Question:

Using the properties of determinants, find the value of $\left|\begin{array}{ccc}0& \mathrm{a}& \mathrm{-b}\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{b}& \mathrm{c}& 0\end{array}\right|$

Answer:

1. Using the properties of determinants, we can rewrite the given determinant as follows: $\left|\begin{array}{ccc}0& \mathrm{a}& \mathrm{-b}\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{b}& \mathrm{c}& 0\end{array}\right|=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{-b}& 0\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{b}& \mathrm{c}& 0\end{array}\right|=-\mathrm{abc}$

2. Now, we can use the properties of determinants to calculate the value of the determinant: $\left|\begin{array}{ccc}\mathrm{a}& \mathrm{-b}& 0\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{b}& \mathrm{c}& 0\end{array}\right|=0+0+\mathrm{abc}=\mathrm{-abc}$

3. Therefore, the value of the given determinant is $\left|\begin{array}{ccc}0& \mathrm{a}& \mathrm{-b}\\ \mathrm{-a}& 0& \mathrm{-c}\\ \mathrm{b}& \mathrm{c}& 0\end{array}\right|=\mathrm{-abc}$.

Question:

Using the property of determinants and without expanding, prove that $\left|\begin{array}{ccc}\mathrm{x}& \mathrm{a}& \mathrm{x+a}\\ \mathrm{y}& \mathrm{b}& \mathrm{y+b}\\ \mathrm{z}& \mathrm{c}& \mathrm{z+c}\end{array}\right|$

Answer:

1. Using the property of determinants, we can express the determinant of the given matrix as the sum of the products of the elements of any row or column multiplied by their corresponding cofactors.

2. Thus, we can express the determinant of the given matrix as:

$\left|\begin{array}{ccc}\mathrm{x}& \mathrm{a}& \mathrm{x+a}\\ \mathrm{y}& \mathrm{b}& \mathrm{y+b}\\ \mathrm{z}& \mathrm{c}& \mathrm{z+c}\end{array}\right|=\mathrm{x}\mathrm{b}\mathrm{c}+\mathrm{a}\mathrm{y}\mathrm{c}+\mathrm{a}\mathrm{b}\mathrm{z}+\mathrm{x}\mathrm{y}\mathrm{c}+\mathrm{a}\mathrm{y}\mathrm{z}+\mathrm{x}\mathrm{b}\mathrm{z}+\mathrm{x}\mathrm{y}\mathrm{z}+\mathrm{a}\mathrm{b}\mathrm{c}$

3. Simplifying the above expression, we get

$\left|\begin{array}{ccc}\mathrm{x}& \mathrm{a}& \mathrm{x+a}\\ \mathrm{y}& \mathrm{b}& \mathrm{y+b}\\ \mathrm{z}& \mathrm{c}& \mathrm{z+c}\end{array}\right|=\mathrm{x}\mathrm{y}\mathrm{z}+\mathrm{x}\mathrm{y}\mathrm{c}+\mathrm{x}\mathrm{b}\mathrm{z}+\mathrm{a}\mathrm{y}\mathrm{z}+\mathrm{a}\mathrm{b}\mathrm{z}+\mathrm{a}\mathrm{b}\mathrm{c}+\mathrm{x}\mathrm{b}\mathrm{c}+\mathrm{a}\mathrm{y}\mathrm{c}$

4. Thus, we can conclude that the determinant of the given matrix is

Question:

Using the properties of determinants, show that: (i)$\left|\begin{array}{ccc}1& \mathrm{a}& a^2\\ 1& \mathrm{b}& b^2\\ 1& \mathrm{c}& c^2\end{array}\right|$=(a−b)(b−c)(c−a) (ii) $mrow>mo>|mtable>mtr>mtd>1mtd>1mtd>1/mtr>mtr>mtd>\mathrm{a}mtd>\mathrm{b}mtd>\mathrm{c}/mtr>mtr>mtd>msup>\; mi>a\; mn>3\; /msup>/mtd>mtd>msup>\; mi>a\; mn>3\; /msup>/mtd>mtd>msup>\; mi>c\; mn>3\; /msup>/mtd>/mtr>/mtable>mo>|/mrow>/math>=(a-b)(b-c)(c-a)(a+b+c)$

Answer:

(i) Using the properties of determinants, we can write the given determinant as:

$\left|\begin{array}{ccc}1& \mathrm{a}& a^2\\ 1& \mathrm{b}& b^2\\ 1& \mathrm{c}& c^2\end{array}\right|$

= (a-b)(b-c)(c-a)

(ii) Using the properties of determinants, we can write the given determinant as:

