Determinant of matrix 1

$\det\left(\begin{array}{cc}1&a\\2&b \end{array}\right)=4$ $\color{#3D99F6}\large \implies$ $b-2a=4$ $\color{#D61F06}(1)$

$\det\left(\begin{array}{cc}1&b\\2&a \end{array}\right)=1$ $\color{#3D99F6}\large \implies$ $a-2b=1$ $\color{#D61F06}(2)$

From $\color{#D61F06}(1)$ , we get

$b=4+2a$

Substitute the above equation in $\color{#D61F06}(2)$ , we have

$a-2(4+2a)=1$

$a-8-4a=1$

$-3a=9$

$a=-3$

It follows that, $a^2=9$ .

Solving for $b$ , we have

$b=4+2(-3)$

$b=4-6$

$b=-2$

It follows that, $b^2=4$

Finally,

$a^2+b^2=9+4=$ $\color{plum}\boxed{\large13}$

(I) -2a + b = 4

(II) 2a - 4b = 2

-3b = 6

b = -2

---

(I) -4a + 2b = 8

(II) a - 2b = 1

-3a = 9

a = -3

---

a² + b²

9 + 4

13

solving determinants give b-2a=4 (1) and a-2b=1 (2). From (1)-(2) we have -3×(a-b)=3 => (a-b)^2=1.and (1)×(2) gives (ab- 2b^2-2a^2+4ab)= 4×1
=> ab-2(a-b)^2= 4
=> ab-2×1=4 => ab=4+2=6 =>2×ab =2×6=12.
Now (a-b)^2=1. {from(1)-(2)} =>a^2 +b^2-2ab=1 => a^2 + b^2 = 1+ 2ab =1+12=13

$\begin{cases} \det \begin{pmatrix} 1 & a \\ 2 & b \end{pmatrix} = 4 & \implies b - 2a = 4 & ...(1) \\ \det \begin{pmatrix} 1 & b \\ 2 & a \end{pmatrix} = 1 & \implies a - 2b = 1 & ...(2) \end{cases}$

$\begin{aligned} 2 \times (2): \quad 2a-4b & = 2 & ...(2a) \\ (1)+(2a): \quad -3b & = 6 \\ \implies b & = -2 \\ (2): \quad a - 2(-2) & = 1 \\ \implies a & = -3 \end{aligned}$

$\implies a^2 + b^2 = (-3)^2 + (-2)^2 = 9 + 4 = \boxed{13}$