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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.

Can you answer this question?

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.
answered Apr 5, 2013