8. What should be the values of a and b so that the system \( \left\{\begin{array}{l}x+y+z=0 \\ 3 x-a y-z=4 \\ x+5 y+5 z=b\end{array}\right. \) has no solution. 9. Find the inverse of \( \mathrm{A}=\left(\begin{array}{cc}-4 & 5 \\ 2 & -3\end{array}\right) \) (if it exists) using adjoint method? 10. Suppose A and B be \( 3 \times 3 \) matrices such that \( \mathrm{A}=\left(\begin{array}{ccc}2 & 0 & 0 \\ -4 & 6 & 0 \\ 0 & -1 & \frac{2}{3}\end{array}\right) \) and \( |B|=\frac{1}{4} \), then find the

Sure, let's address each of your questions step by step. --- ### **Question 8** **Determine the values of** \( a \) **and** \( b \) **so that the system has no solution:** \[ \begin{cases} x + y + z = 0 \\ 3x - a y - z = 4 \\ x + 5y + 5z = b \end{cases} \] **Solution:** To determine when the system has no solution, we need to analyze the consistency of the equations. We'll use the method of **Gaussian elimination** to simplify the system. 1. **Write the Augmented Matrix:** \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 3 & -a & -1 & | & 4 \\ 1 & 5 & 5 & | & b \\ \end{bmatrix} \] 2. **Eliminate \( x \) from Equations 2 and 3:** - **Row 2 (R2) becomes:** R2 - 3 * R1 \[ R2: 3 - 3(1) = 0,\quad -a - 3(1) = -a-3,\quad -1 - 3(1) = -4,\quad 4 - 0 = 4 \] - **Row 3 (R3) becomes:** R3 - R1 \[ R3: 1 - 1 = 0,\quad 5 - 1 = 4,\quad 5 - 1 = 4,\quad b - 0 = b \] The matrix now looks like: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -a-3 & -4 & | & 4 \\ 0 & 4 & 4 & | & b \\ \end{bmatrix} \] 3. **Analyze the Last Two Equations:** \[ \begin{cases} (-a-3) y - 4z = 4 \quad \text{(Equation 2)} \\ 4y + 4z = b \quad \text{(Equation 3)} \end{cases} \] For the system to have **no solution**, these two equations must be **parallel** but **inconsistent**. This happens when the ratio of the coefficients of \( y \) and \( z \) is the same, but the ratio of the constants is different. - **Determine When Coefficients are Proportional:** \[ \frac{-a-3}{4} = \frac{-4}{4} \Rightarrow \frac{-a-3}{4} = -1 \Rightarrow -a - 3 = -4 \Rightarrow a = 1 \] - **Check Consistency of Constants:** From Equation 2 (with \( a = 1 \)): \[ -4y - 4z = 4 \quad \Rightarrow \quad y + z = -1 \] From Equation 3: \[ 4y + 4z = b \quad \Rightarrow \quad y + z = \frac{b}{4} \] For inconsistency: \[ \frac{b}{4} \neq -1 \quad \Rightarrow \quad b \neq -4 \] **Conclusion:** - **\( a = 1 \)** - **\( b \neq -4 \)** When \( a = 1 \) and \( b \) is any real number except \( -4 \), the system has no solution. --- ### **Question 9** **Find the inverse of** \( \mathrm{A} = \begin{pmatrix} -4 & 5 \\ 2 & -3 \end{pmatrix} \) **using the adjoint method (if it exists).** **Solution:** To find the inverse of a \( 2 \times 2 \) matrix using the adjoint (adjugate) method, follow these steps: 1. **Compute the Determinant (\( \det A \)) of Matrix \( A \):** \[ \det A = (-4)(-3) - (5)(2) = 12 - 10 = 2 \] Since \( \det A \neq 0 \), the inverse exists. 2. **Find the Adjugate (\( \text{adj}(A) \)) of Matrix \( A \):** For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the adjugate is: \[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] Applying to \( A \): \[ \text{adj}(A) = \begin{pmatrix} -3 & -5 \\ -2 & -4 \end{pmatrix} \] 3. **Compute the Inverse (\( A^{-1} \)) Using the Formula:** \[ A^{-1} = \frac{1}{\det A} \cdot \text{adj}(A) = \frac{1}{2} \begin{pmatrix} -3 & -5 \\ -2 & -4 \end{pmatrix} = \begin{pmatrix} -\frac{3}{2} & -\frac{5}{2} \\ -1 & -2 \end{pmatrix} \] **Conclusion:** \[ A^{-1} = \begin{pmatrix} -\frac{3}{2} & -\frac{5}{2} \\ -1 & -2 \end{pmatrix} \] Alternatively, it can be expressed as: \[ A^{-1} = \frac{1}{2} \begin{pmatrix} -3 & -5 \\ -2 & -4 \end{pmatrix} \] --- ### **Question 10** **Incomplete Question** It appears that **Question 10** was cut off and incomplete: > 10. Suppose A and B be \( 3 \times 3 \) matrices such that \( \mathrm{A} = \begin{pmatrix} 2 & 0 & 0 \\ -4 & 6 & 0 \\ 0 & -1 & \frac{2}{3} \end{pmatrix} \) and \( |B| = \frac{1}{4} \), then find the ... Could you please provide the complete question for **Question 10**? This will allow me to assist you accurately. --- If you have any further questions or need clarification on the solutions provided, feel free to ask!