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NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

Edited By Ravindra Pindel | Updated on Sep 15, 2022 04:51 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 4 Determinants will help in learning the ways to find the determinant of various square matrices, co-factors, the inverse of matrices etc. NCERT exemplar Class 12 Maths chapter 4 solutions are useful for the students to get a deeper and better look at the matrices and how to solve them uniquely. From the scoring point of view, chapter 4 of NCERT Class 12 Maths Solutions can be very crucial for 12 Class students. Students can use NCERT exemplar Class 12 Maths solutions chapter 4 PDF download and study the topic and the solutions offline as well.

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  1. More About NCERT Exemplar Solutions for Class 12 Maths Chapter 4
  2. NCERT Exemplar Class 12 Maths Solutions
  3. Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 4

Question:1

Using the properties of determinants in evaluate:
\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|

Answer:

\left|\begin{array}{cc} x^{2}-x+1 & x-1 \\ x+1 & x+1 \end{array}\right|
\begin{aligned} &\text { If } A=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right| \\ &\text { The value of determinant of A can then be found by: }\\ &|A|=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|=a d-b c \end{aligned}
\\ \vspace{\baselineskip}= (x\textsuperscript{2} -x + 1) \times (x + 1) -(x + 1) \times (x -1)\\ \\ \vspace{\baselineskip}= x (x\textsuperscript{2} -x + 1) + 1 (x\textsuperscript{2} -x + 1) -(x\textsuperscript{2} -1)\\ \\ \vspace{\baselineskip}[$\because$ (a -b)(a + b) = (a\textsuperscript{2} -b\textsuperscript{2})]\\ \\ \vspace{\baselineskip}= x\textsuperscript{3} -x\textsuperscript{2} + x + x\textsuperscript{2} -x + 1 -x\textsuperscript{2} + 1\\ \\ \vspace{\baselineskip}= x\textsuperscript{3} -x\textsuperscript{2} + 2\\

Question:2

Using the properties of determinants in evaluate:
\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|

Answer:

A=\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|
C_{1} \rightarrow C_{1}+C_{2}+C_{3}
\begin{array}{l} =\left|\begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array}\right| \\\\ =(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ y & a+z \end{array}\right| \end{array}
R_{2} \rightarrow R_{2}-R_{1}
R_{3} \rightarrow R_{3}-R_{\mathrm{1}}
\begin{array}{l} =(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right| \\\\ =(a+x+y+z)\left|\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right|\\\\=a^{2}(a+z+x+y) \end{array}

Question:3

Using the properties of determinants in evaluate:
\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|

Answer:

\text{Let } A=\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|
=x^{2} y^{2} z^{2}\left|\begin{array}{lll} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{array}\right|
C_{2} \rightarrow C_{2}-C_{3}
=x^{2} y^{2} z^{2}\left|\begin{array}{ccc} 0 & 0 & x \\ y & -y & y \\ z & z & 0 \end{array}\right|
Expand IAI along C3 to get:
=x^{2} y^{2} z^{2}(x(y z+y z))
=x^{2} y^{2} z^{2}(2xy z)
=2x^{3} y^{3} z^{3}

Question:4

Using the properties of determinants in evaluate:
\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|

Answer:

Let \: \: A=\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|

Apply - C\textsubscript{1}$ \rightarrow $ C\textsubscript{1} + C\textsubscript{2} + C\textsubscript{3,}\\

\\\begin{aligned} &=\left|\begin{array}{ccc} 3 x-x+y-x+z & -x+y & -x+z \\ x-y+3 y+z-y & 3 y & z-y \\ x-z+y-z+3 z & y-z & 3 z \end{array}\right|\\ &=\left|\begin{array}{ccc} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z \end{array}\right|\\ &\text { Take }(x+y+z) \text { common from } C_{1}\\ &=(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{array}\right| \end{aligned}

\\\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1,} \text { you will get }\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 1-1 & 3 \mathrm{y}-(-\mathrm{x}+\mathrm{y}) & \mathrm{z}-\mathrm{y}-(-\mathrm{x}+\mathrm{z}) \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\\end{aligned}

\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y}+\mathrm{x}-\mathrm{y} & \mathrm{z}-\mathrm{y}+\mathrm{x}-\mathrm{z} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &\text { Now apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1-1 & \mathrm{y}-\mathrm{z}-(-\mathrm{x}+\mathrm{y}) & 3 \mathrm{z}-(-\mathrm{x}+\mathrm{z}) \end{array}\right| \end{aligned}

\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{y}-\mathrm{z}+\mathrm{x}-\mathrm{y} & 3 \mathrm{z}+\mathrm{x}-\mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3},\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}-(-\mathrm{x}+\mathrm{z}) & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-(\mathrm{x}-\mathrm{y}) & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-(2 \mathrm{z}+\mathrm{x}) & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\\end{aligned}

\\\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-\mathrm{x}+\mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-2 \mathrm{z}-\mathrm{x} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & \mathrm{y}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & -3 \mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right| \end{aligned}

Now, expand the determinant along Column 1

\\\vspace{\baselineskip} = (x + y + z) [1$ \times $ $ \{ $ (3y)(2z + x) -(-3z)(x -y)$ \} $ ]\\ \\ \vspace{\baselineskip}= (x + y + z) [6yz + 3yx + (3z)(x -y)]\\ \\ \vspace{\baselineskip}= (x + y + z) [6yz + 3yx + 3zx -3zy]\\ \\ \vspace{\baselineskip}= (x + y + z) [3yz + 3zx + 3yx]\\ \\ \vspace{\baselineskip}= 3(x + y + z)(yz + zx + yx)\\

Question:5

Using the properties of determinants in evaluate:
\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|

Answer:

Let\: \: A=\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|
\\\begin{aligned} &\text { Using } C_{1} \rightarrow C_{1}+C_{2}+C_{3},\\ &=\left|\begin{array}{ccc} x+4+x+x & x & x \\ x+x+4+x & x+4 & x \\ x+x+x+4 & x & x+4 \end{array}\right|\\ &=\left|\begin{array}{ccc} 3 x+4 & x & x \\ 3 x+4 & x+4 & x \\ 3 x+4 & x & x+4 \end{array}\right|\\ &\text { Take }(3 x+4) \text { common form } C \text { , }\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{array}\right|\\ &\mathrm{Apply}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & x & x \\ 0 & 4 & 0 \\ 1-1 & x-x & x+4-x \end{array}\right| \end{aligned}
\\\begin{aligned} &=(3 \mathrm{x}+4)\left|\begin{array}{lll} 1 & \mathrm{x} & \mathrm{x} \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{1}\\ &=(3 x+4)[1 \times\{(16)-0\}]\\ &=(3 x+4)(16)\\ &=16(3 x+4) \end{aligned}

Question:6

Using the properties of determinants in evaluate:
\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|

Answer:

Let\: \: A=\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|
\\\begin{aligned} &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{y}\\ &=\left|\begin{array}{ccc} \mathrm{a}-\mathrm{b}-\mathrm{c}+2 \mathrm{~b}+2 \mathrm{c} & 2 \mathrm{a}+\mathrm{b}-\mathrm{c}-\mathrm{a}+2 \mathrm{c} & 2 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}-\mathrm{a}-\mathrm{b} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &=\left|\begin{array}{ccc} \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{a}+\mathrm{b}+\mathrm{c} \\ 2 \mathrm{~b} & \mathrm{~b}-\mathrm{c}-\mathrm{a} & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & \mathrm{c}-\mathrm{a}-\mathrm{b} \end{array}\right|\\ &\text { Take }(a+b+c) \text { common form the first row }\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|\\ &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \end{aligned}
\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 2 b & b-c-a-2 b & 2 b \\ 2 c & 2 c-2 c & c-a-b \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 b & -b-c-a & 2 b \\ 2 c & 0 & c-a-b \end{array}\right| \end{array}
Now apply \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}
\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 2 b & -(a+b+c) & 2 b-2 b \\ 2 c & 0 & c-a-b-2 c \end{array}\right| \\ =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{array}\right| \end{array}
Expand the determinant along Row 1
\\= (a + b+ c)[1$ \times $ $ \{ $ -(a + b + c)$ \times $ $ \{ $ -(a + b + c)$ \} $ -0$ \} $ ]\\ \\ \vspace{\baselineskip}= (a + b + c)[(a + b + c)\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= (a + b + c)\textsuperscript{3}\\

Question:7

Using the properties of determinants in prove that:
\left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0

Answer:

Taking LHS, \left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|
\\\begin{aligned} &\text { Multiply and divide } \mathrm{R}_{1} \mathrm{R}_{2}, \mathrm{R}_{3} \text { respectively by } \mathrm{x}, \mathrm{y}, \mathrm{z}\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x(y+z) \\ y z^{2} x^{2} & y z x & y(z+x) \\ z x^{2} y^{2} & z x y & z(x+y) \end{array}\right|\\ &=\frac{1}{x y z}\left|\begin{array}{lll} x y^{2} z^{2} & x y z & x y+x z \\ x^{2} y z^{2} & x y z & y z+x y \\ x^{2} y^{2} z & x y z & x z+y z \end{array}\right|\\ &\text { Now, take xyz common from the first and second Column }\\ &=\frac{1}{\mathrm{xyz}} \times \mathrm{xyz} \times \mathrm{xyz}\left|\begin{array}{lll} \mathrm{yz} & 1 & \mathrm{xy}+\mathrm{xz} \\ \mathrm{xz} & 1 & \mathrm{yz}+\mathrm{xy} \\ \mathrm{xy} & 1 & \mathrm{xz}+\mathrm{yz} \end{array}\right|\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z \\ x z & 1 & y z+x y \\ x y & 1 & x z+y z \end{array}\right| \end{aligned}
\\\begin{aligned} &\text { Apply, } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\mathrm{C}_{2}\\ &=\operatorname{xyz}\left|\begin{array}{lll} y z & 1 & x y+x z+y z \\ x z & 1 & y z+x y+x z \\ x y & 1 & x z+y z+x y \end{array}\right|\\ &\text { Take }(\mathrm{xy}+\mathrm{yz}+\mathrm{xz}) \text { common from } \mathrm{C}_{3}\\ &=(\text { xyz })(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\left|\begin{array}{lll} \mathrm{yz} & 1 & 1 \\ \mathrm{xz} & 1 & 1 \\ \mathrm{xy} & 1 & 1 \end{array}\right| \end{aligned}

Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0

Hence, \left|\begin{array}{lll} y^{2} z^{2} & y z & y+z \\ z^{2} x^{2} & z x & z+x \\ x^{2} y^{2} & x y & x+y \end{array}\right|=0

