To solve the problem, we need to compute $P + Q = R$ and determine the values of $a$ and $b$ such that this equality holds.

Matrices:

Given:

• $P = \begin{bmatrix} 13 & 4a + b \\ -2a & -5 \end{bmatrix}$
• $Q = \begin{bmatrix} -6 & a \\ b & 2 \end{bmatrix}$
• $R = \begin{bmatrix} 7 & 10 \\ -8 & -3 \end{bmatrix}$

Addition of Matrices:

$P + Q = \begin{bmatrix} 13 & 4a + b \\ -2a & -5 \end{bmatrix} + \begin{bmatrix} -6 & a \\ b & 2 \end{bmatrix} = \begin{bmatrix} 13 - 6 & (4a + b) + a \\ (-2a) + b & -5 + 2 \end{bmatrix}$

Simplify the terms: $P + Q = \begin{bmatrix} 7 & 4a + b + a \\ -2a + b & -3 \end{bmatrix}$

Equating with $R = \begin{bmatrix} 7 & 10 \\ -8 & -3 \end{bmatrix}$, we get the system of equations:

1. $4a + b + a = 10$ (from the top-right elements)
2. $-2a + b = -8$ (from the bottom-left elements)

Solve the System of Equations:

From the first equation: 5a + b = 10 \tag{1}

From the second equation: -2a + b = -8 \tag{2}

Subtract equation (2) from equation (1): $(5a + b) - (-2a + b) = 10 - (-8)$ $7a = 18 \implies a = \frac{18}{7}$

Substitute $a = \frac{18}{7}$ into equation (1): $5\left(\frac{18}{7}\right) + b = 10$ $\frac{90}{7} + b = 10$ $b = 10 - \frac{90}{7} = \frac{70}{7} - \frac{90}{7} = -\frac{20}{7}$

Compute $a - b$:

$a - b = \frac{18}{7} - \left(-\frac{20}{7}\right) = \frac{18}{7} + \frac{20}{7} = \frac{38}{7}$

Final Answer:

$a - b = \frac{38}{7}$

Let me know if you'd like a more detailed explanation or have any additional questions.

Related Questions:

1. How do you verify if the solution satisfies all equations?
2. What are the properties of matrix addition that make this problem solvable?
3. Can the solution be checked by directly substituting into $P + Q$?
4. What happens if $P$, $Q$, or $R$ are not of the same dimensions?
5. How can this method be applied to larger matrices?

Tip:

When solving systems of linear equations, always double-check your substitutions to avoid arithmetic errors.