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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Let A = $\begin{bmatrix} 1 & -2 & 1\\ -2 & 3 & 1\\ 1 & 1 & 5 \end{bmatrix}$. Verify that \[\] $(ii)\big(A^{-1}\big)^{-1}$

This is part 2 of a 2 part question, split as 2 separate questions here.
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Toolbox:
  • A matrix is said to be singular if |A|=0.
  • A matrix is said to be invertible if |A|$\neq 0$
  • $|A|^{-1}=\frac{1}{|A|}(adj A)$ where (adj A)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
We have already found $A^{-1}=\begin{bmatrix}-14/13 & -11/13 & 5/13\\-11/13 & -4/13 & 3/13\\5/13 & 3/13 & 1/13\end{bmatrix}$
 
Let us find the inverse of $A^{-1}$
 
Determinant of |$A^{-1}$| can be found by expanding along $R_1$
 
$\;\;\;=-14/13[-4/13\times 1/13-3/13\times 3/13]+11/13[-11/13\times 1/13-3/13\times 5/13]+5/13[-11/13\times 3/13-5/13\times 4/13]$
 
$\;\;\;=-14/13[-4/169-9/169]+11/13[-11/169-15/169]+5/13[-33/169+20/169]$
 
$\;\;\;=-14/13[-13/169]+11/13[-26/169]+5/13[-13/169]$
 
$\;\;\;=-14/13\times -1/13+11/13\times -2/13+5/13\times -1/13$
 
$\;\;\;=14/169-22/169-5/169= -13/169= -1/13\neq 0$
 
Hence it is a non-singular matrix.
 
$|A^{-1}=-1/13$
 
Now let us find the adj A$^{-1}$
 
The cofactors of $A^{-1}$ are
 
$M_{11}=\begin{vmatrix}-4/13 & 3/13\\3/13 & 1/13\end{vmatrix}=-4/169-9/169=-13/169=-1/13$
 
$M_{12}=\begin{vmatrix}-11/13 & 3/13\\5/13 & 1/13\end{vmatrix}=-11/169-15/169=-26/169=-2/13$
 
$M_{113}=\begin{vmatrix}-11/13 & -4/13\\5/13 & 3/13\end{vmatrix}=-33/169+20/169=-13/169=-1/13$
 
$M_{21}=\begin{vmatrix}-11/13 & 5/13\\3/13 & 1/13\end{vmatrix}=-11/169-15/169=-26/169=-2/13$
 
$M_{22}=\begin{vmatrix}-14/13 & 5/13\\5/13 & 1/13\end{vmatrix}=-14/169-25/169=-39/169=-3/13$
 
$M_{23}=\begin{vmatrix}-14/13 & -11/13\\5/13 & 3/13\end{vmatrix}=-42/169+55/169=13/169=1/13$
 
$M_{31}=\begin{vmatrix}-11/13 & 5/13\\-4/13 & 3/13\end{vmatrix}=-33/169+20/169=-13/169=-1/13$
 
$M_{32}=\begin{vmatrix}-14/13 & 5/13\\-11/13 & 3/13\end{vmatrix}=-42/169+55/169=-13/169=-1/13$
 
$M_{33}=\begin{vmatrix}-14/13 & -11/13\\-11/13 & -4/13\end{vmatrix}=56/169-121/169=-65/169=-5/13$
 
$A_{11}=(-1)^{1+1}.-1/13=-1/13.$
 
$A_{12}=(-1)^{1+2}.-2/13=2/13.$
 
$A_{13}=(-1)^{1+3}.-1/13=-1/13.$
 
$A_{21}=(-1)^{2+1}.-2/13=2/13.$
 
$A_{22}=(-1)^{2+2}.-3/13=-3/13.$
 
$A_{23}=(-1)^{2+3}.1/13=-1/13.$
 
$A_{31}=(-1)^{3+1}.-1/13=-1/13.$
 
$A_{32}=(-1)^{3+2}.-1/13=1/13.$
 
$A_{33}=(-1)^{3+3}.-5/13=-5/13.$
 
$(adj A^{-1})=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}-1/13 & 2/13 & -1/13\\2/13 & -3/13 & 1/13\\-1/13 & -1/13 & -5/13\end{bmatrix}$
 
$|A|^{-1}=\frac{1}{|A|}(adj A)$
 
$\frac{1}{|A|}=-1/13$
 
Therefore $(A^{-1})^{-1}=-1/13\begin{bmatrix}-1/13 & 2/13 & -1/13\\2/13 & -3/13 & 1/13\\-1/13 & -1/13 & -5/13\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}1 & -2 & 1\\-2 & 3 & -1\\1 & 1 & 5\end{bmatrix}=A$
 
Hence $(A^{-1})^{-1}=A.$
 
Hence verified.

 

answered Mar 2, 2013 by sreemathi.v
 

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