Question Solve for question b and d  Exercise 1.2.8 In each of the following, find (if possi-ble) conditions on a and b such that the system has nosolution, one solution, and infinitely many solutions.a. x- 2y = 1ax + by = 5b. x+ by = -1ах + 2y 3D 5c. x- by = -1x+ ay = 3d. ax + y = 12x + y = b

Step 1(b) Given system in matxin form\left[\begin{array}{ll}1 & b \\a & 2\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{r}-1 \\5\end{array}\right]consider \left[\begin{array}{ll|l}1 & b & -1 \\ a & 2 & 5\end{array}\right]\begin{array}{l}R_{2}-a R_{1} \\{\left[\begin{array}{cc|c}1 & b & -1 \\0 & 2-a b & 5-14\end{array}\right]}\end{array}Ho situhor: if 2-a b=0 \quad and 5+a \neq 0 if ab-2 and a+-5Gne soluhion : if 2-a b \neq 0\text { if } a b+2Infinitely maris soluhin : if 2-a b=0 and a+5=0if 2=a b and a=-5if b=-\frac{2}{5}and a=-5Step 2(d) Matriu form \left[\begin{array}{ll}a & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}1 \\ b\end{array}\right]consider\begin{array}{l}{\left[\begin{array}{cc|c}a & 1 & 1 \\2 & 1 & b\end{array}\right]} \\a R_{2}-2 R_{1} \\{\left[\begin{array}{ll|l}a & 1 & 1 \\0 & a-2 & a b-2\end{array}\right]}\end{array}This is row eachlon form of [A \mid B] No solution: if a-2 \neq 0 and a b-2 \neq 0 if a=2 \quad and a b \neq 2\text { if } a=2 \quad \text { and } b \neq 1one solution: if a-2 \neq 0if a \neq 2Infinitely meny selution if a-2=0 and a b-2=0 if a=2 and b=1 ...