∴ LHS = RHS

Question:8

Using the properties of determinants in prove that:
\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|=4 x y z

Answer:

LHS given,

\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|

Apply, R\textsubscript{1}$ \rightarrow $ R\textsubscript{1} + R\textsubscript{2} + R\textsubscript{3},\\

\\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z}+\mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{z}+\mathrm{x}+\mathrm{x} & \mathrm{y}+\mathrm{x}+\mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ &=\left|\begin{array}{ccc} 2 z+2 y & 2 z+2 x & 2 x+2 y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &2 \text { can be taken common from } \mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ \end{aligned}

\\\begin{aligned} \\&\text { Apply, } R_{1} \rightarrow R_{1}-R_{2}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y}-\mathrm{z} & \mathrm{z}+\mathrm{x}-(\mathrm{z}+\mathrm{x}) & \mathrm{x}+\mathrm{y}-\mathrm{x} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right| \end{aligned}

\\\begin{aligned} &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &\text { Apply, } R_{3} \rightarrow R_{3}-R_{1},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y}-\mathrm{y} & \mathrm{x}-0 & \mathrm{x}+\mathrm{y}-\mathrm{y} \end{array}\right|\\ &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ 0 & x & x \end{array}\right|\\ &\text { Now, Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z}-0 & \mathrm{z}+\mathrm{x}-\mathrm{x} & \mathrm{x}-\mathrm{x} \\ 0 & \mathrm{x} & \mathrm{x} \end{array}\right|\\ &=2\left|\begin{array}{lll} y & 0 & y \\ z & z & 0 \\ 0 & x & x \end{array}\right| \end{aligned}

Take y,z,x common from the R1, R2 and R3 respectively

\begin{aligned} &\text { Expand along Column } 1\\ &\begin{array}{r} |A|=a_{11}(-1)^{1+1}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|+a_{21}(-1)^{2+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right| \\ +a_{31}(-1)^{3+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right| \end{array} \end{aligned}

\\\vspace{\baselineskip}= 2xyz [(1)$ \{ $ (1) -0$ \} $ -(1)$ \{ $ 0 -1$ \} $ + 0$ \} $ ]\\ \\ \vspace{\baselineskip}= 2xyz [1 + 1]\\ \\ \vspace{\baselineskip}= 4xyz\\ \\ \vspace{\baselineskip}= RHS\\ $ \therefore $ LHS = RHS\\ \\

Hence Proved

Question:9

Using the properties of determinants in prove that:
\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^{3}

Answer:

Take LHS

\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|

\\\begin{array}{l} \text { Apply, } R_{1} \rightarrow R_{1}-R_{2} \\ =\left|\begin{array}{ccc} a^{2}+2 a-(2 a+1) & 2 a+1-(a+2) & 1-1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}+2 a-2 a-1 & 2 a+1-a-2 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}-1 & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \end{array}

\\\begin{aligned} &=\left|\begin{array}{ccc} (a-1)(a+1) & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]\\ &\text { Take (a-1) common from R }\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1-3 & a+2-3 & 1-1 \\ 3 & 3 & 1 \end{array} \mid\right|.\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right| \end{aligned}

Take (a-1) common from R_2

\begin{aligned} &=(a-1)^{2}\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{3} \end{aligned}

\\\\ = (a -1)\textsuperscript{2} [1$ \{ $ (a + 1) -2$ \} $ ]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{2} [a + 1 -2]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{3}\\ \\ \vspace{\baselineskip}= RHS\\

Hence,Proved.

Question:10

If A + B + C = 0, then prove that \begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array} =0

Answer:

LHS given: \begin{array}{|ccc|} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{array}

Expand along R_1

\\\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =(1)\left|\begin{array}{cc} 1 & \cos \mathrm{A} \\ \cos \mathrm{A} & 1 \end{array}\right|-\cos \mathrm{C}\left|\begin{array}{cc} \cos \mathrm{C} & \cos \mathrm{A} \\ \cos \mathrm{B} & 1 \end{array}\right|+\cos \mathrm{B}\left|\begin{array}{cc} \cos \mathrm{C} & 1 \\ \cos \mathrm{B} & \cos \mathrm{A} \end{array}\right| \end{array}

\\\vspace{\baselineskip}= [1$ \{ $ 1 -cos\textsuperscript{2}A$ \} $ -cos C $ \{ $ cos C -cos B cos A$ \} $ + cos B $ \{ $ cos C cos A -cos B$ \} $ ]\\ \\ \vspace{\baselineskip}= $ \{ $ 1 -cos\textsuperscript{2}A$ \} $ -$ \{ $ cos\textsuperscript{2}C -cos A cos B cos C$ \} $ + $ \{ $ cos A cos B cos C -cos\textsuperscript{2}B$ \} $ \\ \\ \vspace{\baselineskip}= $ \{ $ sin\textsuperscript{2}A$ \} $ -cos\textsuperscript{2}C + cos A cos B cos C + cos A cos B cos C -cos\textsuperscript{2}B\\ \\ \vspace{\baselineskip}[$\because$ cos\textsuperscript{2}x + sin\textsuperscript{2}x = 1]\\ \\ \vspace{\baselineskip}= sin\textsuperscript{2}A -cos\textsuperscript{2}C -cos\textsuperscript{2}B + 2cos A cos B cos C\\ \\ \vspace{\baselineskip}= -(cos\textsuperscript{2}B -sin\textsuperscript{2}A)-cos\textsuperscript{2}C + 2cos A cos B cos C\\ \\\\\vspace{\baselineskip}= -[cos(B + A)cos(B -A)] + cos C [2 cos A cos B -cos C]\\ \\ \vspace{\baselineskip}[$\because$ cos\textsuperscript{2}B -sin\textsuperscript{2}A = cos(B + A)cos(B -A)]\\ \\ \vspace{\baselineskip}= -[cos(B + A)cos(B -A)] + cos C [cos(A + B) + cos(A -B) -cos C]$ \ldots $ (i)\\ \\ \vspace{\baselineskip}[$\because$ 2cos A cos B = cos(A + B) + cos(A -B)]\\ (A+B+C) = 0, given\\ Which means, A + B = - C\\ \vspace{\baselineskip} = -[cos(- C) cos(B -A)] + cos C [cos(-C) + cos (A -B) -cos C]\\ \\ \vspace{\baselineskip}= -cos C cos(B -A) + cos C[cos C + cos(A -B) -cos C]\\ \\ \vspace{\baselineskip}[$\because$ cos(-C) = cos C]\\ \\ \vspace{\baselineskip}Now, cos(A -B) = cos A cos B + sin A sin B\\ \\ \vspace{\baselineskip}= -cos C$ \{ $ cos B cos A + sin B sin A$ \} $ + cos C [cos A cos B + sin A sin B]\\ \\ \vspace{\baselineskip}= 0 = RHS\\

Question:11

If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are \left (x_1, y_1 \right ), \left (x_2, y_2 \right ), \left (x_3, y_3 \right ), then \left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|^{2}=\frac{3 a^{4}}{4}

Answer:

Area of a triangle with the given vertices will be:

\\\Delta=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\\$ Given: Length of the sides of the equilateral triangle = a \\Thus, the area $=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$ \\$\therefore \frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|=\frac{\sqrt{3}}{4} \mathrm{a}^{2}$\\ Square both sides; \\$\Rightarrow\left(\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\right)^{2}=\left(\frac{\sqrt{3}}{4} \mathrm{a}^{2}\right)^{2}$\\\Rightarrow \frac{1}{4}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|^{2}=\frac{3}{16} \mathrm{a}^{4}$ \\\begin{aligned} &\Rightarrow\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3} \end{array}\right|^{2}=\frac{3}{4} \mathrm{a}^{4}\\ &\text { Hence Proved } \end{aligned}


Question:12

Find the value of θ satisfying \left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0

Answer:

Given:
\left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0
Expand along Row 1
\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =\left|\begin{array}{lll} 1 \end{array}\left|\begin{array}{cc} 3 & \cos 2 \theta \\ -7 & -2 \end{array}\right|-1\left|\begin{array}{cc} -4 & \cos 2 \theta \\ 7 & -2 \end{array}\right|+\sin 3 \theta\left|\begin{array}{cc} -4 & 3 \\ 7 & -7 \end{array}\right|\right. \end{array}

\\\vspace{\baselineskip}$ \Rightarrow $ (1)$ \{ $ -6 -$ \{ $ (-7) cos2$ \theta $ $ \} $ $ \} $ -1$ \{ $ 8 -7cos2$ \theta $ $ \} $ + sin3$ \theta $ $ \{ $ 28 -21$ \} $ = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ -6 + 7cos2$ \theta $ -8 + 7cos2$ \theta $ + 7sin3$ \theta $ = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 14cos2$ \theta $ + 7sin3$ \theta $ -14 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2cos2$ \theta $ + sin3$ \theta $ -2 = 0\\We know,

\\cos 2$ \theta $ = 1 -2sin\textsuperscript{2}$ \theta $ \\ \\ \vspace{\baselineskip}sin 3$ \theta $ = 3sin$ \theta $ -4sin\textsuperscript{3}$ \theta $ \\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2(1 -2sin\textsuperscript{2}$ \theta $ ) + (3sin$ \theta $ -4sin\textsuperscript{3}$ \theta $ ) -2 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2 -4sin\textsuperscript{2}$ \theta $ + 3sin$ \theta $ -4sin\textsuperscript{3}$ \theta $ -2 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ -2 + 4sin\textsuperscript{2}$ \theta $ - 3sin$ \theta $ + 4sin\textsuperscript{3}$ \theta $ + 2 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ sin$ \theta $ (4sin$ \theta $ -3 + 4sin\textsuperscript{2}$ \theta $ ) = 0\\ \\\\\vspace{\baselineskip}$ \Rightarrow $ sin$ \theta $ (4sin\textsuperscript{2}$ \theta $ -6sin$ \theta $ + 2sin$ \theta $ -3) = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ sin$ \theta $ [2sin$ \theta $ (2sin$ \theta $ -3) + 1(2sin$ \theta $ -3)] = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ sin$ \theta $ (2sin$ \theta $ + 1)(2sin$ \theta $ -3) = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ sin$ \theta $ = 0 or 2sin$ \theta $ + 1 = 0 or 2sin$ \theta $ -3 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ $ \theta $ = n$ \pi $ or 2sin$ \theta $ = -1 or 2sin$ \theta $ = 3\\ \begin{array}{l} \Rightarrow \mathrm{or} \sin \theta=-\frac{1}{2} \text { or } \sin \theta=\frac{3}{2} \\ \Rightarrow \theta=\mathrm{n} \pi \text { or } \theta=\mathrm{m} \pi+(-1)^{\mathrm{n}}\left(-\frac{\pi}{6}\right) ; \mathrm{m}, \mathrm{n} \in \mathrm{Z} \end{array}
But it is not possible to have \sin \theta=\frac{3}{2}

Question:13

If \left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0 then find values of x.

