Question

if $\left[ \begin{array} { l l l } 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{array} \right]$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ = $\begin{bmatrix} 4–2 \\ 0–6 \\ –12 \end{bmatrix}$ $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ then (x, y, z) is equal to –

• A. (–4, 2, 2)
• B. (4, 2, 2)
• C. (–4, –2, –2)
• D. (4, –2, –2)

Answer 1

Answer: (A)

(–4, 2, 2)

Answer 2

$\begin{bmatrix} x + 2y + 3z \\ 3x + y + 2z \\ 2x + 3y + z \end{bmatrix}$ = $\begin{bmatrix} 8–2 \\ 0–6 \\ –2 + 2 \end{bmatrix}$

̃ x + 2y + 3z = 6 …(1)

̃ 3x + y + 2z = – 6 …(2)

2x + 3y + z = 0 …(3))

Solving equation (1), (2), (3)) we get

x = – 4, y = 2, z = 2.