1. **Problem:** Given $$ A = \begin{bmatrix} a & a & 0 \\ a & 1 & 0 \\ 0 & 1 & 2 \end{bmatrix} $$ where $a$ satisfies $x^2 - x + 1 = 0$, find $$ \det \left( 3A^3 A^t A^{-1} \right)^{-1} $$ **Step 1:** Find $a$ from the quadratic equation. The roots of $x^2 - x + 1 = 0$ are $$ a = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} $$ which are complex and non-real. However, determinant calculation does not require explicit $a$ if we use properties. **Step 2:** Use determinant properties: $$ \det(3A^3 A^t A^{-1}) = \det(3I) \det(A^3) \det(A^t) \det(A^{-1}) $$ Since $3$ is scalar multiplied to matrix $A^3$, the scalar factor is $3^3$ because $A^3$ is $3 \times 3$ matrix. So, $$ \det(3A^3 A^t A^{-1}) = 3^3 \det(A)^3 \det(A) \det(A)^{-1} = 27 \det(A)^3 \det(A) \det(A)^{-1} = 27 \det(A)^3 $$ **Step 3:** Calculate $\det(A)$: $$ \det(A) = a \begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix} - a \begin{vmatrix} a & 0 \\ 0 & 2 \end{vmatrix} + 0 = a(2 - 0) - a(2a) = 2a - 2a^2 $$ **Step 4:** Use $a^2 = a - 1$ from the equation $a^2 - a + 1 = 0$: $$ 2a - 2a^2 = 2a - 2(a - 1) = 2a - 2a + 2 = 2 $$ **Step 5:** So, $$ \det(A) = 2 $$ **Step 6:** Then, $$ \det(3A^3 A^t A^{-1}) = 27 \times 2^3 = 27 \times 8 = 216 $$ **Step 7:** Finally, $$ \det \left( 3A^3 A^t A^{-1} \right)^{-1} = \frac{1}{216} $$ --- 2. **Problem:** Given non-invertible $2 \times 2$ matrix $A$ satisfying $$ A^{2568} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = A^{2567} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ Find $$ \det \left( A \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} + 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) $$ **Step 1:** Multiply both sides by $A^{-2567}$ (if exists), but $A$ is non-invertible, so $\det(A) = 0$. **Step 2:** Rewrite: $$ A^{2568} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = A^{2567} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ Divide both sides by $A^{2567}$ (assuming $A^{2567} \neq 0$): $$ A \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ **Step 3:** Multiply both sides by $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$: $$ A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} $$ **Step 4:** Calculate determinant of $A$: $$ \det(A) = 0 \times (-1) - 1 \times 1 = -1 $$ **Step 5:** Calculate $$ \det \left( A \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} + 5I \right) $$ Calculate product: $$ A \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 \times (-1) + 1 \times 1 & 0 \times 1 + 1 \times 0 \\ 1 \times (-1) + (-1) \times 1 & 1 \times 1 + (-1) \times 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} $$ Add $5I$: $$ \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ -2 & 6 \end{bmatrix} $$ **Step 6:** Calculate determinant: $$ 6 \times 6 - 0 \times (-2) = 36 $$ --- 3. **Problem:** Given $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} $$ Find $b - c$. **Step 1:** Find inverse of $$ M = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} $$ Determinant: $$ \det(M) = 2 \times 2 - 3 \times 1 = 4 - 3 = 1 $$ Inverse: $$ M^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} $$ **Step 2:** Multiply both sides by $M$: $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} M = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} $$ Calculate: $$ \begin{bmatrix} 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \\ 1 \times 2 + 3 \times 1 & 1 \times 3 + 3 \times 2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ 5 & 9 \end{bmatrix} $$ **Step 3:** So, $$ a=4, b=7, c=5, d=9 $$ **Step 4:** Calculate $$ b - c = 7 - 5 = 2 $$ --- 4. **Problem:** Given $$ A = \begin{bmatrix} 1 & a & b \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} 1 & -1 & -2 \\ 0 & -1 & 1 \\ c & d & 0 \end{bmatrix} $$ Find $a + b + c + d$. **Step 1:** Use the identity $$ A A^{-1} = I $$ Multiply first row of $A$ by columns of $A^{-1}$: - First element: $$ 1 \times 1 + a \times 0 + b \times c = 1 + b c $$ - Second element: $$ 1 \times (-1) + a \times (-1) + b \times d = -1 - a + b d $$ - Third element: $$ 1 \times (-2) + a \times 1 + b \times 0 = -2 + a $$ These must equal the first row of identity matrix: $$ [1, 0, 0] $$ **Step 2:** Set equations: $$ 1 + b c = 1 \Rightarrow b c = 0 $$ $$ -1 - a + b d = 0 \Rightarrow -1 - a + b d = 0 $$ $$ -2 + a = 0 \Rightarrow a = 2 $$ **Step 3:** From $a=2$, substitute into second equation: $$ -1 - 2 + b d = 0 \Rightarrow b d = 3 $$ **Step 4:** From $b c = 0$, either $b=0$ or $c=0$. Multiply second row of $A$ by columns of $A^{-1}$: - First element: $$ 0 \times 1 + 0 \times 0 + 1 \times c = c $$ - Second element: $$ 0 \times (-1) + 0 \times (-1) + 1 \times d = d $$ - Third element: $$ 0 \times (-2) + 0 \times 1 + 1 \times 0 = 0 $$ These equal second row of identity: $$ [0, 1, 0] $$ So, $$ c = 0, d = 1 $$ **Step 5:** From $c=0$, $b c = 0$ is true. From $b d = 3$ and $d=1$, we get $$ b = 3 $$ **Step 6:** Calculate sum: $$ a + b + c + d = 2 + 3 + 0 + 1 = 6 $$ --- 5. **Problem:** Solve systems by row operations. **5.1** $$ \begin{cases} x + 3z + 4w = 5 \\ 2y + 2w = 4 \\ 3z + 8w = 9 \end{cases} $$ **Step 1:** Variables are $x,y,z,w$. From second equation: $$ 2y + 2w = 4 \Rightarrow y + w = 2 \Rightarrow y = 2 - w $$ From third equation: $$ 3z + 8w = 9 \Rightarrow z = \frac{9 - 8w}{3} $$ From first equation: $$ x + 3z + 4w = 5 \Rightarrow x = 5 - 3z - 4w = 5 - 3 \times \frac{9 - 8w}{3} - 4w = 5 - (9 - 8w) - 4w = 5 - 9 + 8w - 4w = -4 + 4w $$ **Step 2:** Solution: $$ x = -4 + 4w, \quad y = 2 - w, \quad z = \frac{9 - 8w}{3}, \quad w = w $$ **5.2** $$ \begin{cases} x - 2y + z = 0 \\ 2x + y - z = 4 \\ 5x - 15y + 8z = 4 \end{cases} $$ **Step 1:** Write augmented matrix: $$ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 2 & 1 & -1 & 4 \\ 5 & -15 & 8 & 4 \end{bmatrix} $$ **Step 2:** Eliminate below first pivot: Row2 = Row2 - 2*Row1: $$ (2 - 2*1, 1 - 2*(-2), -1 - 2*1, 4 - 2*0) = (0, 5, -3, 4) $$ Row3 = Row3 - 5*Row1: $$ (5 - 5*1, -15 - 5*(-2), 8 - 5*1, 4 - 5*0) = (0, -5, 3, 4) $$ **Step 3:** Add Row2 and Row3: $$ (0, 5 + (-5), -3 + 3, 4 + 4) = (0, 0, 0, 8) $$ This implies $0 = 8$, contradiction, so no solution. **5.3** $$ \begin{cases} x - 2y + z = 0 \\ 2x + y - z = 4 \\ x - y + z = 2 \end{cases} $$ **Step 1:** Augmented matrix: $$ \begin{bmatrix} 1 & -2 & 1 & 0 \\ 2 & 1 & -1 & 4 \\ 1 & -1 & 1 & 2 \end{bmatrix} $$ **Step 2:** Eliminate below first pivot: Row2 = Row2 - 2*Row1: $$ (0, 5, -3, 4) $$ Row3 = Row3 - Row1: $$ (0, 1, 0, 2) $$ **Step 3:** Eliminate below second pivot: Row3 = Row3 - (1/5)*Row2: $$ (0, 0, 3/5, 6/5) $$ **Step 4:** Back substitution: $$ z = \frac{6/5}{3/5} = 2 $$ $$ 5y - 3z = 4 \Rightarrow 5y - 6 = 4 \Rightarrow 5y = 10 \Rightarrow y = 2 $$ $$ x - 2y + z = 0 \Rightarrow x - 4 + 2 = 0 \Rightarrow x = 2 $$ **Step 5:** Solution: $$ (x,y,z) = (2,2,2) $$ **5.4** $$ \begin{cases} x + 8y - 3z = 4 \\ 2x + 16y - 6z = 8 \\ 3x + 14y - 9z = 9 \end{cases} $$ **Step 1:** Note second equation is twice the first: $$ 2(x + 8y - 3z) = 2x + 16y - 6z = 8 $$ So second equation is dependent. **Step 2:** Use first and third equations: Multiply first equation by 3: $$ 3x + 24y - 9z = 12 $$ Subtract third equation: $$ (3x + 24y - 9z) - (3x + 14y - 9z) = 12 - 9 \Rightarrow 10y = 3 \Rightarrow y = \frac{3}{10} $$ **Step 3:** Substitute $y$ into first equation: $$ x + 8 \times \frac{3}{10} - 3z = 4 \Rightarrow x + \frac{24}{10} - 3z = 4 \Rightarrow x - 3z = 4 - 2.4 = 1.6 $$ **Step 4:** Express $x$: $$ x = 1.6 + 3z $$ **Step 5:** Infinite solutions parameterized by $z$: $$ (x,y,z) = (1.6 + 3z, 0.3, z) $$ --- **Final answers:** 1. $\frac{1}{216}$ 2. $36$ 3. $2$ 4. $6$ 5.1 $x = -4 + 4w, y = 2 - w, z = \frac{9 - 8w}{3}, w = w$ 5.2 No solution 5.3 $(2,2,2)$ 5.4 $y=0.3$, $x=1.6+3z$, infinite solutions