question_answer

Use matrix product

to solve the system of equations

\[x-y+2z=1,\] \[2y-3z=1\]

and       \[3x-2y+4z=2.\]

OR

The sum of three numbers is 6. Twice the third number when added to the first number gives 7. On adding the sum of the second and third numbers to thrice the first number, we get 12. Find the numbers, using matrix method.

Answer:

Let and  The  product AB is given by

[multiplying row by column]

\[\Rightarrow \]   \[AB={{l}_{3}}\]

On multiplying both sides by \[{{A}^{-1}},\] we get

\[{{A}^{-1}}AB={{A}^{-1}}{{l}_{3}}\]

\[\Rightarrow \]   \[B={{A}^{-1}}\]                    \[[\because \,\,A{{A}^{-1}}=l\,\text{and}\,lA=Al]\]

\[\therefore \]

The given system of equations can be written in matrix form as AX = C

where,  and

Now, \[AX=C\,\,\,\Rightarrow \,\,\,X={{A}^{-1}}C\]

\[\Rightarrow \]

\[\Rightarrow \]

[multiplying row by column]

\[\Rightarrow \]

On comparing corresponding elements from both sides, we get

x = 0, y = 5 and z = 3

OR

Let the first, second and third numbers be x, y, z respectively. Then,

\[x+y+z=6\]                             ?(i)

\[x+2z=7\]                                ?(ii)

\[3x+y+z=12\]                          ?(iii)

Let and

Then given system in matrix form is AX = B.

Now,

\[\left[ \begin{align}   & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\  & {{R}_{3}}\to {{R}_{3}}-3{{R}_{1}} \\ \end{align} \right]\]

\[=1\cdot (2+2)=4\ne 0\]

\[\therefore \] A is invertible.

So, the given system has a unique solution, \[X={{A}^{-1}}B.\] The minors of the elements of |A| are

\[{{M}_{11}}=-\,2,\] \[{{M}_{12}}=-\,5,\] \[{{M}_{13}}=1;\]

\[{{M}_{21}}=0,\] \[{{M}_{22}}=-\,2,\] \[{{M}_{23}}=-\,2;\]

\[{{M}_{31}}=2,\] \[{{M}_{32}}=1,\] \[{{M}_{33}}=-\,1\]

The cofactors of the elements of |A| are

\[{{A}_{11}}=-\,2,\] \[{{A}_{12}}=5,\] \[{{A}_{13}}=1,\]

\[{{A}_{21}}=0,\] \[{{A}_{22}}=-\,2,\] \[{{A}_{23}}=2;\]

\[{{A}_{31}}=2,\] \[{{A}_{32}}=-\,1,\] \[{{A}_{33}}=-\,1.\]

\[\therefore \] (adj A)

\[\Rightarrow \]

\[\Rightarrow \]   \[X={{A}^{-1}}B\]

\[\Rightarrow \]

\[\Rightarrow \]   x = 3, y = 1, z = 2.

Hence, the required numbers are 3, 1, 2.