Answer:

Given:
\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0
\\\begin{aligned} &\text { Apply, } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{ccc} 4-x+4+x+4+x & 4+x & 4+x \\ 4+x+4-x+4+x & 4-x & 4+x \\ 4+x+4+x+4-x & 4+x & 4-x \end{array}\right|=0\\ &\text { Take, }(12+\mathrm{x}) \text { common from Row } 1\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 4+x+4+x & 4+x \\ 1 & 4-x+4+x & 4+x \\ 1 & 4+x+4-x & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1 & 8 & 4+x \\ 1 & 8 & 4-x \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 1-1 & 8-8 & 4+x-(4-x) \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1 & 8 & 4-x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 1-1 & 8-8-2 x & 4-x-4-x \end{array}\right|=0\\ &\Rightarrow(12+x)\left|\begin{array}{ccc} 1 & 2(4+x) & 4+x \\ 0 & 0 & 2 x \\ 0 & -2 x & -2 x \end{array}\right|=0 \end{aligned}
Expand along Column 1

\\\vspace{\baselineskip} $ \Rightarrow $ (12 + x)[(1)$ \{ $ 0 -(2x)(-2x)$ \} $ ] = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (12 + x)(4x\textsuperscript{2}) = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 12 + x = 0 or 4x\textsuperscript{2} = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x = -12 or x = 0\\ \\ \vspace{\baselineskip}\text{Hence, the value of x }= -12 \: \: and\: \: 0\\

Question:14

If a1, a2, a3, ..., ar are in G.P., then prove that the determinant \left|\begin{array}{ccc} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{array}\right| is independent of r.

Answer:

\\\vspace{\baselineskip} a\textsubscript{1}, a\textsubscript{2}$ \ldots $ , a\textsubscript{r} are in G.P\\ \vspace{\baselineskip} \text{We know that, } a\textsubscript{r+1} = AR\textsuperscript{(r+1)-1} = AR\textsuperscript{r} $ \ldots $ (i)\\ \\ \vspace{\baselineskip}[$\because$ a\textsubscript{n} = ar\textsuperscript{n-1}, \text{where a = first term and r = common ratio}]\\ \vspace{\baselineskip}A is the first term of G.PR is the common ratio of G.P.
\\\therefore\left|\begin{array}{lll}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right|=\left|\begin{array}{ccc}A R^{r} & A R^{r+4} & A R^{r+8} \\ A R^{r+6} & A R^{r+10} & A R^{r+14} \\ A R^{r+10} & A R^{r+16} & A R^{r+20}\end{array}\right|_{\ldots[f r o m(i)]}$\\Taking AR", \\\mathrm{AR}^{\mathrm{r}+6}$ and $\mathrm{AR}^{\mathrm{r}+10}$ common from $\mathrm{R}_{1}, \mathrm{R}_{2}$ and $\mathrm{R}_{3}$ \\\\ =\mathrm{AR}^{\mathrm{r}} \times \mathrm{AR}^{\mathrm{r}+6} \times \mathrm{AR}^{\mathrm{r}+10}\left|\begin{array}{ccc} 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{6} & \mathrm{AR}^{10} \end{array}\right|
Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0
Rows 1 and 2 are identical
\\ \therefore\left|\begin{array}{lll}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right|=0$\\ Hence Proved

Question:15

Show that the points (a + 5, a – 4), (a – 2, a + 3) and (a, a) do not lie on a straight line for any value of a.

Answer:

(a + 5, a – 4), (a – 2, a + 3) and (a, a) is given.

We need to prove that they don’t line in a straight line for any value of a
This can be done by proving the points to be vertices of triangle.
Area of triangle:-
\\\begin{array}{l} \Delta=\frac{1}{2}\left|\begin{array}{ccc} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array}\right| \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{array}\right| \\ \text { Apply } R_{2} \rightarrow R_{2}-R_{1} \\ =\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ a-2-a-5 & a+3-a+4 & 1-1 \\ a & a & 1 \end{array}\right| \\\\ =\frac{1}{2}\left| \begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a & a & 1 \end{array}\right| \end{array}
\\\begin{aligned} &\text { Apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ a-a-5 & a-a+4 & 1-1 \end{array}\right|\\ &=\frac{1}{2}\left|\begin{array}{ccc} a+5 & a-4 & 1 \\ -7 & 7 & 0 \\ -5 & 4 & 0 \end{array}\right|\\ &\text { Expand along Column } 3\\ &=\frac{1}{2}[(1)(-28-(7)(-5))]\\ &=\frac{1}{2}(-28+35)=\frac{7}{2} \neq 0 \end{aligned}
This proves that the given points form a triangle therefore do not lie on a straight line.

Question:16

Show that the Δ ABC is an isosceles triangle if the determinant

\Delta=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C \end{array}\right]=0
Answer:

\\\begin{aligned} &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\\ &=\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 1+\cos A & 1+\cos B-1-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)(\cos B+\cos A)+(\cos B-\cos A) & \cos ^{2} C+\cos C \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\begin{array}{|ccc|} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)[(\cos B+\cos A)+1] & \cos ^{2} C+\cos C \end{array} \mid=0\\ &\text { Take, cosB-cosA common from column } 2\\ &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & 1 & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B+\cos A)+1 & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\\ &\cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 1+\cos A & 1 & 1+\cos C-1-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C+\cos C-\cos ^{2} A-\cos A \end{array}\right|=0 \end{aligned}
\\\begin{aligned} &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & \cos C-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C-\cos ^{2} A+\cos C-\cos A \end{array}\right|=0\\ &\Rightarrow(\cos B-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & (\cos C-\cos A) \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & (\cos C-\cos A)(\cos C+\cos A+1) \end{array}\right|=0\\ &\text { Take cosC-cosA common from Column } 3\\ &\Rightarrow(\cos B-\cos A)(\cos C-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & 1 \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos C+\cos A+1 \end{array}\right|=0 \end{aligned}

Expand along Row 1
\\$ \Rightarrow $ (cos B -cos A)(cos C -cos A)[(1)$ \{ $ cos C + cos A + 1 -(cos B + cos A + 1)$ \} $ ] = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (cos B -cos A)(cos C -cos A)[cos C + cos A + 1 -cos B -cos A -1] = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (cos B -cos A)(cos C -cos A)(cos C -cos B) = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ cos B -cos A = 0 \or\ cos C -cos A = 0 \or \ cos C -cos B = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ cos B = cos A \ or\ cos C = cos A \ or\ cos C = cos B\\ \\ \vspace{\baselineskip}$ \Rightarrow $ B = A\ or\ C = A\ or\ C = B\\ \\ Hence, ΔABC is an isosceles triangle.

Question:17

Find A^{-1} if A=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] and show that A^{-1}=\frac{A^{2}-3 I}{2}

Answer:

\\A=\left|\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$ $A^{-1}=\frac{\operatorname{adj} A}{|A|}$ Find IAl Expand IAl along Column 1 $ \\ |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc}\mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33}\end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc}\mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33}\end{array}\right|$ $+\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc}\mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23}\end{array}\right|$ $|\mathrm{A}|=(0)\left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|-1\left|\begin{array}{cc}1 & 1 \\ 1 & 0\end{array}\right|+1\left|\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right|$ \\\begin{aligned} &=0-1\{0-1\}+1\{1-0\}\\ &=1+1\\ &=2\\ &\text { To find adj } \mathrm{A} \end{aligned}
\\\begin{array}{l} a_{11}=\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1 \\ a_{12}=-\left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right|=-(0-1)=1 \\ a_{13}=\left|\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right|=1-0=1 \\ a_{21}=-\left|\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right|=-(0-1)=1 \\ a_{22}=\left|\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right|=0-1=-1 \end{array}
\\\begin{array}{l} \mathrm{a}_{23}=-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right|=-(0-1)=1 \\ \mathrm{a}_{31}=\left|\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right|=1-0=1 \\ \mathrm{a}_{32}=-\left|\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right|=-(0-1)=1 \\ \therefore \mathrm{adj} \mathrm{A}=\left[\begin{array}{ccc} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right] \end{array}

\\\begin{aligned} &\therefore \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\frac{\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]}{2}=\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\\ &A^{-1}=\frac{A^{2}-31}{2}\\ &\text { Now, we need to prove that }\\ &A^{2}=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right] \times\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]=\left[\begin{array}{lll} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{array}\right]\\ &A^{2}=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]\\ &\therefore \frac{A^{2}-3 I}{2}=\frac{1}{2}\left\{\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]-3\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\right\} \end{aligned}
\\\begin{array}{l} =\frac{1}{2}\left\{\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]\right\} \\ =\frac{1}{2}\left\{\left[\begin{array}{ccc} 2-3 & 1 & 1 \\ 1 & 2-3 & 1 \\ 1 & 1 & 2-3 \end{array}\right]\right\} \\ \left.=\frac{1}{2}\left[\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]\right\} \\ =A^{-1} \end{array}
Hence Proved

Question:18

If A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] find A^{-1}. Using A^{-1}, solve the system of linear equations x-2y = 10, 2x -y -z = 8, -2y + z = 7

Answer:

\\A=\left|\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right|$ $\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$ Find IAI Expand IAl along Column 1
\\\begin{aligned} &\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(1)\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} 0 & 0 \\ -1 & 1 \end{array}\right|+0\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right| \\ =(-1+2)+2(0)+0 \\ =1 \end{array}\\ &\text { To find adj } A \end{aligned}
\\\begin{array}{l} a_{11}=\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|=-1-2=-3 \\ a_{12}=-\left|\begin{array}{cc} -2 & -2 \\ 0 & 1 \end{array}\right|=-(-2+0)=2 \\ a_{13}=\left|\begin{array}{cc} -2 & -1 \\ 0 & -1 \end{array}\right|=2+0=2 \\ a_{21}=-\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-(2+0)=-2 \\ a_{22}=\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \\ a_{23}=-\left|\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right|=-(-1)=1 \end{array}
\\\begin{array}{l} a_{31}=\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right|=-4 \\ a_{32}=-\left|\begin{array}{cc} 1 & 0 \\ -2 & -2 \end{array}\right|=-(-2)=2 \\ a_{33}=\left|\begin{array}{ccc} 1 & 2 \\ -2 & -1 \end{array}\right|=-1+4=3 \\ \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 3 \end{array}\right]}{1}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \end{array}
According to linear equation:
\\x -2y = 10\\ \\ \vspace{\baselineskip}2x -y -z = 8\\ \\ \vspace{\baselineskip}-2y + z = 7\\ \\
We know that, AX = B
A=\left[\begin{array}{ccc} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{array}\right]
Here,
So, transpose of \\ A^{-1}
\\ A^{-1}=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]$ \\\begin{array}{l} \Rightarrow X=A^{-1} B \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{l} 10 \\ 8 \\ 7 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right] \\ \therefore x=0, y=-5 \text { and } z=-3 \end{array}

Question:19

Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.

