Name: Properties of Determinants (Part 3)
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Suppose we multiply a row of a determinant with a particular value, then what would be the change in the value of the determinant? Watch this video to learn...

Question

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ . Also show that $A^{- 1} = \frac{A^{2} - 3 I}{2}$

Accepted Answer

Given matrix

$A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$

Expanding along first row, we have

$|A| = 0 - 1 (0 - 1) + 1 (1 - 0)$

$= 1 + 1 = 2 \neq 0$

$\Rightarrow A$ is non-singular. $∴ A^{- 1}$ exists.

Co-factors of $A$ are

$C_{11} = 0 - 1 = - 1,$ $C_{12} = - (0 - 1) = 1,$

$C_{13} = 1 - 0 = 1,$ $C_{21} = - (0 - 1) = 1,$

$C_{22} = 0 - 1 = - 1,$ $C_{23} = - (0 - 1) = 1,$

$C_{31} = 1 - 0 = 1,$ $C_{32} = - (0 - 1) = 1,$

$C_{33} = 0 - 1 = - 1$

$AdjA = {[\begin{matrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{32} \end{matrix}]}^{'} = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'}$

$= [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$

$∴ A^{- 1} = \frac{1}{|A|} adjA$

$= \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$ .......................... $(i)$

$A^{2} = AA = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$

$= [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}]$

$= [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}]$

Now

$\frac{A^{2} - 3 I}{2} = \frac{1}{2} [[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]]$

$= \frac{1}{2} [\begin{matrix} 2 - 3 & 1 - 0 & 1 - 0 \\ 1 - 0 & 2 - 3 & 1 - 0 \\ 1 - 0 & 1 - 0 & 2 - 3 \end{matrix}]$

$= \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]$ ........................... $(ii)$

From equation $(i)$ and $(ii)$ , It is proved

$A^{- 1} = \frac{A^{2} - 3 I}{2}$ .

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ . Also show that $A^{- 1} = \frac{A^{2} - 3 I}{2}$

Important Questions on Determinants

Learn and Prepare for any exam you want !

Find A-1 if A=011101110. Also show that A-1=A2-3I2

Important Questions on Determinants

Learn and Prepare for any exam you want !

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ . Also show that $A^{- 1} = \frac{A^{2} - 3 I}{2}$