Question

If the system of equations

$$\begin{gathered} x-2y+3z=9 \\ 2x+y+z=b \\ x-7y+az=24, \end{gathered}$$

has infinitely many solutions, then $a-b$ is equal to:

Answer 1

Step 1: Find the value of $a$.

As the given system

$$\begin{gathered} x-2y+3z=9 \\ 2x+y+z=b \\ x-7y+az=24, \end{gathered}$$

has infinitely many solutions, so the value of the determinant is equal to zero, so

$$\begin{gathered} \left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0 \\ \Rightarrow 1\left(a+7\right)+2\left(2a-1\right)+3\left(-14-1\right)=0 \\ \Rightarrow a+7+4a-2-45=0 \\ \Rightarrow 5a=40 \\ \Rightarrow a=8 \end{gathered}$$

Step 2. Find the value of $b$.

Again the value of the determinant $D_3=0$

$$\begin{gathered} \left|\begin{array}{ccc}1& -2& 9\\ 2& 1& b\\ 1& -7& 24\end{array}\right|=0 \\ \Rightarrow 1\left(24+7b\right)+2\left(48-b\right)+9\left(-14-1\right)=0 \\ \Rightarrow 24+7b+96-2b-135=0 \\ \Rightarrow 5b=15 \\ \Rightarrow b=3 \end{gathered}$$

Step 3. Find the value of $a-b$.

As $a=8$ and $b=3$, so the value of $a-b$ is $8-3=5$.

Thus, the value of $a-b$ is $5$.

Hence, the answer is $5$.