Answer:

\\Given,\\ \vspace{\baselineskip} 3x + 2y -2z = 3,\\ \\ \vspace{\baselineskip}x + 2y + 3z = 6,\\ \\ \vspace{\baselineskip}2x -y + z = 2\\ \vspace{\baselineskip} We know,\\ \vspace{\baselineskip} AX = B\\
\\\left[\begin{array}{ccc}3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]$\\\therefore \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\ \\\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$ \\Find IAl Expand IAl along Column 1 \begin{array}{c} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(3)\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|-(1)\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|+2\left|\begin{array}{cc} 2 & -2 \\ 2 & 3 \end{array}\right| \end{array}

\\\begin{aligned} &=3(2+3)-1(2-2)+2(6+4)\\ &=3(5)+2(10)\\ &=15+20\\ &=35\\ &\text { To find adj } A\\ &a_{11}=\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|=2+3=5\\ &a_{12}=-\left|\begin{array}{ll} 1 & 3 \\ 2 & 1 \end{array}\right|=-(1-6)=5\\ &a_{13}=\left|\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right|=-1-4=-5 \end{aligned}

\\\begin{array}{l} a_{21}=-\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|=-(2-2)=0 \\ a_{22}=\left|\begin{array}{cc} 3 & -2 \\ 2 & 1 \end{array}\right|=3+4=7 \\ a_{23}=-\left|\begin{array}{cc} 3 & 2 \\ 2 & -1 \end{array}\right|=-(-3-4)=-7 \\ a_{32}=-\left|\begin{array}{cc} 3 & -2 \\ 1 & 3 \end{array}\right|=-(9+2)=-11 \\ a_{33}=\left|\begin{array}{cc} 3 & 2 \\ 1 & 2 \end{array}\right|=6-2=4 \end{array}
\\\begin{array}{l} \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 5 & -5 \\ 0 & 7 & -7 \\ 10 & -11 & 4 \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]}{35}=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \text { Now, } x=A^{-1} B \end{array}
\\\begin{array}{l} {\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 6 \\ 2 \end{array}\right]} \\\\ \Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{c} 15+20 \\ 15+42-22 \\ -15+42+8 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{l} 35 \\ 35 \\ 35 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \end{array}

∴ x = 1, y = 1 and z = 1

Question:20

Given A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] find BA and use this to solve the system of equations y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17.

Answer:

\\\begin{array}{l} A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ \therefore B A=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ =\left[\begin{array}{ccc} 2+4 & 2-2 & -4+4 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ -4+4 & 2-2 & -4+10 \end{array}\right] \end{array}
\begin{array}{l}\\\ =\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\\\ =6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{array}
BA = 6I $ \ldots $ (i)\\
Now, given system of equations is:
\\ \vspace{\baselineskip}y + 2z = 7,\\ \\ \vspace{\baselineskip}x -y = 3,\\ \\ \vspace{\baselineskip}2x + 3y + 4z = 17\\ \\ \vspace{\baselineskip}So,\\

\\ \begin{aligned} &\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & -1 & 0 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 7 \\ 3 \\ 17 \end{array}\right]\\ &\text { Apply, } \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}\\ &\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3}\\ &\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right]\\ &\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]^{-1}\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right]_{\ldots(i i)}\\ &\text { So, } B A=6 I \text { [from eg(i)] } \end{aligned}
\\\Rightarrow \mathrm{B}^{-1}=\frac{\mathrm{A}}{6}$\\ and $A=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$ $\\\Rightarrow \mathrm{B}^{-1}=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$\\ Put the value of $\mathrm{B}^{-1}$ in equation 2 $\\\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$ $\\\\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6+34-28 \\ -12+34-28 \\ 6-17+25\end{array}\right]$ \\\begin{array}{l} \left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] =\frac{1}{6}\left[\begin{array}{l} 12 \\ -6 \\ 24 \end{array}\right] \\\\ \left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] ={\left[\begin{array}{c} 2 \\ -1 \\ 4 \end{array}\right]} \\ \therefore \mathrm{x}=2, \mathrm{y}=-1 \text { and } \mathrm{z}=4 \end{array}

Question:21

If a + b + c ≠ 0 and \left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=0 then prove that a = b = c.

Answer:

\\ A=\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|
\\\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &=\left|\begin{array}{lll} a+b+c & b & c \\ b+a+c & c & a \\ c+a+b & a & b \end{array}\right|\\ &\text { Take }(a+b+c) \text { common from Column } 1\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right| \end{aligned}
Expand along Column 1
\\\vspace{\baselineskip} = (a + b + c)[(1)(bc -a\textsuperscript{2}) -(1)(b\textsuperscript{2} -ac) + (1)(ba -c\textsuperscript{2})]\\ \\ \vspace{\baselineskip}= (a + b + c)[bc -a\textsuperscript{2} -b\textsuperscript{2} + ac + ab -c\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= (a + b + c)[-(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2} -ab -bc -ac)]\\
\\ =-\frac{1}{2}(a+b+c)\left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 a c\right) \\ =-\frac{1}{2}(a+b+c)\left[\left(a^{2}+b^{2}-2 a b\right)+\left(b^{2}+c^{2}-2 b c\right)+\left(c^{2}+a^{2}-2 a c\right)\right] \\ =-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \\ {\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]}
Given that Δ = 0
\Rightarrow-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0
\\\vspace{\baselineskip}$ \Rightarrow $ (a + b + c)[(a -b)\textsuperscript{2} + (b -c)\textsuperscript{2} + (c -a)\textsuperscript{2}] = 0\\ \\ \vspace{\baselineskip}\text{Either }(a + b + c) = 0 or (a -b)\textsuperscript{2} + (b -c)\textsuperscript{2} + (c -a)\textsuperscript{2} = 0\\ \\ \text{but it is given that }(a + b + c) $ \neq $ 0\\ \\ $ \therefore $ (a -b)\textsuperscript{2} + (b -c)\textsuperscript{2} + (c -a)\textsuperscript{2} = 0\\ \\ $ \Rightarrow $ a -b = b -c = c -a = 0\\ \\ $ \Rightarrow $ a = b = c\\ Hence Proved

Question:22

Prove that \left|\begin{array}{ccc} \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right| is divisible by a + b + c and find the quotient.

Answer:

\left|\begin{array}{ccc} \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right| is given.
Apply,R\textsubscript{1}$ \rightarrow $ R\textsubscript{1} -R\textsubscript{2},\\ \\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{b} c-\mathrm{a}^{2}-\mathrm{ca}+\mathrm{b}^{2} & \mathrm{ca}-\mathrm{b}^{2}-\mathrm{ab}+\mathrm{c}^{2} & \mathrm{ab}-\mathrm{c}^{2}-\mathrm{bc}+\mathrm{a}^{2} \\ \mathrm{ca}-\mathrm{b}^{2} & \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} \\ \mathrm{ab}-\mathrm{c}^{2} & \mathrm{bc}-\mathrm{a}^{2} & \mathrm{ca}-\mathrm{b}^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} (b c-c a)+\left(b^{2}-a^{2}\right) & (c a-a b)+\left(c^{2}-b^{2}\right) & (a b-b c)+\left(a^{2}-c^{2}\right) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} c(b-a)+(b-a)(b+a) & a(c-b)+(c-b)(c+b) & b(a-c)+(a-c)(a+c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} (b-a)(c+b+a) & (c-b)(a+c+b) & (a-c)(b+a+c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Take (a+b+c) common from Column 1
\\\begin{aligned} &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ c a-b^{2} & a b-c^{2} & b c-a^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ c a-b^{2}-a b+c^{2} & a b-c^{2}-b c+a^{2} & b c-a^{2}-c a+b^{2} \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=(a+b+c)\left|\begin{array}{ccc} (b-a) & (c-b) & (a-c) \\ (c-b)(a+b+c) & (a-c)(a+b+c) & (b-a)(a+b+c) \\ a b-c^{2} & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Take (a+b+c) common from Column 2
\\\begin{aligned} &=(a+b+c)^{2}\left|\begin{array}{ccc} b-a+c-b+a-c & (c-b) & (a-c) \\ c-b+a-c+b-a & (a-c) & (b-a) \\ a b-c^{2}+b c-a^{2}+c a-b^{2} & b c-a^{2} & c a-b^{2} \end{array}\right|\\ &=(a+b+c)^{2}\left|\begin{array}{ccc} 0 & (c-b) & (a-c) \\ 0 & (a-c) & (b-a) \\ a b+b c+c a-\left(a^{2}+b^{2}+c^{2}\right) & b c-a^{2} & c a-b^{2} \end{array}\right| \end{aligned}
Expand along Column 1

\\= (a + b + c)\textsuperscript{2}[ \{ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})$ \} \{ $ (c -b)(b -a) -(a -c)\textsuperscript{2} \} ]\\ \\ \vspace{\baselineskip}= (a + b + c)\textsuperscript{2}[\{ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})$\} \{ $ (cb -ac -b\textsuperscript{2} + ab -(a + c\textsuperscript{2} -2ac)$ \} ]\\ \\ \vspace{\baselineskip}= (a + b + c)\textsuperscript{2}[\{ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})$\} \{ $ (cb -ac -b\textsuperscript{2} + ab -a - c\textsuperscript{2} + 2ac) \} ]\\ \\ \vspace{\baselineskip}= (a + b + c)\textsuperscript{2}[\{ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})$\} \{ $ ac + bc + ab -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})$ \} $ ]\\ \\
\\\vspace{\baselineskip}=(a + b + c)\textsuperscript{2}[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]\textsuperscript{2}\\
\\\vspace{\baselineskip}=(a + b + c)(a + b + c)[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]^2\\
The determinant is divisible by (a+b+c) and the quotient is \\(a + b + c)[ab + bc + ca -(a\textsuperscript{2} + b\textsuperscript{2} + c\textsuperscript{2})]^2\\

Question:23

If x + y + z = 0, prove that \left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|=x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|

Answer:

