91Ó°ÊÓ

The given system of equations is:

In determinant $D$all the coefficients are taken.

So,

$$\begin{gathered} D=\left|\begin{array}{ccc}2& 5& 0\\ 0& 3& -1\\ 4& 0& 3\end{array}\right| \\ D=2(9-0)-5(0+4)+0 \\ D=18-20 \\ D=-2 \end{gathered}$$

In determinant $D_x$, we take the constants in place of coefficients of

So,

$$\begin{gathered} D_x=\left|\begin{array}{ccc}4& 5& 0\\ 3& 3& -1\\ -3& 0& 3\end{array}\right| \\ {D}_x=4(9-0)-5(9-3)+0 \\ {D}_x=36-30 \\ {D}_x=6 \end{gathered}$$

In the determinant $D_y$, we take the constant in place of coefficients of

So,

$$\begin{gathered} D_y=\left|\begin{array}{ccc}2& 4& 0\\ 0& 3& -1\\ 4& -3& 3\end{array}\right| \\ {D}_y=2(9-3)-4(0+4) \\ {D}_y=12-16 \\ {D}_y=-4 \end{gathered}$$

In the determinant $D_z$, we take the constant in place of coefficients of

So,

$$\begin{gathered} D_z=\left|\begin{array}{ccc}2& 5& 4\\ 0& 3& 3\\ 4& 0& -3\end{array}\right| \\ {D}_z=2(-9-0)-5(0-12)+4(0-12) \\ {D}_z=-18+60-48 \\ {D}_z=-6 \end{gathered}$$

For $x$

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{6}{-2} \\ x=-3 \end{gathered}$$

For $y$

$$\begin{gathered} y=\frac{D_z}{D} \\ y=\frac{-4}{-2} \\ y=2 \end{gathered}$$

For $z$

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{-6}{-2} \\ z=3 \end{gathered}$$

The solution of the system in the ordered triad is$\left(-3,2,3\right)$

Substituting $\left(-3,2\right)$in the equation, we get:

$$\begin{gathered} 2x+5y=4 \\ 2(-3)+5\left(2\right)=4 \\ -6+10=4 \\ 4=4 \end{gathered}$$

This is true

Substituting $\left(-3,2,3\right)$in the equation, we get:

$$\begin{gathered} 3y−z=3 \\ 3\left(2\right)-3=3 \\ 3=3 \end{gathered}$$

This is true.

Substituting $\left(-3,2,3\right)$in the equation we get:

$$\begin{gathered} 4x+3z=−3 \\ 4(-3)+3\left(3\right)=-3 \\ -12+9=-3 \\ -3=-3 \end{gathered}$$

The solution of the system of equations are the ordered triad is$(-3,2,3)$