Question

Using Cramer’s rule to solve the equation:

Answer 1

`\large\boxed{x=(5)/(2),\ y=(11)/(6),\ z=(5)/(6)}`

Explanation:

Cramer's rule:

`\left\{\begin{array}{ccc}a_1x+b_1y+c_1z=d_1\\a_2x+b_2y+c_2z=d_2\\a_3x+b_3y+c_3z=d_3\end{array}\right\\\\A=\left[\begin{array}{ccc}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{array}\right],\ A_x=\left[\begin{array}{ccc}d_1&b_1&c_1\\d_2&b_2&c_2\\d_3&b_3&c_3\end{array}\right],\\A_y=\left[\begin{array}{ccc}a_1&d_1&c_1\\a_2&d_2&c_2\\a_3&d_3&c_3\end{array}\right] ,\ A_z=\left[\begin{array}{ccc}a_1&b_1&d_1\\a_2&b_2&d_2\\a_3&b_3&d_3\end{array}\right]`

`\left\{\begin{array}{ccc}x=(\det A_x)/(\det A)\\y=(\det A_y)/(\det A)\\z=(\det A_z)/(\det A)\end{array}\right`

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We have the system of equations:

`\left\{\begin{array}{ccc}2x-y+z=4\\x-y-2z=-1\\y-z=1\end{array}\right\\\\A=\left[\begin{array}{ccc}2&-1&1\\1&-1&-2\\0&1&-1\end{array}\right] \\\\\det A=\\(2)(-1)(-1)+(0)(-1)(-2)+(1)(1)(1)+(0)(-1)(1)-(2)(1)(-2)-(-1)(1)(-1)\\=2+0+1+0+4-1=6`

`A_x=\left[\begin{array}{ccc}4&-1&1\\-1&-1&-2\\1&1&-1\end{array}\right] \\\\\det A_x=\\(4)(-1)(-1)+(1)(-1)(-2)+(1)(-1)(1)-(1)(-1)(1)-(4)(1)(-2)-(-1)(-1)(-1)\\=4+2-1+1+8+1=15`

`A_y=\left[\begin{array}{ccc}2&4&1\\1&-1&-2\\0&1&-1\end{array}\right] \\\\\det A_y=\\(2)(-1)(-1)+(0)(4)(-2)+(1)(1)(1)-(0)(-1)(1)-(-1)(1)(4)-(2)(1)(-2)\\=2+0+1+0+4+4=11`

`A_z=\left[\begin{array}{ccc}2&-1&4\\1&-1&-1\\0&1&1\end{array}\right] \\\\\det A_z=\\(2)(-1)(1)+(0)(-1)(-1)+(4)(1)(1)-(0)(-1)(4)-(1)(1)(-1)-(2)(1)(-1)\\=-2-0+4+0+1+2=5`

`x=(15)/(6)=(5)/(2)\\\\y=(11)/(6)\\\\z=(5)/(6)`