Question

Solve using Cramer’s rule
3x + y + z = 3
2x + 2y + 5z = -1
x - 3y - 4z = 2

Answer 1

Answer:

To find the Cramer's rule formula for a $3 \times 3$ matrix, we need to consider the system of $3$ equation with three variable.

Consider

$a_1x+b_1y+c_1=d_1\\a_2x+b_2y+c_2=d_2\\a_3x+b_2y+c_3=d_3$

Let us write these equation in the form $Axcb.$

$\begin{bmatrix}a_1 &b_1 &c_1 \\ a_2& b_2&c_2 \\a_3 &b_3&c_3 \\\end{bmatrix}\begin{bmatrix}x \\ y\\z \\\end{bmatrix} =\begin{bmatrix}d_1 \\ d_2\\d_3\\\end{bmatrix}$

Explanation:

Given:

$3x+y+z=3\\2x+2y+5z=-1\\x-3y-4z=2$

Write the equation in the form $Ax=B$

$\begin{bmatrix}3 &1 &1 \\ 2& 2&5 \\1 &-3&-4 \\\end{bmatrix}\begin{bmatrix}x \\ y\\z \\\end{bmatrix} =\begin{bmatrix}3 \\ -1\\2\\\end{bmatrix}$

$D=(A)=\begin{vmatrix}3 &1 &1 \\ 2& 2&5 \\1 &-3&-4 \\\end{vmatrix}=3[-8+15]-1[-8+15]-1[-8-5]+1[-6-2]$

$=3[7]+13-8\\=21+13-8\\=26$

$D \neq 0,$ so the given system of equation has unique solution,

$D_x=\begin{vmatrix}3 &1 &1 \\ -1& 2&5 \\2 &-3&-4 \\\end{vmatrix}=3[-8+15]-1[+4-10]+1[3-4]$

$=21+6-1\\=26$

$D_y=\begin{vmatrix}3 &3 &1 \\ 2&-1&5 \\1 &2&-4 \\\end{vmatrix}=3[4-10]-3[-8-5]-1[4+1]\\=-18+39+5\\=26$

$D_z=\begin{vmatrix}3 &1 &3 \\ 2&2&-1 \\1 &-3&2 \\\end{vmatrix}=3[4-3]-1[4+1]+3[-6-2]$

$=3-5-24\\=-26$

Thus,

$x=\frac {Dx}{D}=\frac {26}{26}=1\\y=\frac {Dy}{D}=\frac {26}{26}=1\\z=\frac {Dz}{D}=\frac {-26}{26}=-1$

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