} 0 + c - b + b - c & 0 + c + a + a - c & 0 + a - b + a + b \\ b + c + 0 + b - a & c - a + 0 + a - c & b - a + 0 + a + b \\ b + c + c - b + 0 & c - a + c + a + 0 & b - a + a - b + 0 \end{array}] [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] \\ = \frac{1}{4} [\begin{array}{ccc} 0 & 2 a & 2 a \\ 2 b & 0 & 2 b \\ 2 c & 2 c & 0 \end{array}] [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}] \\ = \frac{1}{4} [\begin{array}{ccc} 0 + 2 a + 2 a & 0 - 2 a + 2 a & 0 + 2 a - 2 a \\ - 2 b + 0 + 2 b & 2 b + 0 + 2 b & 2 b + 0 - 2 b \\ - 2 c + 2 c + 0 & 2 c - 2 c + 0 & 2 c + 2 c + 0 \end{array}] \\ = \frac{1}{4} [\begin{array}{ccc} 4 a & 0 & 0 \\ 0 & 4 b & 0 \\ 0 & 0 & 4 c \end{array}] \end{array}$
$= [\begin{array}{lll} a 0 0 \\ 0 b 0 \\ 0 0 c \end{array}]$

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