Exercise 6A

### Evaluate Solution: $\left|\begin{array}{cc}\sqrt{3} & \sqrt{5} \\ -\sqrt{5} & 3 \sqrt{3}\end{array}\right| \cdot=3 \sqrt{3} \times \sqrt{3}-(-\sqrt{5} \times \sqrt{5})$ $=14$

### Evaluate Solution: $\begin{array}{l} \left|\begin{array}{cc} 14 & 9 \\ -8 & -7 \end{array}\right|=14 \times(-7)-9 \times(-8) \\ =-26 \end{array}$

### For what value of , the given matrix is a singular matrix?

Solution: For $A$ to be singular matrix its determinant should be equal to 0 . $\begin{array}{l} 0=(3-2 x) \times 4-(x+1) \times 2 \\ 0=12-8 x-2 x-2 \\ 0=10-10 x \\ x=1 \end{array}$

### Evaluate Solution: After finding determinant we will get a trigonometric identity. $\begin{array}{l} \cos ^{2} \alpha+\sin ^{2} \alpha \\ =1 \end{array}$ $\because \sin ^{2} \theta+\cos ^{2} \theta=1$

### If , write the value of .

Solution: This question is having the same logic as above. $\begin{array}{l} 2 x^{2}-40=18+14 \\ \Rightarrow 2 x^{2}=72 \\ \Rightarrow x^{2}=36 \\ \Rightarrow x=\pm 6 \end{array}$

### Evaluate Solution: Theorem: This evaluation can be done in two different ways either by taking out the common things anc then calculating the determinants or simply take determinant. I will prefer first...

### If is the cofactor of the element of then write the value of .
Solution: Theorem: $A_{i j}$ is found by deleting $j^{t h}$ rowand $j^{t h}$ column, the determinant of left matrix is called cofactor with multiplied by $(-1)^{(i+j)}$ Given: $\mathrm{j}=3$ and...