Determinants

### Solve the following equations:

Solution: Operating $R 1 \rightarrow R 1+R_{2}+R_{3}$ $0=\left|\begin{array}{ccc} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|$ Taking $(x+9)$ common from...

### Solve the following equations:

Solution: Operating $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $0=\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|$ Taking $(x+9)$ common...

### Using properties of determinants prove that:

Solution: $\left|\begin{array}{ccc}a^{2} & b^{2} & c^{2} \\ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \\ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}\end{array}\right|$...

### Using properties of determinants prove that:

Solution: $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$...

### Evaluate

Solution: We know that expansion of determinant with respect to first row is $a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$. $0(3 \times 6-5 \times 4)-2(2 \times 6-4 \times 4)+0(2 \times 5-4 \times 3)$...

### Evaluate .

Solution: Find determinant $\begin{array}{l} \sqrt{6} \times \sqrt{24-\sqrt{2}} 20 \times \sqrt{5} \\ \sqrt{1} 144-\sqrt{1} 100 \\ =12-10 \\ =2 \end{array}$

### Evaluate .

Solution: It is determinant multiplied by a scalar number 2 , just find determinant of matrix and multiply it by 2 . $\begin{array}{l} 2 \times(35-20) \\ 2 \times 15=30 \end{array}$

### If , find the value of .

Solution: Find the determinant of $A$ and then multiply it by 3 $\begin{array}{l} |A|=2 \\ 3|A|=3 \times 2 \\ =6 \end{array}$

Solution: This we can very simply go through directly. $\begin{array}{l} ((a+i b)(a-i b))-((-c+i d)(c+i d)) \\ \Rightarrow\left(a^{2}+b^{2}\right)-\left(-c^{2}-d^{2}\right) \\ \Rightarrow... read more ### Evaluate Solution: Theorem: This evaluation can be done in two different ways either by taking out the common things anc then calculating the determinants or simply take determinant. I will prefer first... read more ### If is the cofactor of the element of then write the value of . Solution: Theorem:$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$and... read more ### Let be a square matrix of order 3, write the value of , where . Solution: Theorem: If$A$be$k \times k$matrix then$|p A|=p^{k}|A|$. Given:$p=2, k=3$and$|A|=4\begin{array}{l} |2 A|=2^{3} \times|A| \\ =8 \times 4 \\ =32 \end{array}$read more ### If is a matrix such that and then write the value of . Solution: Theorem: If Let$A$be$k \times k$matrix then$|p A|=p^{k}|A|$. Given:$\mathrm{k}=3$and$\mathrm{p}=3$.$\begin{array}{l} |3 \mathrm{~A}|=3^{3} \times|\mathrm{A}| \\ =27|\mathrm{~A}|...
Solution: Theorem: If $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given, $\mathrm{p}=4, \mathrm{k}=2$ and $|\mathrm{A}|=5$. \$\begin{array}{l} |4 \mathrm{~A}|=4^{2} \times 5 \\ =16 \times 5 \\...