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

Question

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

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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!
**Question 8:** - **Values for $$ a $$ and $$ b $$ that make the system have no solution:** - $$ a = 1 $$ - $$ b \neq -4 $$ **Question 9:** - **Inverse of Matrix $$ A $$:** $$ A^{-1} = \begin{pmatrix} -\frac{3}{2} & -\frac{5}{2} \ -1 & -2 \end{pmatrix} $$ **Question 10:** - **Incomplete question. Please provide the full question for assistance.**

What should be the values of a and b so that the system has no
solution.

Find the inverse of (if it exists) using adjoint method?

Suppose A and B be matrices such that and , then find the

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