Question

With the help of matrices, the solution of the equations

3x + y + 2z = 3, 2x – 3y – z = –3, x + 2y + z = 4 are given by

• A. x = 1, y = 2, z = -1
• B. x = -1, y = 2, z = 1
• C. x = 1, y = -2, z = -1
• D. x = -1, y = -2, z = 1

Answer 1

Answer: (A)

x = 1, y = 2, z = -1

Answer 2

We can write the given equations as

AX = B …(1)

Where, $A = \begin{bmatrix} 3 & 1 & 2 \ 2 & - 3 & - 1 \ 1 & 2 & 1 \end{bmatrix},X = \begin{bmatrix} x \ y \ z \end{bmatrix},B = \begin{bmatrix} 3 \

• 3 \ 4 \end{bmatrix}$

Since, $|A| = \left| \begin{matrix} 3 & 1 & 2 \\ 2 & - 3 & - 1 \\ 1 & 2 & 1 \end{matrix} \right|$

= 3 (-3 + 2 ) –1 (2 + 1) + 2 (4 +3) = -3 –3 + 14 = 8 $\neq 0$

From (1), we have X = $A^{- 1}$B …(2)

Now,

}{A_{21} = 3, A_{22} = 1, A_{23} = - 5 }$$$A_{31} = 5, A_{32} = 7 A_{33} = - 11$Let, $C = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} - 1 & - 3 & 7 \\ 3 & 1 & - 5 \\ 5 & 7 & - 11 \end{bmatrix}$ adj $A = \begin{bmatrix} - 1 & 3 & 5 \\ - 3 & 1 & 7 \\ 7 & - 5 & - 11 \end{bmatrix}$ $ $A^{- 1} = \frac{1}{8}\begin{bmatrix} - 1 & 3 & 5 \\ - 3 & 1 & 7 \\ 7 & - 5 & - 11 \end{bmatrix}$ $ $A^{- 1}B = \frac{1}{8}\begin{bmatrix} - 1 & 3 & 5 \\ - 3 & 1 & 7 \\ 7 & - 5 & - 11 \end{bmatrix}.\begin{bmatrix} 3 \\ - 3 \\ 4 \end{bmatrix} = \frac{1}{8}\begin{bmatrix} - 3 - 9 + 20 \\ - 9 - 3 + 28 \\ 21 + 15 - 44 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ - 1 \end{bmatrix}$Hence, from (2) $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ - 1 \end{bmatrix} \Rightarrow x = 1,y = 2,z = - 1.$$