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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 \[\] $(i)\ \; \left [adj A \right ]^{-1} = adj(A^{-1}) \;\;\;\; $

This is part 1 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 $|A|\neq 0.$
  • $A^{-1}=\frac{1}{|A|}(adj\;A)$
  • (adj A)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
  • Where $A_{11},A_{12}....A_{33}$ are cofactors.
Let us find the determinant of A by expanding along $R_1$
|A|=$1(3\times 5-1\times 1)-(-2)(-2\times 5-1\times 1)+1(-2\times 1-3\times 1)$
$\;\;\;=(15-1)+2(-10-1)+(-2-3)$
$\;\;\;=14-22-5=-13\neq 0$
Hence A is a non-singular matrix.
Therefore $A^{-1}$ exists.
Let us now find the adjoint of A,by finding its cofactors.
$M_{11}=\begin{vmatrix}3 & 1\\1 & 5\end{vmatrix}$=15-1=14.
$M_{12}=\begin{vmatrix}-2 & 1\\1 & 5\end{vmatrix}$=-10-1= - 11.
$M_{13}=\begin{vmatrix}-2 & 3\\1 & 1\end{vmatrix}$=-2-3= - 5.
$M_{21}=\begin{vmatrix}-2 & 1\\1 & 5\end{vmatrix}$=-10-1= - 11.
$M_{22}=\begin{vmatrix}1 & 1\\1 & 5\end{vmatrix}$=5-1= 4.
$M_{23}=\begin{vmatrix}1 & -2\\1 & 1\end{vmatrix}$=1+2= 3.
$M_{31}=\begin{vmatrix}-2 & 1\\3 & 1\end{vmatrix}$=-2-3= - 5.
$M_{32}=\begin{vmatrix}1 & 1\\-2 & 1\end{vmatrix}$=1+2= 3.
$M_{33}=\begin{vmatrix}1 & -2\\-2 & 3\end{vmatrix}$=3-4= - 1.
$A_{11}=(-1)^{1+1}.14=14$
$A_{12}=(-1)^{1+2}.(-11)=11$
$A_{13}=(-1)^{1+3}-5=-5$
$A_{21}=(-1)^{2+1}.(-11)=11$
$A_{22}=(-1)^{2+2}.4=4$
$A_{23}=(-1)^{2+3}.(3)= - 3$
$A_{31}=(-1)^{3+1}.(5)=-5$
$A_{32}=(-1)^{3+1}.(3)= -3 $
$A_{33}=(-1)^{3+3}.(-1)=-1$
(adj A)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}14 & 11 & -5\\11 & 4 & -3\\-5 & -3 & -1\end{bmatrix}$
|A|= - 13.
Hence $A^{-1}=\frac{1}{|A|}(adj\;A)$
$\qquad\quad=\frac{1}{-13}\begin{bmatrix}14 & 11 & -5\\11 & 4 & -3\\-5 & -3 & -1\end{bmatrix}$
$\qquad\quad=\frac{1}{13}\begin{bmatrix}-14 & -11 & 5\\-11 & -4 & 3\\5 & 3 & 1\end{bmatrix}$
Let us verify $[adj A]^{-1}=adj( A^{-1})$
|adj A]=14(-4-9)-11(-11-15)-5(-33+20)
$\;\;\;\;\;\;\;=14(-13)-11(-26)-5(-13)$
$\;\;\;\;\;\;\;=-182+286+65=169$
Next let us find adj (adj A)
adj A=$\begin{bmatrix}14 & 11 & -5\\11 & 4 & -3\\-5 & -3 & -1\end{bmatrix}$
$M_{11}=\begin{vmatrix}4 & -3\\-3 & -1\end{vmatrix}$=-4-9=-13
$M_{12}=\begin{vmatrix}11 & -3\\-5 & -1\end{vmatrix}$=-11-15= - 26.
$M_{13}=\begin{vmatrix}11 & 4\\-5 & -3\end{vmatrix}$=-33+20= - 13.
$M_{21}=\begin{vmatrix}11 & -5\\-3 & -1\end{vmatrix}$=-11-15= - 26.
$M_{22}=\begin{vmatrix}14 & -5\\-5 & -1\end{vmatrix}$=-14-25= -39.
$M_{23}=\begin{vmatrix}14 & 11\\-5 & -3\end{vmatrix}$=-42+55= 13.
$M_{31}=\begin{vmatrix}-11 & 5\\-4 & 3\end{vmatrix}$=-33+20= - 13.
