The problem has two parts: (a) involves linear transformations and matrices, and (b) involves polynomials and remainders. (a) (i) We are given two linear transformations $P$ and $Q$ and asked to write down the matrices $P$ and $Q$ which represent these transformations. The transformations are: $P(x, y) = (5x + 3y, 6x + 4y)$ $Q(x, y) = (4x - 3y, -6x + 5y)$ (ii) Find the matrix $QP$. (iii) Find the inverse of matrix $Q$, $Q^{-1}$. (b) Two polynomials are given: $x^3 + 4x^2 - 19x - 6$ and $x^3 - 3x^2 + 5x - 15$. When these polynomials are divided by $(x + m)$, they have the same remainder. Find the value(s) of $m$.

1. Problem Description

The problem has two parts: (a) involves linear transformations and matrices, and (b) involves polynomials and remainders.

(a)

(i) We are given two linear transformations $P$ and $Q$ and asked to write down the matrices $P$ and $Q$ which represent these transformations. The transformations are:

$P(x, y) = (5x + 3y, 6x + 4y)$

$Q(x, y) = (4x - 3y, -6x + 5y)$

(ii) Find the matrix $QP$.

(iii) Find the inverse of matrix $Q$, $Q^{-1}$.

(b)

Two polynomials are given: $x^3 + 4x^2 - 19x - 6$ and $x^3 - 3x^2 + 5x - 15$. When these polynomials are divided by $(x + m)$, they have the same remainder. Find the value(s) of $m$.

2. Solution Steps

(a)

(i)

The matrix for transformation $P$ can be directly written from the coefficients of $x$ and $y$ in the transformation $P(x, y) = (5x + 3y, 6x + 4y)$. Thus, the matrix $P$ is:

$P = \begin{bmatrix} 5 & 3 \\ 6 & 4 \end{bmatrix}$

Similarly, the matrix for transformation $Q$ can be written from the coefficients of $x$ and $y$ in the transformation $Q(x, y) = (4x - 3y, -6x + 5y)$. Thus, the matrix $Q$ is:

$Q = \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix}$

(ii)

To find the matrix $QP$, we need to multiply matrix $Q$ by matrix $P$.

$QP = \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 6 & 4 \end{bmatrix} = \begin{bmatrix} (4*5 + (-3)*6) & (4*3 + (-3)*4) \\ (-6*5 + 5*6) & (-6*3 + 5*4) \end{bmatrix} = \begin{bmatrix} (20 - 18) & (12 - 12) \\ (-30 + 30) & (-18 + 20) \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

(iii)

To find the inverse of matrix $Q$, $Q^{-1}$, we use the formula for the inverse of a $2 \times 2$ matrix. If $Q = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $Q^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

For $Q = \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix}$, we have $a = 4$, $b = -3$, $c = -6$, and $d = 5$.

The determinant of $Q$ is $ad - bc = (4)(5) - (-3)(-6) = 20 - 18 = 2$.

Therefore, $Q^{-1} = \frac{1}{2} \begin{bmatrix} 5 & 3 \\ 6 & 4 \end{bmatrix} = \begin{bmatrix} 5/2 & 3/2 \\ 3 & 2 \end{bmatrix}$

(b)

According to the Remainder Theorem, the remainder when a polynomial $f(x)$ is divided by $(x - c)$ is $f(c)$.

Let $f(x) = x^3 + 4x^2 - 19x - 6$ and $g(x) = x^3 - 3x^2 + 5x - 15$. We are given that when $f(x)$ and $g(x)$ are divided by $(x + m)$, they have the same remainder. Thus, $f(-m) = g(-m)$.

$f(-m) = (-m)^3 + 4(-m)^2 - 19(-m) - 6 = -m^3 + 4m^2 + 19m - 6$

$g(-m) = (-m)^3 - 3(-m)^2 + 5(-m) - 15 = -m^3 - 3m^2 - 5m - 15$

Setting these equal:

$-m^3 + 4m^2 + 19m - 6 = -m^3 - 3m^2 - 5m - 15$

$7m^2 + 24m + 9 = 0$

We can solve this quadratic equation for $m$ using the quadratic formula or by factoring. Let's try factoring:

$(7m + 3)(m + 3) = 0$

So, $7m + 3 = 0$ or $m + 3 = 0$.

If $7m + 3 = 0$, then $m = -\frac{3}{7}$.

If $m + 3 = 0$, then $m = -3$.

3. Final Answer

(a)

(i) $P = \begin{bmatrix} 5 & 3 \\ 6 & 4 \end{bmatrix}$, $Q = \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix}$

(ii) $QP = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

(iii) $Q^{-1} = \begin{bmatrix} 5/2 & 3/2 \\ 3 & 2 \end{bmatrix}$

(b) $m = -3$ or $m = -\frac{3}{7}$