Given LHS,
\left|\begin{array}{lll} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{array}\right|
Expand along Row 1
\\\\ = xa$ \{ $ (za)(ya) -(xc)(xb)$ \} $ -(yb)$ \{ $ (yc)(ya) -(zb)(xb)$ \} $ + (zc)$ \{ $ (yc)(xc) -(zb)(za)$ \} $ \\ \\ \vspace{\baselineskip}= xa$ \{ $ a\textsuperscript{2}yz -x\textsuperscript{2}bc$ \} $ -yb$ \{ $ y\textsuperscript{2}ac -b\textsuperscript{2}xz$ \} $ + zc$ \{ $ c\textsuperscript{2}xy -z\textsuperscript{2}ab$ \} $ \\ \\ \vspace{\baselineskip}= a\textsuperscript{3}xyz -x\textsuperscript{3}abc -y\textsuperscript{3}abc + b\textsuperscript{3}xyz + c\textsuperscript{3}xyz -z\textsuperscript{3}abc\\ \\ \vspace{\baselineskip}= xyz(a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3}) -abc(x\textsuperscript{3} + y\textsuperscript{3} + z\textsuperscript{3})\\

Given x+y+z = 0
\\\vspace{\baselineskip} $ \Rightarrow $ x\textsuperscript{3} + y\textsuperscript{3} + z\textsuperscript{3} = 3xyz\\ \\ \vspace{\baselineskip}= xyz(a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3}) -abc (3xyz)\\ \\ \vspace{\baselineskip}= xyz(a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3} -3abc)\\ \\ =x y z\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|

Question:24

If \left|\begin{array}{cc} 2 \mathrm{x} & 5 \\ 8 & \mathrm{x} \end{array}\right|=\left|\begin{array}{cc} 6 & -2 \\ 7 & 3 \end{array}\right| then value of x is
A. 3
B. ± 3
C. ± 6
D. 6

Answer:

Given:
\left|\begin{array}{cc} 2 \mathrm{x} & 5 \\ 8 & \mathrm{x} \end{array}\right|=\left|\begin{array}{cc} 6 & -2 \\ 7 & 3 \end{array}\right|
\\\begin{aligned} &A=\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|\\ &\text { Then the determinant of } \mathrm{A} \text { is }\\ &|\mathrm{A}|=\left|\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right|=\mathrm{ad}-\mathrm{b} \mathrm{c} \end{aligned}
\\ \vspace{\baselineskip}$ \Rightarrow $ (2x)(x) -(5)(8) = (6)(3) -(7)(-2)\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2x\textsuperscript{2} -40 = 18 -(-14)\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2x\textsuperscript{2} -40 = 18 + 14\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x\textsuperscript{2} -20 = 9 + 7\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x\textsuperscript{2} -20 = 16\\ $ \Rightarrow $ x\textsuperscript{2} = 16 + 20\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x\textsuperscript{2} = 36\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x = $ \surd$ 36\\ \\ \vspace{\baselineskip}$ \Rightarrow $ x = $ \pm $ 6\\

Question:25

The value of determinant \left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|
A. a^3 + b^3 + c^3
B. 3 bc
C. a^3 + b^3 + c^3-3abc
D. none of these

Answer:

C)
Given:
\left|\begin{array}{lll} a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c \end{array}\right|
Apply C2→ C2 + C3
\\\begin{aligned} &=\left|\begin{array}{lll} a-b & a+b+c & a \\ b-c & c+a+b & b \\ c-a & a+b+c & c \end{array}\right|\\ &\text { Take }(a+b+c) \text { common from Column } 2\\ &=(a+b+c)\left|\begin{array}{lll} a-b & 1 & a \\ b-c & 1 & b \\ c-a & 1 & c \end{array}\right|\\ &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\ &=(a+b+c)\left|\begin{array}{lll} a-b-a & 1 & a \\ b-c-b & 1 & b \\ c-a-c & 1 & c \end{array}\right|\\ &=(a+b+c)\left|\begin{array}{ccc} -b & 1 & a \\ -c & 1 & b \\ -a & 1 & c \end{array}\right| \end{aligned}
Expand along Row 1
\\ \vspace{\baselineskip} = (a + b + c)[(-b)$ \{ $ c -b$ \} $ -(1)$ \{ $ -c\textsuperscript{2} -(-ab)$ \} $ + a$ \{ $ -c -(-a)$ \} $ ]\\ \\ \vspace{\baselineskip}= (a + b + c)(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2})\\ \\ \vspace{\baselineskip}= a(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2}) + b(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2}) + c(-bc + b\textsuperscript{2} + c\textsuperscript{2} -ab -ac + a\textsuperscript{2})\\

\\ \\ \vspace{\baselineskip}= -abc + ab\textsuperscript{2} + ac\textsuperscript{2} -a\textsuperscript{2}b -a\textsuperscript{2}c + a\textsuperscript{3} -b\textsuperscript{2}c + b\textsuperscript{3} + bc\textsuperscript{2} -ab\textsuperscript{2} -abc + a\textsuperscript{2}b -bc\textsuperscript{2} + b\textsuperscript{2}c + c\textsuperscript{3} -abc -ac\textsuperscript{2}+ a\textsuperscript{2}c\\ \\ \vspace{\baselineskip}= a\textsuperscript{3} + b\textsuperscript{3} + c\textsuperscript{3} -3abc\\

Question:26

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be
A. 9
B. 3
C. – 9
D. 6

Answer:

B)
\\ \begin{aligned} &\text { The area of a triangle with vertices }\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) \text { is given by }\\ &\Delta=\frac{1}{2}\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3} & 1 \end{array}\right|\\ &\Delta=\frac{1}{2}\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=9\\ &\Delta=\left|\begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & \mathrm{k} & 1 \end{array}\right|=18 \end{aligned}
Expand along Column 2
\\ \vspace{\baselineskip} $ \Rightarrow $ -(k) $ \{ $ -3 -3$ \} $ = 18\\ \\ \vspace{\baselineskip}$ \Rightarrow $ -k (-6) = 18\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 6k = 18\\ \\ \vspace{\baselineskip}$ \Rightarrow $ k = 3\\

Question:27

The determinant \left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right| equals

A. abc (b–c) (c – a) (a – b)
B. (b–c) (c – a) (a – b)
C. (a + b + c) (b – c) (c – a) (a – b)
D. None of these

Answer:

D)
Given:
\left|\begin{array}{ccc} b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2} \end{array}\right|
\\=\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right|$ \\Take (b-a) common from both Column 1 and 3\\ $=(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right|$ \\Apply $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\$ $=(\mathrm{b}-\mathrm{a})(\mathrm{b}-\mathrm{a})\left|\begin{array}{lll}\mathrm{b}-\mathrm{c} & \mathrm{b}-\mathrm{c} & \mathrm{c} \\ \mathrm{a}-\mathrm{b} & \mathrm{a}-\mathrm{b} & \mathrm{b} \\ \mathrm{c}-\mathrm{a} & \mathrm{c}-\mathrm{a} & \mathrm{a}\end{array}\right|
Whenever any two columns or rows in any determinant are equal, its value becomes = 0
Here Column 1 and 2 are identical
\\\therefore\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2}\end{array}\right|=0$

Question:28

The number of distinct real roots of \left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0 in the interval -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} is

A. 0

B. –1
C. 1
D. None of these

Answer:

C)
Given
\left|\begin{array}{ccc} \sin x & \cos & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0
\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\\ &\Rightarrow\left|\begin{array}{lll} \sin x+\cos x+\cos x & \cos x & \cos x \\ \cos x+\sin x+\cos x & \sin x & \cos x \\ \cos x+\cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\Rightarrow\left|\begin{array}{ccc} 2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x \end{array}\right|=0\\ &\text { Take }(2 \cos x+\sin x) \text { common from Column } 1\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}\\ &\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 1-1 & \sin x-\cos x & \cos x-\cos x \\ 1 & \cos x & \sin x \end{array}\right|=0 \end{aligned}
\\\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1 & \cos x & \sin x \end{array}\right|=0 \\ \text { Apply } R_{3} \rightarrow R_{3}-R_{1} \\ \Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 1-1 & \cos x-\cos x & \sin x-\cos x \end{array}\right|=0 \\

\Rightarrow 2 \cos x+\sin x\left|\begin{array}{ccc} 1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x \end{array}\right|=0
Expand along Column 1
\\\vspace{\baselineskip} (2cos X + sin X) [(1)$ \{ $ (sin X -cos X)(sin X -cos X)$ \} $ ]\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (2cos X + sin X)(sin X -cos X)\textsuperscript{2} = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2cos X = -sin X or (sin X -cos X)\textsuperscript{2} = 0\\\\ \Rightarrow 2=-\frac{\sin \mathrm{x}}{\cos \mathrm{x}}$ or $\sin \mathrm{x}=\cos \mathrm{x}$\\ $\Rightarrow \tan x=-2$ or tan $x=1\left[\because \tan x=\frac{\sin x}{\cos x}\right]$ \\but tan $x=-2$ is not possible as for $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ \\So, $\tan x=1$ $\\\therefore x=\frac{\pi}{4}$ Only one real and distinct root occurs.

Question:29

If A, B and C are angles of a triangle, then the determinant \begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid is equal to
A. 0

B. –1
C. 1
D. None of these

Answer:

A)
Given:
\begin{array}{|ccc|} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array} \mid
Expand along Column 1
\begin{array}{r} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{ll} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{ll} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \end{array}
\Delta=(-1)\left|\begin{array}{cc} -1 & \cos A \\ \cos A & -1 \end{array}\right|-\cos C\left|\begin{array}{cc} \cos C & \cos B \\ \cos A & -1 \end{array}\right|+\cos B\left|\begin{array}{cc} \mid \cos C & \cos B \\ -1 & \cos A \end{array}\right|
\\ \vspace{\baselineskip}= [(-1)$ \{ $ 1 -cos\textsuperscript{2}A$ \} $ -cos C$ \{ $ -cos C -cos Acos B$ \} $ + cos B$ \{ $ cos A cos C + cos B$ \} $ ]\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}C + cos A cos B cos C + cos A cos B cos C + cos\textsuperscript{2}B\\ \\ \vspace{\baselineskip}= -1 + cos\textsuperscript{2}A + cos\textsuperscript{2}B + cos\textsuperscript{2}C + 2cos A cos B cos C\\ \vspace{\baselineskip} \text{Using the formula}\\ \vspace{\baselineskip} 1 + cos2A = 2cos\textsuperscript{2}A\\
\\ \begin{aligned} &=-1+\frac{1+\cos 2 \mathrm{~A}}{2}+\frac{1+\cos 2 \mathrm{~B}}{2}+\frac{1+\cos 2 \mathrm{C}}{2}+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}\\ &\text { Taking L.C.M, we get }\\ &=\frac{-2+1+\cos 2 A+1+\cos 2 B+1+\cos 2 C+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+(\cos 2 A+\cos 2 B)+\cos 2 C+4 \cos C \cos A \cos B}{2}\\ &\text { Now use: } \cos (A+B) \cos (A-B)=2 \cos A \cos B\\ &so, \cos 2 A+\cos 2 B=2 \cos (A+B) \cos (A-B) \end{aligned}

