Solution:

A = $\left[\begin{array}{cccc} 2 & -2 & -5 \\ 4 & -1 & 1 \\ 3 & -2 & 3 \\ 1 & -3 & 7 \end{array}\right]$

X = $\left[\begin{array}{cccc} x \\ y \\ z \\ \end{array}\right]$

B = $\left[\begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$

AX = B

$\left[\begin{array}{cccc} 2 & -2 & -5 \\ 4 & -1 & 1 \\ 3 & -2 & 3 \\ 1 & -3 & 7 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R1 ⇔ R4 $\left[\begin{array}{cccc} 1 & -3 & 7 \ 4 & -1 & 1 \ 3 & -2 & 3 \ 2 & -2 & -5 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R2 ⇒ R2 - 4(R1) R3 ⇒ R3 - 3(R1) R4 ⇒ R4 - 2(R1) $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 11 & -27 \ 0 & 7 & -18 \ 0 & 4 & -19 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R4 ⇒ R4 + R3 $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 11 & -27 \ 0 & 7 & -18 \ 0 & 1 & -12 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R2 ⇔ R4 $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 1 & -12 \ 0 & 7 & -18 \ 0 & 11 & -27 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R3 ⇒ R3 - 7(R2) R4 ⇒ R4 - 11(R2) $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 1 & -12 \ 0 & 0 & 66 \ 0 & 0 & 105 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ R3 ⇒ R3/66 $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 1 & -12 \ 0 & 0 & 1 \ 0 & 0 & 105 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right] $ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right] $ R4 ⇒ R4 - 182(R3) $\left[\begin{array}{cccc} 1 & -3 & 7 \ 0 & 1 & -12 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{array}\right]$$\left[\begin{array}{cccc} x \ y \ z \ \end{array}\right]$ = $\left[\begin{array}{cccc} 0 \ 0 \ 0 \ 0 \end{array}\right]$ By matrix multiplication , z = 0 $y -12z = 0$ Substituting the value of z, we get: ⇒y = 0 Substituting the value of y, we get: $x -3y+7z = 0$

⇒x = 0