The problem involves finding the matrix representation of the linear transformation $T$ relative to the basis $B = \{ b_1(x), b_2(x), b_3(x) \}$. Let’s break it into steps:

Step 1: Recall the basis and the transformation

The basis is: $B = \{ b_1(x) = x^2 + 2x - 4, \; b_2(x) = x - 2, \; b_3(x) = 1 \}.$

The transformation is given as: $T(a + bx + cx^2) = 3b + (a + 4c)x + bx^2.$

Step 2: Represent $T(b_1(x)), T(b_2(x)),$ and $T(b_3(x))$ as linear combinations of $b_1(x), b_2(x), b_3(x)$

To find the matrix representation of $T$, we apply $T$ to each basis vector $b_1(x), b_2(x),$ and $b_3(x)$, and express the result in terms of the basis $B$.

Step 2.1: Apply $T$ to $b_1(x) = x^2 + 2x - 4$

Substitute $a = -4$, $b = 2$, $c = 1$ (coefficients of $1, x, x^2$ in $b_1(x)$) into $T(a + bx + cx^2)$: $T(b_1(x)) = T(-4 + 2x + 1x^2) = 3(2) + (-4 + 4(1))x + 2x^2.$ Simplify: $T(b_1(x)) = 6 + (-4 + 4)x + 2x^2 = 6 + 2x^2.$

Now express $T(b_1(x)) = 6 + 2x^2$ in terms of $B$: $6 + 2x^2 = 2b_1(x) - 2b_2(x) + 6b_3(x).$ Thus, the coefficients are $[2, -2, 6]$.

Step 2.2: Apply $T$ to $b_2(x) = x - 2$

Substitute $a = -2$, $b = 1$, $c = 0$ into $T(a + bx + cx^2)$: $T(b_2(x)) = T(-2 + 1x + 0x^2) = 3(1) + (-2 + 4(0))x + 1x^2.$ Simplify: $T(b_2(x)) = 3 - 2x + x^2.$

Now express $T(b_2(x)) = 3 - 2x + x^2$ in terms of $B$: $3 - 2x + x^2 = b_1(x) - 3b_2(x) + 3b_3(x).$ Thus, the coefficients are $[1, -3, 3]$.

Step 2.3: Apply $T$ to $b_3(x) = 1$

Substitute $a = 1$, $b = 0$, $c = 0$ into $T(a + bx + cx^2)$: $T(b_3(x)) = T(1 + 0x + 0x^2) = 3(0) + (1 + 4(0))x + 0x^2.$ Simplify: $T(b_3(x)) = x.$

Now express $T(b_3(x)) = x$ in terms of $B$: $x = -\frac{1}{2}b_1(x) + \frac{3}{2}b_2(x) + 0b_3(x).$ Thus, the coefficients are $[-1/2, 3/2, 0]$.

Step 3: Construct the matrix representation

The columns of the matrix are the coefficients of $T(b_1(x)), T(b_2(x)), T(b_3(x))$ relative to $B$: