Question

find x, y, z by method of cramer's rule​

Answer 1

Answer:

Cramer's Rule: 5. Use Cramer's Rule to find x,y and z for the following system of equations. X 2 7x + 2y - z= -1 ។ 6x + 5y + z = 16 -5x - 4y + 3z = -5 2 : 2 a. Write the coefficient matrix first for the system above. Call it matrix D. 7 2 5 L-8-4 3 1 14 ] = 0 b. Find the determinant of the coefficient matrix (det(D)).

Answer 2

The following system of equations can be represented in the matrix equation,

$\small\longrightarrow\left[\begin{array}{ccc}1&1&1\\2&2&3\\1&4&9\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\6\\3\end{array}\right]$

To solve it using Cramer's rule, we need to find out the following determinants.

$\small\longrightarrow D=\left|\begin{array}{ccc}1&1&1\\2&2&3\\1&4&9\end{array}\right|$

Performing operations $\smallC_2-C_1\to C_2$ and $\smallC_3-C_1\to C_3,$

$\small\longrightarrow D=\left|\begin{array}{ccc}1&0&0\\2&0&1\\1&3&8\end{array}\right|$

Expanding along R₁ we get,

$\small\longrightarrow D=-3$

and,

$\small\longrightarrow D_x=\left|\begin{array}{ccc}1&1&1\\6&2&3\\3&4&9\end{array}\right|$

Performing operations $\smallC_2-C_1\to C_2$ and $\smallC_3-C_1\to C_3,$

$\small\longrightarrow D_x=\left|\begin{array}{ccc}1&0&0\\6&-4&-3\\3&1&6\end{array}\right|$

Expanding along R₁ we get,

$\small\longrightarrow D_x=-21$

and,

$\small\longrightarrow D_y=\left|\begin{array}{ccc}1&1&1\\2&6&3\\1&3&9\end{array}\right|$

Performing operations $\smallC_2-C_1\to C_2$ and $\smallC_3-C_1\to C_3,$

$\small\longrightarrow D_y=\left|\begin{array}{ccc}1&0&0\\6&4&1\\3&2&8\end{array}\right|$

Expanding along R₁ we get,

$\small\longrightarrow D_y=30$

and,

$\small\longrightarrow D_z=\left|\begin{array}{ccc}1&1&1\\2&2&6\\1&4&3\end{array}\right|$

Performing operations $\smallC_2-C_1\to C_2$ and $\smallC_3-C_1\to C_3,$

$\small\longrightarrow D_z=\left|\begin{array}{ccc}1&0&0\\2&0&4\\1&3&2\end{array}\right|$

Expanding along R₁ we get,

$\small\longrightarrow D_z=-12$

Now,

$\small\longrightarrow x=\dfrac{D_x}{D}=\dfrac{-21}{-3}=7$

$\small\longrightarrow y=\dfrac{D_y}{D}=\dfrac{30}{-3}=-10$

$\small\longrightarrow z=\dfrac{D_z}{D}=\dfrac{-12}{-3}=4$

Hence,

$\small\text{ \longrightarrow\underline{\underline{\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}7\\-10\\4\end{array}\right]}} }$