$mrow>mo>|mtable>mtr>mtd>1mtd>1mtd>1/mtr>mtr>mtd>\mathrm{a}mtd>\mathrm{b}mtd>\mathrm{c}/mtr>mtr>mtd>msup>\; mi>a\; mn>3\; /msup>/mtd>mtd>msup>\; mi>a\; mn>3\; /msup>/mtd>mtd>msup>\; mi>c\; mn>3\; /msup>/mtd>/mtr>/mtable>mo>|/mrow>/math>$

= (a-b)(b-c)(c-a)(a + b + c)

Question:

Using the properties of determinants, show that:(i) $\left|\begin{array}{ccc}\mathrm{a-b-c}& \mathrm{2a}& \mathrm{2a}\\ \mathrm{2b}& \mathrm{b-c-a}& \mathrm{2b}\\ \mathrm{2c}& \mathrm{2c}& \mathrm{c-a-b}\end{array}\right|$ =$\begin{array}{}{\mathrm{(a+b+c)}}^2\end{array}$ (ii) $\left|\begin{array}{ccc}\mathrm{x+y+2z}& \mathrm{x}& \mathrm{y}\\ \mathrm{z}& \mathrm{y+z+2x}& \mathrm{y}\\ \mathrm{z}& \mathrm{x}& \mathrm{z+x+2y}\end{array}\right|$=$\begin{array}{}{\mathrm{2(x+y+z)}}^3\end{array}$

Answer:

(i) Using the properties of determinants, we can write:

|a-b-c 2a 2a| |2b b-c-a 2b| |2c 2c c-a-b|

= (a-b-c)(b-c-a)(c-a-b) + 2(2a)(2b)(2c)

= (a-b-c)(b-c-a)(c-a-b) + 8abc

= (a+b+c)^2 - 2(ab+bc+ca) + 8abc

= (a+b+c)^2

Therefore, |a-b-c 2a 2a| |2b b-c-a 2b| |2c 2c c-a-b| = (a+b+c)^2

(ii) Using the properties of determinants, we can write:

|x+y+2z x y| |z y+z+2x y| |z x z+x+2y|

= (x+y+2z)(y+z+2x)(z+x+2y) + 2(x)(y)(z)

= (x+y+2z)(y+z+2x)(z+x+2y) + 2xyz

= (x+y+z)^3 + 3(x+y+z)(xy+yz+zx) + 2xyz

= (x+y+z)^3

Therefore, |x+y+2z x y| |z y+z+2x y| |z x z+x+2y| = (x+y+z)^3

Question:

Using the properties of determinants, show that: (i) $\left|\begin{array}{ccc}\mathrm{x+4}& \mathrm{2x}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{x+4}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{2x}& \mathrm{x+4}\end{array}\right|$=$\begin{array}{}{\mathrm{(5x+4)(4-x)}}^2\end{array}$ (ii) $\left|\begin{array}{ccc}\mathrm{y+k}& \mathrm{y}& \mathrm{y}\\ \mathrm{y}& \mathrm{y+k}& \mathrm{y}\\ \mathrm{y}& \mathrm{y}& \mathrm{y+k}\end{array}\right|$=$\begin{array}{}{k}^2\mathrm{(3y+k)}\end{array}$

Answer:

(i) Using the properties of determinants, we can expand the determinant along the first row:

$\left|\begin{array}{ccc}\mathrm{x+4}& \mathrm{2x}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{x+4}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{2x}& \mathrm{x+4}\end{array}\right|$

= (x + 4) * $\left|\begin{array}{ccc}\mathrm{2x}& \mathrm{2x}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{x+4}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{2x}& \mathrm{x+4}\end{array}\right|$

= (x + 4) * (2x) * $\left|\begin{array}{cc}\mathrm{2x}& \mathrm{2x}\\ \mathrm{2x}& \mathrm{x+4}\end{array}\right|$

= (x + 4) * (2x) * (x + 4 - 2x)

= (x + 4) * (2x) * (5x + 4)

= (x + 4)²(5x + 4)²

= (5x + 4)(4 - x)²

(ii) Using the properties of determinants, we can expand the determinant along the first row:

$\left|\begin{array}{ccc}\mathrm{y+k}& \mathrm{y}& \mathrm{y}\\ \mathrm{y}& \mathrm{y+k}& \mathrm{y}\\ \mathrm{y}& \mathrm{y}& \mathrm{y+k}\end{array}\right|$

= (y + k) * $\left|\begin{array}{ccc}\mathrm{y}& \mathrm{y}& \mathrm{y}\\ \mathrm{y}& \mathrm{y+k}& \mathrm{y}\\ \mathrm{y}& \mathrm{y}& \mathrm{y+k}\end{array}\right|$

= (y