\\ \begin{aligned} &=\frac{1+\cos 2 C+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &=\frac{1+2 \cos ^{2} C-1+\{2 \cos (A+B) \cos (A-B)\}+4 \cos A \cos B \cos C}{2}\\ &.=\frac{2 \cos ^{2} \mathrm{C}+[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})\}+4 \cos \mathrm{A} \cos \mathrm{B} \cos C}{2} \ldots (\mathrm{i})\\ &\text { We know that } A, B, C \text { are angles of triangle }\\ &\Rightarrow A+B+C=\pi\\ &\Rightarrow A+B=\pi-C \end{aligned}
\\ =\frac{2 \cos ^{2} C+\{2 \cos (\pi-C) \cos (A-B)\}+4 \cos A \cos B \cos C}{2} \\ =\frac{2 \cos ^{2} C+\{-2 \cos C \cos (A-B)\}+4 \cos A \cos B \cos C}{2} [\because \cos (\pi-x)=-\cos x] \\ =\frac{-2 \cos C\{\cos (A-B)-\cos C\}+4 \cos A \cos B \cos C}{2}
\\ \vspace{\baselineskip}= -cos C$ \{ $ cos(A -B) -cos C$ \} $ + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[cos(A -B) -cos$ \{ $ $ \pi $ -(A + B)$ \} $ ] + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[cos(A -B) + cos(A + B)] + + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= -cos C[2cos Acos B] + 2cos Acos Bcos C\\ \\ \vspace{\baselineskip}= 0\\

Question:30

Let f(t)=\left|\begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{array}\right| then \lim _{t \rightarrow 0} \frac{f(t)}{t^{2}} is equal to
A. 0
B. –1
C. 2
D. 3

Answer:

A)
Given:
f(t)=\left|\begin{array}{ccc} \cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t \end{array}\right|
\\ \begin{aligned} &\text { Divide } \mathrm{R}_{2} \text { and } \mathrm{R}_{3} \text { by } t\\ &f(t)=t^{2}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{t} & \frac{t}{t} & \frac{2 t}{t} \\ \frac{\operatorname{sint}}{t} & \frac{t}{t} & \frac{t}{t} \end{array}\right|\\ &f(t)=t^{2}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{ t} & 1 & 2 \\ \frac{\sin t}{t} & 1 & 1 \end{array}\right| \end{aligned}
\\ \Rightarrow \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^{2}}=\frac{\mathrm{t}^{2}}{\mathrm{t}^{2}}\left|\begin{array}{ccc} \cos t & t & 1 \\ \frac{2 \sin t}{ t} & 1 & 2 \\ \frac{\sin t}{t} & 1 & 1 \end{array}\right|
\\\Rightarrow \lim _{t \rightarrow 0} \frac{\mathrm{f}(\mathrm{t})}{\mathrm{t}^{2}}=\left|\begin{array}{lll} \lim _{t \rightarrow 0} \cos t & \lim _{t \rightarrow 0} t & \lim _{t \rightarrow 0} 1 \\ \lim _{t \rightarrow 0} \frac{2 \sin t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 2 \\ \lim _{t \rightarrow 0} \frac{\sin t}{t} & \lim _{t \rightarrow 0} 1 & \lim _{t \rightarrow 0} 1 \end{array}\right|
=\left|\begin{array}{lll} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 1 & 1 \end{array}\right|\left(\because \lim _{t \rightarrow 0} \frac{\sin t}{t}=1\right)
Expand along Row 1
\\ = (1)(1 -2) + (1)(2 -1)\\ \\ \vspace{\baselineskip}= - 1 + 1\\ \vspace{\baselineskip}= 0\\

Question:31

The maximum value of \\\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right|_{\text {is }(\theta \text { is a real number })}\\$
\\A. \frac{1}{2}$ \\B. $\frac{\sqrt{3}}{2}$ \\C. $\sqrt{2}$\\ D. $\frac{2 \sqrt{3}}{4}$

Answer:

A)
\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\\ &\Rightarrow \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & \sin \theta & 1 \\ \cos \theta & 0 & 1 \end{array}\right| \end{aligned}
\\ \vspace{\baselineskip}= 0 -0 + 1 (sin \theta . cos \theta )\\ \\ \vspace{\baselineskip} \text{Multiply and divide by 2},\\ \\ \vspace{\baselineskip}= 1/2 (2sin \theta cos \theta )\\ \vspace{\baselineskip} \text{We already know that } {2 sin \theta cos \theta = sin 2 \theta }\\ = 1/2 (sin 2 \theta )\\ \vspace{\baselineskip}The maximum value of :sin 2$ \theta $ is 1, $ \theta $ = 45$ ^{\circ} $ .\\\\ $ \therefore $ $ \Delta $ = 1/2 (sin 2(45$ ^{\circ} $ ))\\ \\ \vspace{\baselineskip}= 1/2 sin 90$ ^{\circ} $ \\ \\ \vspace{\baselineskip}= 1/2 (1)\\ \\\\\vspace{\baselineskip}$ \therefore $ $ \Delta $ = 1/2\\

Question:32

If f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|

A. f (a) = 0
B. f (b) = 0
C. f (0) = 0
D. f (1) = 0

Answer:

C)
We have:
f(x)=\left|\begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array}\right|
\\ \begin{aligned} &\text { If we put } x=a\\ &f(a)=\left|\begin{array}{ccc} 0 & a-a & a-b \\ a+a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right|\\ &=0\left|\begin{array}{cc} 0 & a-c \\ a+c & 0 \end{array}\right|-0\left|\begin{array}{cc} 2 a & a-c \\ a+b & 0 \end{array}\right|+(a-b)\left|\begin{array}{cc} 2 a & 0 \\ a+b & a+c \end{array}\right| \end{aligned}
\\ \vspace{\baselineskip}= 0 -0 + (a -b) [2a (a + c) -0 (a + b)]\\ \\ \vspace{\baselineskip}= (a -b) [2a\textsuperscript{2} + 2ac -0]\\ \\ \vspace{\baselineskip}= (a -b) (2a\textsuperscript{2} + 2ac) $ \neq $ 0\\ \vspace{\baselineskip} If x = b\\

\\ \begin{array}{l} f(b)=\left|\begin{array}{ccc} 0 & b-a & b-b \\ b+a & 0 & b-c \\ b+b & b+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & b-c \\ b+c & 0 \end{array}\right|-(b-a)\left|\begin{array}{cc} b+a & b-c \\ 2 b & 0 \end{array}\right|+0\left|\begin{array}{cc} b+a & 0 \\ 2 b & b+c \end{array}\right| \end{array}
\\ \vspace{\baselineskip}= 0 -(b - a) [(b + a) (0) -(b -c) (2b)] + 0\\ \\ \vspace{\baselineskip}= - (b -a) [0 - 2b\textsuperscript{2} + 2bc]\\ \\ \vspace{\baselineskip}= (a -b) (2b\textsuperscript{2} -2bc) $ \neq $ 0\\ \vspace{\baselineskip} If x = 0 according to the given question:
\\ \begin{array}{l} f(0)=\left|\begin{array}{ccc} 0 & 0-a & 0-b \\ 0+a & 0 & 0-c \\ 0+b & 0+c & 0 \end{array}\right| \\ =0\left|\begin{array}{cc} 0 & -c \\ c & 0 \end{array}\right|-(-a)\left|\begin{array}{cc} a & -c \\ b & 0 \end{array}\right|+(-b)\left|\begin{array}{ll} a & 0 \\ b & c \end{array}\right| \end{array}
\\ \vspace{\baselineskip}= 0 + a [a (0) -(-bc)] -b [ac -b (0)]\\ \\ \vspace{\baselineskip}= a [bc] -b [ac]\\ \\ \vspace{\baselineskip}= abc -abc = 0\\ \vspace{\baselineskip}
Then the condition is satisfied.

Question:33

If A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]

then A^{-1} exists if
A. λ = 2
B. λ ≠ 2
C. λ ≠ -2
D. None of these

Answer:

D)
We have
A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]
\\ \vspace{\baselineskip}$ \Rightarrow $ $ \vert $ A$ \vert $ = 2 (6 -5) - $ \lambda $ (0 -5) + (-3) (0 -2)\\ \\ \vspace{\baselineskip}= 2 + 5$ \lambda $ + 6\\ \\ \vspace{\baselineskip}= 5$ \lambda $ + 8\\ \text{Inverse of A exists only if A is non singular} i.e. $ \vert $ A$ \vert $ $ \neq $ 0.\\ $ \therefore $ 5$ \lambda $ + 8 $ \neq $ 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 5$ \lambda $ $ \neq $ -8\\ \therefore \lambda \neq-\frac{8}{5}
So, A^{-1} exists if and only if \lambda \neq-\frac{8}{5}

Question:34

If A and B are invertible matrices, then which of the following is not correct?
A. adj A = |A|. A^{-1}

B. det (A)^{-1} = [det (A)]^{-1}

C. (AB)^{-1} = B^{-1} A^{-1}
D. (A + B)^{-1} = B^{-1} + A^{-1}

Answer:

D)
We know, A and B are invertible matrices
\\ \vspace{\baselineskip} Consider (AB) B\textsuperscript{-1} A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (AB) B\textsuperscript{-1} A\textsuperscript{-1} = A(BB\textsuperscript{-1}) A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}= AIA\textsuperscript{-1} = (AI) A\textsuperscript{-1}\\ \\ \vspace{\baselineskip}= AA\textsuperscript{-1} = I\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (AB)\textsuperscript{-1} = B\textsuperscript{-1} A\textsuperscript{-1} $ \ldots $ option (C)\\

\\ \\ \vspace{\baselineskip}Also AA\textsuperscript{-1} = I\\ \\ \vspace{\baselineskip}$ \Rightarrow $ $ \vert $ AA\textsuperscript{-1}$ \vert $ = $ \vert $ I$ \vert $ \\ \\ \vspace{\baselineskip}$ \Rightarrow $ $ \vert $ A$ \vert $ $ \vert $ A\textsuperscript{-1}$ \vert $ = 1\\ \\ \begin{aligned} &\Rightarrow|\mathrm{A}|^{-1}=\frac{1}{|\mathrm{~A}|}\\ &\therefore \operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1} \ldots(B)\\ &\text { We know that } \frac{|\mathrm{A}|^{-1}}{ }=\frac{\operatorname{adj} \mathrm{A}}{|\vec{A}|}\\ &\Rightarrow \text { adj } A=|A| \cdot A^{-1} \ldots \text { option }(A)\\ &\Rightarrow(\mathrm{A}+\mathrm{B})^{-1}=\frac{1}{|\mathrm{~A}+\mathrm{B}|} \operatorname{adj}(\mathrm{A}+\mathrm{B})\\ &{\text { But }}{\mathrm{B}}^{-1}+\mathrm{A}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}+\frac{1}{|\mathrm{~A}|} \text { adj } \mathrm{A} \end{aligned}
\therefore $ (A + B)\textsuperscript{-1} $ \neq $ B\textsuperscript{-1} + A\textsuperscript{-1}\\

