Question

If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third numbers, we get 8. If three times the first number is added to the sum of second and third numbers, we get 4. Find the numbers using matrices.

Answer 1

Let the three numbers be x, y, z.

According to the first condition,

x + y + z = 2

According to the second condition,

(x + z) − 2y = 8

i.e. x − 2y + z = 8

According to the third condition,

3x + y + z = 4

Matrix form of the given system of equations is

${\displaystyle \left[\begin{array}{ccc}1& 1& 1\\ 1& -2& 1\\ 3& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 8\\ 4\end{array}\right]}$

Applying R₂ → R₂ − R₁ and R₃ → R₃ − 3R₁

${\displaystyle \left[\begin{array}{ccc}1& 1& 1\\ 0& -3& 0\\ 0& -2& -2\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 6\\ -2\end{array}\right]}$

Applying R₂ → ${\displaystyle (-\frac{1}{3})}$ R₂ and R₃ → ${\displaystyle (-\frac{1}{2})}$ R₃ 

${\displaystyle \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 1\end{array}\right]}$

Applying R₃ → R₃ − R₂,

${\displaystyle \left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]}$

Hence, the original matrix is reduced to an upper triangular matrix

∴ By equality of matrices, we get

x + y + z = 2    .......(i)

y = − 2       .......(ii)

z = 3       .......(iii)

Putting y = −2 and z = 3 in equation (i), we get

x − 2 + 3 = 2

∴ x = 1

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