Exercise 4.2 Archives - Noon Academy

Exercise 4.2

Choose the corrrect answer Which of the following is correct (A) Determinant is a square matrix. (B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix. (D) None of these

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

The correct option is Option (C) Since, Determinant is a number associated to a square matrix.

Choose the corrrect answer Let $A$ be a square matrix of order $3 \times 3$ , then $|k A|$ is equal to (A) $k|A|(B) k^{2}|A| \mid$ (C) $k^{3}|\mathbf{A}|$ (D) $3 \mathrm{k}|\mathbf{A}|$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

The correct option is Option (C). Let A = be a square matrix of order 3 x 3. ……….(i) = [From eq. (i)]

By using properties of determinants, show that: $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

LHS $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|$ Multiplying by respectively and then dividing the...

By using properties of determinants, show that: $\left| \begin{matrix} 1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\ 2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\ 2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\ \end{matrix} \right|$ ${{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{3}}$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

LHS $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-b \mathrm{C}_{3}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+a \mathrm{C}_{3}$ $=\left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -2 b \\ 0...

By using properties of determinants, show that:
$\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|=\left(1-x^{3}\right)^{2}$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

L.H.S. = = = = = = = = = = = = = R.H.S. Proved.

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

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

LHS $\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$ $\mathrm{R}_{1} \rightarrow...

By using properties of determinants, show that:
$\text { (i) }\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|=(5 x+4)(4-x)^{2}$

$\text { (ii) }\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|=k^{2}(3 y+k)$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

(i) LHS = [operating and ] = = = R.H.S. (ii) L.H.S. = = = = [operating and ] = = = = R.H.S. Proved.

By using properties of determinants, show that: $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

LHS= $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$ Mulitiplying $R_{1}, R_{2}, R_{3}$ by $x, y, z$ respectively...

By using properties of determinants, show that:(i) $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a)$ (ii) $\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

(i) LHS: $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$ $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and...

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

Solving L.H.S $\left| \begin{matrix} -{{a}^{2}} & ab & ac \\ ba & -{{b}^{2}} & bc \\ ca & cb & -{{c}^{2}} \\ \end{matrix} \right|$ Taking a common from${{R}_{1}}$,b common...

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

Let $\Delta=\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|$ Taking (-1) common from all the 3 rows. Again, interchanging rows and columns,...

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{lll}(b+c) & q+r & y+z \\ (c+a) & r+p & z+x \\ (a+b) & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

LHS: Applying: $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ $\left|\begin{array}{ccc}b+c+c+a+a+b & q+r+r+p+p+q & y+z+z+x+x+y \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$...

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

$\left|\begin{array}{ccc}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}+\mathrm{C}_{2}$...

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$ Applying: $\mathrm{C}_{3}->\mathrm{C}_{3}-\mathrm{C}_{1}$ $\left|\begin{array}{lll}2...

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{ccc}a-b & b-c & a-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

$\left|\begin{array}{ccc}a-b & b-c & a-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ Applying: $\mathrm{C}_{1} \rightarrow...

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

CBSE, Class 12, Determinants, Exercise 4.2, Maths, NCERT

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$ L.H.S. $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z &...

Choose the corrrect answer Which of the following is correct (A) Determinant is a square matrix. (B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix. (D) None of these

Choose the corrrect answer Let $A$ be a square matrix of order $3 \times 3$ , then $|k A|$ is equal to (A) $k|A|(B) k^{2}|A| \mid$ (C) $k^{3}|\mathbf{A}|$ (D) $3 \mathrm{k}|\mathbf{A}|$

By using properties of determinants, show that: $\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

By using properties of determinants, show that: $\left| \begin{matrix} 1+{{a}^{2}}-{{b}^{2}} & 2ab & -2b \\ 2ab & 1-{{a}^{2}}+{{b}^{2}} & 2a \\ 2b & -2a & 1-{{a}^{2}}-{{b}^{2}} \\ \end{matrix} \right|$ ${{\left( 1+{{a}^{2}}+{{b}^{2}} \right)}^{3}}$

By using properties of determinants, show that:
$\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|=\left(1-x^{3}\right)^{2}$

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

By using properties of determinants, show that:
$\text { (i) }\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|=(5 x+4)(4-x)^{2}$

$\text { (ii) }\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|=k^{2}(3 y+k)$

By using properties of determinants, show that: $\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

By using properties of determinants, show that:(i) $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a)$ (ii) $\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

Using the properties of determinants and without expanding, prove that $\left|\begin{array}{lll}(b+c) & q+r & y+z \\ (c+a) & r+p & z+x \\ (a+b) & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{ccc}a-b & b-c & a-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

Using the properties of determinants and without expanding, prove that: $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

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