Trilogy of Equations

The conditions $a, b, y$ are real and $b^3+b^2=2$ yields $b=1$

Solution to the equation $a^2-2ay+6a+y^2-6y+9=0$ is $y=a+3$

$A$ will be ill-defined if determinant of the matrix is zero. That is,

$8a(a+1)(a+2)=15\implies a=\dfrac {1}{2}\implies y=\dfrac {1}{2}+3=\dfrac {7}{2}$

So, $u=7,v=2$ and $u+v=\boxed 9$ .

We need to write $a$ in terms of $y$ . We can easily see that $b$ will be a real number, and that it is independent of $y$ . Let's start by solving the first equation in terms of $a$ . We have: $\\ { a }^{ 2 }-2ay+6a+{ y }^{ 2 }-6y+9=0 \\ { a }^{ 2 }-2a(y-3)+{ (y-3) }^{ 2 }=0 \\ { (a-(y-3)) }^{ 2 }=0 \\ a=y-3 \\$ Now let's solve for $b$ . $\\ { b }^{ 3 }+{ b }^{ 2 }=2 \\ { b }^{ 3 }+{ b }^{ 2 }=1+1 \\ { b }^{ 3 }-1+{ b }^{ 2 }-1=0 \\ (b-1)({ b }^{ 2 }+{ b }+1)+(b-1)(b+1)=0 \\ (b-1)({ b }^{ 2 }+2b+2)=0 \\$ We can easily check that the quadratic trinomial here has no real roots, and thus the only solution of this equation is $b=1$ . Plugging the following values into the matrix $A$ , we get: $\\ { \begin{bmatrix} y-3 & 1 \\ \frac { 15 }{ 8 } & (y-2)(y-1) \end{bmatrix} }^{ -1 }\\$ We know that a matrix has an inverse iff it is a square matrix and its determinant is not equal to zero. Since the existence of an inverse matrix depends on the choice of $y$ in this case, we will find the value for which it is nonexistent by making the determinant of the "original" matrix equal to zero. $\\$ Let $B=A^{-1}$ . The determinant of $B$ is equal to: $\\ \begin{vmatrix} y-3 & 1 \\ \frac { 15 }{ 8 } & (y-2)(y-1) \end{vmatrix}=(y-3)(y-2)(y-1)-\frac { 15 }{ 8 }.\\$ Now we have to solve for: $\\(y-3)(y-2)(y-1)-\frac { 15 }{ 8 }=0.\\$ We will simplify this equation. Let $u=y-2$ . We have: $\\$ $(u-1)u(u+1)-\frac { 15 }{ 8 } =0\\$ ${ u }^{ 3 }-u-\frac { 15 }{ 8 } =0\\$ Let's try to solve this equation using the intermediate value theorem from calculus. We can take the derivative of the polynomial to find that it is monotonically increasing from negative infinity to $-\frac { 1 }{ \sqrt { 3 } }$ and from $\frac { 1 }{ \sqrt { 3 } }$ to infinity. Since the function only starts increasing indefinitely for these (positive) $u$ values, and we can check that its left local extreme value is still negative, we can conclude that the root of the function only exists for positive $u$ values. For $u=1$ we get $-\frac { 15 }{ 8 }$ and for $u=2$ we get $\frac { 33 }{ 8 }$ , so we can see that our desired value is between $1$ and $2$ . For $u=\frac { 3 }{ 2 }$ we get 0, so that is the value we were looking for. Substituting back to get $y$ , we get that $y=\frac { 3 }{ 2 } +2=\frac { 7 }{ 2 }$ , so the answer is 9.