Question

Find zby Cramer's rule:

$\{\begin{array}{c}\begin{array}{c}(a-b)x-(a-b)y+3b^{2}z=3ab\\ (a+2b)x-(a+2b)y-\left(3ab+3b^2\right)z=3b^2\\ bx+ay-\left(2b^2+a^2\right)z=0\end{array}\end{array}$

Short answer

The required value is $z=1$.

Step-by-step solution

Formula and concept used

If $\mathit{A}\mathit{X}\mathbf{}\mathbf{=}\mathbf{}\mathit{B}$is the system of simultaneous linear equations, $\mathit{D}$the determinant of coefficient matrix $\mathit{A}$, and ${\mathit{D}}_{\mathbf{z}}$is the determinant of matrix obtained by replacing third column of matrix $\mathit{A}$, then the value of$\mathit{z}$is$\mathit{Z}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{z}}}{\mathbf{D}}$.

Computing the value of the determinant

The given system of linear equations may be written as

$\left[\begin{array}{ccc}(a-b)& -(a-b)& 3b^2\\ (a+2b)& -(a+2b)& -(3ab+3b^2)\\ b& a& -(2b^2+a^2)\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3ab\\ 3b^2\\ 0\end{array}\right]$or $AX=B$

The determinant of the coefficient matrix is,

role="math" localid="1664277759275" $D=\left|\begin{array}{ccc}(a-b)& -(a-b)& 3b^2\\ (a+2b)& -(a+2b)& -(3ab+3b^2)\\ b& a& -(2b^2+a^2)\end{array}\right|$

Compute the value of by row and column operations as:

role="math" localid="1664278361879" $$\begin{gathered} D=\left|\begin{array}{ccc}(a-b)& -(a-b)& 3b^2\\ (a+2b)& -(a+2b)& -(3ab+3b^2)\\ b& a& -(2b^2+a^2)\end{array}\right|\stackrel{C_1\to C_1+C_2}{\to }\left|\begin{array}{ccc}0& -(a-b)& 3b^2\\ 0& -(a+2b)& -(3ab+3b^2)\\ a+b& a& -(2b^2+a^2)\end{array}\right| \\ \stackrel{R_2\to R_2-R_1}{\to }\left|\begin{array}{ccc}0& -(a-b)& 3b^2\\ 0& -3b& -(3ab+3b^2)\\ a+b& a& -(2b^2+a^2)\end{array}\right| \end{gathered}$$

Simplify further.

$$\begin{gathered} D=(a+b)\left|\begin{array}{ccc}0& -(a-b)& -3b^2\\ 0& -3b& -(3ab+6b^2)\\ 1& a& -(2b^2+a^2)\end{array}\right| \\ =(a+b)\left[(a-b)\left(3ab+6b^2\right)+9b^3\right] \\ =(a+b)\left[3a^{2}b+6a{b}^2-3a{b}^2-6b^3+9b^3\right] \\ =(a+b)\left[3a^{2}b+3a{b}^2+3b^3\right] \\ =3b(a+b)\left[a^2+ab+b^2\right] \end{gathered}$$

Thus, $D=3b(a+b)\left[a^2+ab+b^2\right]$.

Now, find $D_z$.

$D_z=\left|\begin{array}{ccc}(a-b)& -(a-b)& 3ab\\ (a+2b)& -(a+2b)& 3b^2\\ b& a& 0\end{array}\right|\stackrel{C_1\to C_1+C_2}{\to }\left|\begin{array}{ccc}0& -(a-b)& 3ab\\ 0& -(a+2b)& 3b^2\\ a+b& a& 0\end{array}\right|$

Evaluate it further.

$$\begin{gathered} D_z=3b(a+b)\left|\begin{array}{ccc}0& -(a-2b)& a\\ 0& -(a+2b)& b\\ 1& a& 0\end{array}\right| \\ =3b(a+b)[-b(a-b)+a(a+2b\left)\right] \\ =3b(a+b)\left(a^2+ab+b^2\right) \end{gathered}$$

Thus,$D_z=3b(a+b)\left[a^2+ab+b^2\right]$ .

Hence,$z=\frac{D_z}{D}=\frac{3b(a+b)\left[a^2+ab+b^2\right]}{3b(a+b)\left[a^2+ab+b^2\right]}=1$ .

Thus, the value of $z=1$.

The solution of linear equation for one variable $z=1$.