solve these linear equations by Cramer's Rules Xj=det Bj / det A:(a) 2x1 + 5x2 =1 x1+4x2=2(b) 2x1+x2=1 x1+2x2+x3=0 x2+2x3=0

Answer:(a)[tex]x_1=-2,x_2=1[/tex](b)[tex]x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}[/tex]Step-by-step explanation:(a) For using Cramer's rule you need to find matrix [tex]A[/tex] and the matrix [tex]B_j[/tex] for each variable. The matrix [tex]A[/tex] is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get [tex]A[/tex] more easily.[tex]2x_1+5x_2=1\\x_1+4x_2=2[/tex][tex]\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right][/tex]To get [tex]B_1[/tex], replace in the matrix A the 1st column with the results of the equations:[tex]B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right][/tex]To get [tex]B_2[/tex], replace in the matrix A the 2nd column with the results of the equations:[tex]B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right][/tex]Apply the rule to solve [tex]x_1[/tex]:[tex]x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2[/tex]In the case of B2,  the determinant is going to be zero. Instead of using the rule, substitute the values ​​of the variable [tex]x_1[/tex] in one of the equations and solve for [tex]x_2[/tex]:[tex]2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1[/tex](b) In this system, follow the same steps,ust remember [tex]B_3[/tex] is formed by replacing the 3rd column of A with the results of the equations:[tex]2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0[/tex][tex]\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right][/tex][tex]B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right][/tex][tex]B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right][/tex][tex]B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right][/tex][tex]x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}[/tex][tex]x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}[/tex][tex]x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}[/tex]