Question:35

If x, y, z are all different from zero and ,\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0, then value of x^{-1} + y^{-1} + z^{-1} is
\\A. x y z\\B. x^{-1} y^{-1} z^{-1}\\C. -x -y -z\\D. -1

Answer:

We have
\left|\begin{array}{ccc} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{array}\right|=0
\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}
\left|\begin{array}{ccc} x & 0 & 1 \\ 0 & y & 1 \\ -z & -z & 1+z \end{array}\right|=0
Expand along Row 1
\\ \vspace{\baselineskip} $ \Rightarrow $ x [y (1 + z) + z] -0 + 1 (yz) = 0\\ xy + xyz + xz + yz = 0\\ \vspace{\baselineskip} Divide both side by xyz
\\ \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\ \therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x^{-1}+y^{-1}+z^{-1}=-1

Question:36

The value of the determinant \left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right| is
A. 9x^2 (x + y)
B. 9y^2 (x + y)
C. 3y^2 (x + y)
D. 7x^2 (x + y)

Answer:

B)
\\\begin{aligned} &\text { Matrix given: }\\ &\left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|\\ &=\mathrm{x}\left|\begin{array}{cc} \mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{x}+2 \mathrm{y} & \mathrm{x} \end{array}\right|-(\mathrm{x}+\mathrm{y})\left|\begin{array}{cc} \mathrm{x}+2 \mathrm{y} & \mathrm{x}+\mathrm{y} \\ \mathrm{x}+\mathrm{y} & \mathrm{x} \end{array}\right|+(\mathrm{x}+2 \mathrm{y})\left|\begin{array}{cc} \mathrm{x}+2 \mathrm{y} & \mathrm{x} \\ \mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} \end{array}\right| \end{aligned}
\\\vspace{\baselineskip}= x [x\textsuperscript{2} -(x + y) (x + 2y)] -(x + y) [(x + 2y) (x) -(x + y)\textsuperscript{2}] + (x + 2y) [(x + 2y)\textsuperscript{2} -x (x + y)]\\ \\ \vspace{\baselineskip}= x [x\textsuperscript{2} -x\textsuperscript{2} -3xy -2y\textsuperscript{2}] -(x + y) [x\textsuperscript{2} + 2xy -x\textsuperscript{2} -2xy -y\textsuperscript{2}] + (x + 2y) [x\textsuperscript{2} + 4xy + 4y\textsuperscript{2} -x\textsuperscript{2} -xy]\\ \\ \vspace{\baselineskip}= x [-3xy -2y\textsuperscript{2}] -(x + y) [-y\textsuperscript{2}] + (x + 2y) [3xy + 4y\textsuperscript{2}]\\ \\ \vspace{\baselineskip}= -3x\textsuperscript{2}y -2xy\textsuperscript{2} + xy\textsuperscript{2} + y\textsuperscript{3} +3x\textsuperscript{2}y + 4xy\textsuperscript{2} + 6xy\textsuperscript{2} + 8y\textsuperscript{3}\\ \\ \vspace{\baselineskip}= 9y\textsuperscript{3} + 9xy\textsuperscript{2}\\ \\ \vspace{\baselineskip}= 9y\textsuperscript{2} (x + y)\\

Question:37

There are two values of a which makes determinant, \Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86 ,then sum of these number is
A. 4
B. 5
C. -4
D. 9

Answer:

C)
\\ \begin{aligned} &\text { We have: }\\ &\Delta=\left|\begin{array}{ccc} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86\\ &1\left|\begin{array}{cc} \mathrm{a} & -1 \\ 4 & 2 \mathrm{a} \end{array}\right|-(-2)\left|\begin{array}{cc} 2 & -1 \\ 0 & 2 \mathrm{a} \end{array}\right|+5\left|\begin{array}{cc} 2 & \mathrm{a} \\ 0 & 4 \end{array}\right|=86 \end{aligned}
\\\vspace{\baselineskip} 1 [2a\textsuperscript{2} -(-4)] + 2 [4a -0] + 5 [8 -0] = 86\\ \\ \vspace{\baselineskip} 1 [2a\textsuperscript{2} + 4] + 2 [4a] + 5 [8] = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 4 + 8a + 40 = 86\\ \\ \vspace{\baselineskip} 2a\textsuperscript{2} + 8a + 44 = 86\\ \\ \vspace{\baselineskip}2a\textsuperscript{2} + 8a = 42\\ \\ \vspace{\baselineskip}2 (a\textsuperscript{2} + 4a) = 42\\

\\ \\ \vspace{\baselineskip} (a\textsuperscript{2} + 4a) = 21\\ \\ \vspace{\baselineskip}$ \Rightarrow $ a\textsuperscript{2} + 4a -21 = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ (a + 7) (a -3) = 0\\ \\ \vspace{\baselineskip}$ \therefore $ a = -7 \ or\ 3\\ \vspace{\baselineskip} Sum of -7 and 3 = -4

Question:38

Fill in the blanks
If A is a matrix of order 3 × 3, then |3A| = ___.

Answer:

If A is a matrix of order 3 \times 3, then \vert 3A \vert = {27 \vert A \vert }.\\we know:if A = [a\textsubscript{ij}]_{3 \times 3}, then \vert k.A \vert = k\textsuperscript{3} \vert A \vert \\ \\ \vspace{\baselineskip}\therefore $ $ \vert $ 3A$ \vert $ = 3\textsuperscript{3} $ \vert $ A$ \vert $ = 27 $ \vert $ A$ \vert $ \\

Question:39

Fill in the blanks
If A is invertible matrix of order 3 × 3, then |A^{-1}|= ____.

Answer:

If A is invertible matrix of order 3 $ \times $ 3, then $ \vert $ A\textsuperscript{-1}$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{-1}\\GivenA = invertible matrix 3x3
\\ \vspace{\baselineskip} AA\textsuperscript{-1} = I.\\ \vspace{\baselineskip} \Rightarrow \vert A \vert \vert A\textsuperscript{-1} \vert = 1\\ \\ \therefore\left|A^{-1}\right|=\frac{1}{|A|} \\

Question:40

Fill in the blanks
If x, y, z ∈ R, then the value of determinant \left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| is equal to ___.

Answer:

\\ \begin{aligned} &\text { Given }\\ &\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\\ &=\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Apply Formula: }(a+b)^{2}-(a-b)^{2}=4 a b \text { . } \end{aligned}
\\ \begin{aligned} &=\left|\begin{array}{lll} 4\left(2^{x}\right)\left(2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4\left(3^{x}\right)\left(3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4\left(4^{x}\right)\left(4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} 4 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Column } 1 \text { and } 3 \text { thus become proportional }\\ &\therefore\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0 \end{aligned}

Question:41

Fill in the blanks

If cos 2θ = 0, then \left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=

Answer:

We know

\\ \vspace{\baselineskip} cos 2$ \theta $ = 0\\ \\ \vspace{\baselineskip}$ \Rightarrow $ cos 2$ \theta $ = cos $ \pi $ /2\\ \\ \vspace{\baselineskip}$ \Rightarrow $ 2$ \theta $ = $ \pi $ /2\\ \\ \vspace{\baselineskip}$ \therefore $ $ \theta $ = $ \pi $ /4\\ \\ \begin{aligned} &\therefore \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\\ &\text { Then }\\ &\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\left|\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right|^{2}\\ &\Rightarrow\left[0-\frac{1}{\sqrt{2}}\left(\frac{1}{2}\right)+\frac{1}{\sqrt{2}}\left(-\frac{1}{2}\right)\right]^{2}=\left[\frac{-2}{2 \sqrt{2}}\right]^{2}=\left[-\frac{1}{\sqrt{2}}\right]^{2}\\ &\therefore\left|\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right|^{2}=\frac{1}{2} \end{aligned}

Question:42

Fill in the blanks
If A is a matrix of order 3 × 3, then (A^2)^{-1 }= ____.

Answer:

For matrix A is of order 3X3
\begin{aligned} \left(A^{2}\right)^{-1} &=(A \cdot A)^{-1} \\ &=A^{-1} \cdot A^{-1} \\ &=\left(A^{-1}\right)^{2} \end{aligned}

Question:43

If A is a matrix of order 3 × 3, then number of minors in determinant of A are

Answer:

If matrix A is of order 3X3 then
Number of Minors of IAI = 9 as there are 9 elements in a 3x3 matrix

Question:44

Fill in the blanks
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ___.

Answer:

If, A=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|

\\ \vspace{\baselineskip}then $ \vert $ A$ \vert $ = a\textsubscript{11}C\textsubscript{11} + a\textsubscript{12}C\textsubscript{12} + a\textsubscript{13}C\textsubscript{13}\\ \vspace{\baselineskip} We know that the determinant is equal to sum of corresponding co factors of any row or column

Question:45

Fill in the blanks
If x = -9 is a root of \left|\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|=0 , then other two roots are ___.

