Find h in linear transformation

Since the matrix A is transforming the vector v, the final (transformed) vector will be Av:

$\begin{pmatrix} 1 & 2 \\ 3 & h \end{pmatrix}$ $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ = $\begin{pmatrix} (1*1) + (2*1) \\ (3*1) + (h*1) \end{pmatrix}$ = $\begin{pmatrix} 3 \\ 3 + h \end{pmatrix}$ = Av

To find the value of h that makes the length of vector Av equal to 4, we must use the distance formula:

Length of Av = $\sqrt{(v_1)^2 + (v_2)^2}$ , where $\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ = $\begin{pmatrix} 3 \\ 3 + h \end{pmatrix}$

So, then, plugging v 1 and v 2 into the formula above, we get: $\sqrt{(3)^2 + (3+h)^2}$ as the length of Av . To solve the problem, we need to set the length equal to 4, and then solve algebraically for h.

• $3^2 + (3+h)^2 = 4^2$
• $(3+h)^2 = 16 - 9 = 7$
• $3+h = \sqrt{7}$
• $h = \sqrt{7} - 3$

The question asks for h to be solved in the form $h = \sqrt{a} - b$ , which is exactly what we've done here. We can see that:

$a=7$ and $b=3$

so, the value of $ab$ is simply $21$ .

The transformation gives a new vector $\begin{pmatrix} 1 & 2 \\ 3 & h \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 + h \end{pmatrix}$ , which has a length of $\sqrt{3^2 + (3 + h)^2} = 4$ , and this solves to $h = \pm \sqrt{7} - 3$ .

Therefore, $a = 7$ , $b = 3$ , and $ab = \boxed{21}$ .