Matrix Problems
1. Problem 16: Find matrix \( B \) given \( A \) and \( AB \).
Step 1: Write down the matrices.
\[
A = \begin{pmatrix} 2 & -2 \\ -1 & 3 \end{pmatrix}, \quad AB = \begin{pmatrix} 4 & -2 \\ 0 & 7 \end{pmatrix}
\]
Step 2: To find \( B \), multiply both sides by \( A^{-1} \) on the left: \( B = A^{-1} AB = A^{-1} (AB) = B \).
Step 3: Find \( A^{-1} \).
\[
|A| = (2)(3) - (-2)(-1) = 6 - 2 = 4
\]
\[
A^{-1} = \frac{1}{|A|} \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix}
\]
Step 4: Multiply \( A^{-1} \) and \( AB \):
\[
B = A^{-1} (AB) = \frac{1}{4} \begin{pmatrix} 3 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 4 & -2 \\ 0 & 7 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 3 \times 4 + 2 \times 0 & 3 \times (-2) + 2 \times 7 \\ 1 \times 4 + 2 \times 0 & 1 \times (-2) + 2 \times 7 \end{pmatrix}
\]
\[
= \frac{1}{4} \begin{pmatrix} 12 & -6 + 14 \\ 4 & -2 + 14 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 12 & 8 \\ 4 & 12 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix}
\]
--
2. Problem 17: Find matrix \( X \) satisfying:
\[
|P| X^T - 2P^{-1} + RS + 2 \text{diag}(1,-2) - 5I = 0
\]
Step 1: Write matrices:
\[
P = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix},
R = \begin{pmatrix} 1 & 1 & 0 \\ 2 & -1 & 1 \end{pmatrix},
S = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix}
\]
Step 2: Calculate \(|P|\):
\[
|P| = 1 \times 3 - 2 \times 1 = 3 - 2 = 1
\]
Step 3: Calculate \( P^{-1} \):
\[
P^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix}
\]
Step 4: Calculate \( RS \): Note \( R \) is \(2 \times 3\), \( S \) is \(3 \times 2\), product \(2 \times 2\).
\[
RS = \begin{pmatrix} 1 & 1 & 0 \\ 2 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 \times 1 + 1 \times 1 + 0 \times 0 & 1 \times 0 + 1 \times 1 + 0 \times 1 \\ 2 \times 1 + (-1) \times 1 + 1 \times 0 & 2 \times 0 + (-1) \times 1 + 1 \times 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}
\]
Step 5: Calculate \( 2 \text{diag}(1,-2) \):
\[
2 \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & -4 \end{pmatrix}
\]
Step 6: Identity matrix \( I \):
\[
I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
Step 7: Write the original equation and solve for \( X^T \):
\[
|P| X^T - 2P^{-1} + RS + 2 \text{diag}(1,-2) - 5I = 0 \implies
X^T = \frac{1}{|P|} \left( 2P^{-1} - RS - 2 \text{diag}(1,-2) + 5I \right)
\]
Step 8: Substitute values (\(|P|=1\)):
\[
X^T = 2 \begin{pmatrix} 3 & -2 \\ -1 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & -4 \end{pmatrix} + 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
Calculate stepwise:
\[
2P^{-1} = \begin{pmatrix} 6 & -4 \\ -2 & 2 \end{pmatrix}
\]
Sum:
\[
\begin{pmatrix} 6 & -4 \\ -2 & 2 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 4 & -5 \\ -3 & 2 \end{pmatrix}
\]
\[
\begin{pmatrix} 4 & -5 \\ -3 & 2 \end{pmatrix} - \begin{pmatrix} 2 & 0 \\ 0 & -4 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -3 & 6 \end{pmatrix}
\]
\[
\begin{pmatrix} 2 & -5 \\ -3 & 6 \end{pmatrix} + \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 7 & -5 \\ -3 & 11 \end{pmatrix}
\]
Step 9: Since \( X^T = \begin{pmatrix} 7 & -5 \\ -3 & 11 \end{pmatrix} \), transpose to find \( X \):
\[
X = \begin{pmatrix} 7 & -3 \\ -5 & 11 \end{pmatrix}
\]
--
3. Problem 18: Given:
\[
A = \begin{pmatrix} x & 1 \\ y & 2 \end{pmatrix}, \quad A = 3 A^{-1}
\]
Find \( x + y \).
Step 1: Multiply both sides by \( A \):
\[
A^2 = 3I
\]
Step 2: Calculate \( A^2 \):
\[
A^2 = \begin{pmatrix} x & 1 \\ y & 2 \end{pmatrix} \begin{pmatrix} x & 1 \\ y & 2 \end{pmatrix} = \begin{pmatrix} x^2 + y & x + 2 \\ yx + 2y & y + 4 \end{pmatrix}
\]
Step 3: Set equal to \( 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \):
Matching elements:
\[
x^2 + y = 3 \quad (1)
\]
\[
x + 2 = 0 \implies x = -2 \quad (2)
\]
\[
yx + 2y = 0 \implies y(x + 2) = 0 \quad (3)
\]
\[
y + 4 = 3 \implies y = -1 \quad (4)
\]
Step 4: From (2), \( x = -2 \).
From (4), \( y = -1 \).
Step 5: Check equation (1):
\[
(-2)^2 + (-1) = 4 -1 = 3 \quad \checkmark
\]
Step 6: Find \( x + y = -2 + (-1) = -3 \).
--
4. Problem 19: Find \( X \) such that:
\[
|A| X - \text{tr}(B) A^{-1} = C^T B
\]
Given:
\[
A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}, C = \begin{pmatrix} 2 & 0 \\ 1 & 1 \end{pmatrix}
\]
Step 1: Compute \(|A|\):
\[
|A| = 2 \times 2 - 1 \times 3 = 4 - 3 = 1
\]
Step 2: Compute \( \text{tr}(B) = 1 + 2 = 3 \).
Step 3: Find \( A^{-1} \):
\[
A^{-1} = \frac{1}{1} \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix}
\]
Step 4: Compute \( C^T B \):
\[
C^T = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}
\]
\[
C^T B = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times 1 + 1 \times 0 & 2 \times (-1) + 1 \times 2 \\ 0 \times 1 + 1 \times 0 & 0 \times (-1) + 1 \times 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}
\]
Step 5: Form equation:
\[
1 \times X - 3 \times A^{-1} = C^T B
\]
\[
X = 3 A^{-1} + C^T B = 3 \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 6 & -3 \\ -9 & 6 \end{pmatrix} + \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 8 & -3 \\ -9 & 8 \end{pmatrix}
\]
Final answers:
\[
B = \begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix}, \quad X = \begin{pmatrix} 7 & -3 \\ -5 & 11 \end{pmatrix}, \quad x+y = -3, \quad X = \begin{pmatrix} 8 & -3 \\ -9 & 8 \end{pmatrix}
\]