$M_{32}=\begin{vmatrix}-14& 5\\-11 & 3\end{vmatrix}$=-42+55= 13.
$M_{33}=\begin{vmatrix}-14 & -14\\-11 & -4\end{vmatrix}$=56-121= - 65.
$A_{11}=(-1)^{1+1}. - 13= -13$
$A_{12}=(-1)^{1+2}.(-26)=26$
$A_{13}=(-1)^{1+3}-13=- 13$
$A_{21}=(-1)^{2+1}.(-26)=26$
$A_{22}=(-1)^{2+2}.-39= - 39 $
$A_{23}=(-1)^{2+3}.(13)= - 13$
$A_{31}=(-1)^{3+1}.(-13)=- 13$
$A_{32}=(-1)^{3+2}.(13)= -13 $
$A_{33}=(-1)^{3+3}.(-65)=-65$
adj (adj A)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}- 13 & -26 & -13\\-26 & -39 & -13\\-13 & -13 & -65\end{bmatrix}$
$(adj A)^{-1}=\frac{1}{|adj A|}.adj(adj A)$we know |adj A|=169.
$(adj A)^{-1}=\frac{1}{169}\begin{bmatrix}-13 & -26 & -13\\-26 & -39 & -13\\-13 & -13 & -65\end{bmatrix}=\begin{bmatrix}-13/169 & -26/169 & -13/169\\-26/169 & -39/169 & -13/169\\-13/169 & -13/169 & -65/169\end{bmatrix}$
$\;\;\;=\begin{bmatrix}-1/13 & 2/13 & -1/13\\2/13 & -3/13 & -1/13\\-1/13 & -1/13 & -5/13\end{bmatrix}$-----LHS
Let us now find ${adj A}^{-1}$
We know $A^{-1}=\frac{1}{13}\begin{bmatrix}-14 & -11 & 5\\-11 & -4 & 3\\5 & 3 & 1\end{bmatrix}$
(adj A) is
$M_{11}=\frac{1}{169}\begin{vmatrix}-4 & 3\\3 & 1\end{vmatrix}$={-4-9}/169=-13/169
$M_{12}=\frac{1}{169}\begin{vmatrix}-11 & 3\\5 & 1\end{vmatrix}$={-11-15}/169=-13/169
$M_{13}=\frac{1}{169}\begin{vmatrix}-11 & -4\\5 & 3\end{vmatrix}$={-33+20}/169=-13/169
$M_{21}=\frac{1}{169}\begin{vmatrix}-11 & 5\\3 & 1\end{vmatrix}$={-11-15}/169=-26/169
$M_{22}=\frac{1}{169}\begin{vmatrix}-14 & 5\\5 & 1\end{vmatrix}$={-14-25}/169=-39/169
$M_{23}=\frac{1}{169}\begin{vmatrix}-14 & -11\\5 & 3\end{vmatrix}$={-42+55}/169=13/169
$M_{31}=\frac{1}{169}\begin{vmatrix}-11 & 5\\-4 & 3\end{vmatrix}$={-33+20}/169=-13/169
$M_{32}=\frac{1}{169}\begin{vmatrix}-14 & 5\\-11 & 3\end{vmatrix}$={-42+55}/169=13/169
$M_{33}=\frac{1}{169}\begin{vmatrix}-14 & -11\\-11 & -4\end{vmatrix}$={56-121}/169=-65/169
$A_{11}=(-1)^{1+1}.(-13)=-13/169=-1/13.$
$A_{12}=(-1)^{1+2}.(-26)=26/169=2/13.$
$A_{13}=(-1)^{1+3}.(-13)=-13/169=-1/13.$
$A_{21}=(-1)^{2+1}.(-26)=26/169=2/13.$
$A_{22}=(-1)^{2+2}.(-39)=-39/169=-3/13.$
$A_{23}=(-1)^{2+3}.(13)=-13/169=-1/13.$
$A_{31}=(-1)^{3+1}.(-13)=-13/169=-1/13.$
$A_{32}=(-1)^{3+2}.(13)=-13/169=-1/13.$
$A_{33}=(-1)^{3+3}.(-65)=-65/169=-5/13.$
adj $\big(A^{-1}\big)$=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}- 1/13 & 2/13 & -1/13\\2/13 & -3/13 & -1/13\\-1/13 & -1/13 & -5/13\end{bmatrix}$-----RHS
Hence LHS=RHS
$\Rightarrow [adj A]^{-1}=adj \big(A^{-1}\big)$
answered Mar 5, 2013 by vijayalakshmi_ramakrishnans
 

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