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Questions  >>  CBSE XII  >>  Math  >>  Determinants
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The sum of three numbers is 6.If we multiply the third number by 2 and add the first number to the result,we get 7.By adding second and third numbers to three times the first number we get 12.Use determinants to find the numbers.

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A)
Toolbox:
  • If |A|$\neq 0$,then the matrix is non-singular and hence inverse exists.
  • $A^{-1}=\frac{1}{|A|}(adj A)$
  • AX=B$\Rightarrow x=A^{-1}B.$
Step 1:Let the numbers be x,y and z.
From the given condition we get,
x+y+z=6
x+2z=7
3x+y+z=12
This is of the form AX=B.
$(i.e)\begin{bmatrix}1 & 1& 1\\1 & 0 & 2\\3 & 1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\7\\12\end{bmatrix}$
Where $A=\begin{bmatrix}1 & 1 & 1\\1 & 0 & 2\\3 & 1 & 1\end{bmatrix},X=\begin{bmatrix}x\\y\\z\end{bmatrix},B=\begin{bmatrix}6\\7\\12\end{bmatrix}$
Now by expanding along $R_1$ let us find |A|
|A|=1(0-2)-1(1-6)+1(1-0)
$\;\;=-2+5+1$
$\;\;=4\neq 0$
Hence it is non-singular and $A^{-1}$ exists.
Step 2:Now let us find the adj A.
$A_{11}=(-1)^{1+1}\begin{vmatrix}0 & 2\\1 & 1\end{vmatrix}$=0-2=-2
$A_{12}=(-1)^{1+2}\begin{vmatrix}1 & 2\\3 & 1\end{vmatrix}$=-(1-6)=5
$A_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0\\3 & 1\end{vmatrix}$=1-0=1
$A_{21}=(-1)^{2+1}\begin{vmatrix}1 & 1\\1 & 1\end{vmatrix}$=-(1-1)=0
$A_{22}=(-1)^{2+2}\begin{vmatrix}1 & 1\\3 & 1\end{vmatrix}$=1-3=-2
$A_{23}=(-1)^{2+3}\begin{vmatrix}1 & 1\\3 & 1\end{vmatrix}$=-(1-3)=2
$A_{31}=(-1)^{3+1}\begin{vmatrix}1 & 1\\0 & 2\end{vmatrix}$=2-0=2
$A_{32}=(-1)^{3+2}\begin{vmatrix}1 & 1\\1 & 2\end{vmatrix}$=-(2-1)=-1
$A_{33}=(-1)^{3+3}\begin{vmatrix}1 & 1\\1 & 0\end{vmatrix}$=0-1=-1
Step 3:Hence adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 &2 &-1\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}adj (A)$ we know |A|=4.
$A^{-1}=\frac{1}{4}\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 & 2 & -1\end{bmatrix}$
Step 4:We know X=$A^{-1}B$
Now substituting for X,$A^{-1}$,B we get
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-2 & 0 & 2\\5 & -2 & -1\\1 & 2 & -1\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-12+0+24\\30-14-12\\16+14-12\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\frac{12}{4}\\\frac{4}{4}\\\frac{8}{4}\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}$
Therefore x=3,y=1,z=2.
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