Answer:

\\ \begin{aligned} &\text { We know that }\\ &x=-9 \text { is a root of }\left|\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|=0\\ &\Rightarrow \mathrm{x}\left|\begin{array}{ll} \mathrm{x} & 2 \\ 6 & \mathrm{x} \end{array}\right|-3\left|\begin{array}{ll} 2 & 2 \\ 7 & \mathrm{x} \end{array}\right|+7\left|\begin{array}{ll} 2 & \mathrm{x} \\ 7 & 6 \end{array}\right|=0 \end{aligned}
\\ $ \Rightarrow $ x [x\textsuperscript{2} -12] -3 [2x -14] + 7 [12 -7x] = 0\\ \\ $ \Rightarrow $ x\textsuperscript{3} -12x -6x + 42 + 84 -49x = 0\\ \\ $ \Rightarrow $ x\textsuperscript{3} -67x + 126 =0\\(x+9)(2-x)(7-x)=0 \\ Here, 126 $ \times $ 1 = 9 $ \times $ 2 $ \times $ 7\\ \\ For x = 2,\\ \\ $ \Rightarrow $ 2\textsuperscript{3} -67 (2) + 126 = 134 -134 = 0\\ \\ $ \therefore $ x = 2 \text{is one root}.\\ \\ For x = 7,\\ \\ $ \Rightarrow $ 7\textsuperscript{3} -67 (7) + 126 = 469 -469 = 0\\ \\ $ \therefore $ x = 7 \text{will be another root}.\\

Question:46

Fill in the blanks

\left|\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right|=

Answer:

\\ Matrix is: \left|\begin{array}{ccc}0 & \text { xyz } & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|$ \\Apply, $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3}\\$ $\Rightarrow\left|\begin{array}{ccc}0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|=\left|\begin{array}{ccc}z-x & x y z & x-z \\ z-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|$\\ Now take (z-x) common from first row \\$\Rightarrow\left|\begin{array}{ccc}z-x & x y z & x-z \\ z-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|=(z-x)\left|\begin{array}{ccc}1 & x y z & x-z \\ 1 & 0 & y-z \\ 1 & z-y & 0\end{array}\right|$ (z-x)\left[\begin{array}{cc} \left.1\left|\begin{array}{cc} 0 & \mathrm{y}-\mathrm{z} \\ \mathrm{z}-\mathrm{y} & 0 \end{array}\right|-(\mathrm{xyz})\left|\begin{array}{cc} 1 & \mathrm{y}-\mathrm{z} \\ 1 & 0 \end{array}\right|+(\mathrm{x}-\mathrm{z}) \mid \begin{array}{cc} 1 & 0 \\ 1 & \mathrm{z}-\mathrm{y} \end{array} \|\right] \end{array}\right.
\\ \vspace{\baselineskip}= (z -x) [1 [0 -(y -z) (z -y)] - (xyz) [0 - (y - z)] + (x -z) [(z -y) -0]]\\ \\ \vspace{\baselineskip}= (z -x) (z -y) (-y + z -xyz + x -z)\\ \\ \vspace{\baselineskip}= (z -x) (z -y) (x -y -xyz)\\ \\ \vspace{\baselineskip}= (z -x) (y -z) (y -x + xyz)\\ \vspace{\baselineskip}

Question:47

If f(x)=\left|\begin{array}{lll} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{array}\right|=A+B x+C x^{2}+\ldots, then A = ____.

Answer:

\\ Given f(x)=\left|\begin{array}{lll}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right|\\\Rightarrow \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{17}(1+\mathrm{x})^{23}(1+\mathrm{x})^{41}\left|\begin{array}{lll}1 & (1+\mathrm{x})^{2} & (1+\mathrm{x})^{6} \\ 1 & (1+\mathrm{x})^{6} & (1+\mathrm{x})^{11} \\ 1 & (1+\mathrm{x})^{2} & (1+\mathrm{x})^{6}\end{array}\right|$\\ We can see that row 1 and 3 are identical \\$\therefore \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{17}(1+\mathrm{x})^{23}(1+\mathrm{x})^{41}\left|\begin{array}{lll}1 & (1+\mathrm{x})^{2} & (1+\mathrm{x})^{6} \\ 1 & (1+\mathrm{x})^{6} & (1+\mathrm{x})^{11} \\ 1 & (1+\mathrm{x})^{2} & (1+\mathrm{x})^{6}\end{array}\right|=0$ $$ \therefore \mathrm{A}=0

Question:48

State True or False for the statements
(A^3)^{-1} = (A^{-1})^3, where A is a square matrix and |A| ≠ 0.

Answer:

\\ \vspace{\baselineskip} (A\textsuperscript{3})\textsuperscript{-1} = (A\textsuperscript{-1})\textsuperscript{3}\\ Because, (A\textsuperscript{n})\textsuperscript{-1} = (A\textsuperscript{-1})\textsuperscript{n}, where n $ \in $ N.\\

Question:49

State True or False for the statements
(aA)^{-1} = (1/a) A^{-1}, where a is any real number and A is a square matrix.

Answer:

For a non singular matrix, aA is invertible such that
\\ (\mathrm{aA})\left(\frac{1}{\mathrm{a}} \mathrm{A}^{-1}\right)=\left(\mathrm{a} \cdot \frac{1}{\mathrm{a}}\right)\left(\mathrm{AA}^{-1}\right) \\ \text {i.e. } \quad(\mathrm{aA})^{-1}=\frac{1}{\mathrm{a}} \mathrm{A}^{-1} \\
here a = any non-zero scalar. Here A should be a non-singular matrix which is not given in the statement, thus the statement given in question in false.

Question:50

State True or False for the statements
|A^{-1}| \neq |A|^{-1}, where A is non-singular matrix.

Answer:

We know A is a non singular Matrix

In that case: AA\textsuperscript{-1} = I.\\ \\

\\\vspace{\baselineskip}$ \Rightarrow $ $ \vert $ A$ \vert $ $ \vert $ A\textsuperscript{-1}$ \vert $ = 1\\ \\ \vspace{\baselineskip}$ \therefore $ $ \vert $ A\textsuperscript{-1}$ \vert $ = 1/$ \vert $ A$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{-1}\\ \vspace{\baselineskip} Thus the statement is false.

Question:51

State True or False for the statements
If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.

Answer:

We know that:
\\ \vspace{\baselineskip} \vert AB \vert = \vert A \vert . \vert B \vert and if A = [a\textsubscript{ij}]_{3 \times 3}, then \vert k.A \vert = k\textsuperscript{3} \vert A \vert .

\\ \\ \\ \vspace{\baselineskip}$ \therefore $ $ \vert $ 3A$ \vert $ = 27 $ \vert $ AB$ \vert $ \\ \\ \vspace{\baselineskip}= 27 $ \vert $ A$ \vert $ $ \vert $ B$ \vert $ \\ \\ \vspace{\baselineskip}= 27 (5) (3)\\ \\ \vspace{\baselineskip}= 405\\ \vspace{\baselineskip} Hence the statement given in question is true.

Question:52

State True or False for the statements
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.

Answer:

\\ \vspace{\baselineskip} Given: $ \therefore $ $ \vert $ A$ \vert $ = 12\\ \vspace{\baselineskip}For any square matrix of order n, adj A$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{n-1}\\ \\

\\ \vspace{\baselineskip}For n = 3, $ \vert $ adj A$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{3-1}\\ \\ \vspace{\baselineskip}= $ \vert $ A$ \vert $ \textsuperscript{2}\\ \\ \vspace{\baselineskip}= 12\textsuperscript{2} = 144\\ \vspace{\baselineskip} Thus the given statement is true.

Question:53

State True or False for the statements
\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0, where a, b, c are in A.P.

Answer:

\\ \begin{aligned} &\text { since } a, b, c \text { are in } A P, 2 b=a+c .\\ &\therefore\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0\\ &A p p l y_{i}, R_{1} \rightarrow R_{1}+R_{3}\\ &\Rightarrow\left|\begin{array}{ccc} 2 x+4 & 2 x+6 & 2 x+a+c \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0 \end{aligned}
Since 2b = a + c,
\Rightarrow\left|\begin{array}{ccc} 2(x+2) & 2(x+3) & 2(x+b) \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0
We can see that Row 1 and 3 are proportional

Thus determinant = 0

Question:54

State True or False for the statements
|adj. A| = |A|^2, where A is a square matrix of order two.

Answer:

For any square matrix of order n, adj A$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{n-1}\\Here n =2,\\ \\ $ \Rightarrow $ $ \vert $ adj A$ \vert $ = $ \vert $ A$ \vert $ \textsuperscript{n-1} = $ \vert $ A$ \vert $ \\ Thus the statement given in question is false

Question:55

State True or False for the statements

The determinant \left|\begin{array}{ccc} \sin A & \cos A & \sin A+\cos A \\ \sin B & \cos B & \sin B+\cos B \\ \sin C & \cos C & \sin C+\cos C \end{array}\right| is equal to zero.
Answer:

\left|\begin{array}{ccc} \sin A & \cos A & \sin A+\cos A \\ \sin B & \cos B & \sin B+\cos B \\ \sin C & \cos C & \sin C+\cos C \end{array}\right|
\\=\left|\begin{array}{ccc} \sin A & \cos A & \sin A \\ \sin B & \cos B & \sin B \\ \sin C & \cos C & \sin C \end{array}\right|+\left|\begin{array}{ccc} \sin A & \cos A & \cos A \\ \sin B & \cos B & \cos B \\ \sin C & \cos C & \cos C \end{array}\right|
We can see that columns are identical in both the matrix on Right hand side

Thus Determinant = 0

Statement in question is therefore true

Question:56

State True or False for the statements

If the determinant \left|\begin{array}{ccc} x+a & p+u & 1+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right| splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8

Answer:

\\ Given \left|\begin{array}{ccc}x+a & p+u & 1+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$ \\Split row 1\\ $\Rightarrow\left|\begin{array}{ccc}\mathrm{x}+\mathrm{a} & \mathrm{p}+\mathrm{u} & 1+\mathrm{f} \\ \mathrm{y}+\mathrm{b} & \mathrm{q}+\mathrm{v} & \mathrm{m}+\mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h}\end{array}\right|=\left|\begin{array}{ccc}\mathrm{x} & \mathrm{p} & 1 \\ \mathrm{y}+\mathrm{b} & \mathrm{q}+\mathrm{v} & \mathrm{m}+\mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h}\end{array}\right|+\left|\begin{array}{ccc}\mathrm{a} & \mathrm{u} & \mathrm{f} \\ \mathrm{y}+\mathrm{b} & \mathrm{q}+\mathrm{v} & \mathrm{m}+\mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h}\end{array}\right|$ Split row 2
\\ \begin{array}{ccc} \left|\begin{array}{ccc} \mathrm{x} & \mathrm{p} & 1 \\ \mathrm{y} & \mathrm{q} & \mathrm{m} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right|+\left|\begin{array}{ccc} \mathrm{a} & \mathrm{u} & \mathrm{f} \\ \mathrm{y} & \mathrm{q} & \mathrm{m} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right|+\left|\begin{array}{ccc} \mathrm{x} & \mathrm{p} & \mathrm{l} \\ \mathrm{b} & \mathrm{v} & \mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right| \\ & +\left|\begin{array}{ccc} \mathrm{a} & \mathrm{u} & \mathrm{f} \\ \mathrm{b} & \mathrm{v} & \mathrm{g} \\ \mathrm{z}+\mathrm{c} & \mathrm{r}+\mathrm{w} & \mathrm{n}+\mathrm{h} \end{array}\right| \end{array}
We can split all the rows in the same way. Thus the statement given in the question is true.

Question:57

State True or False for the statements

Let \Delta=\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|=16 ,then \Delta_{1}=\left|\begin{array}{lll} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{array}\right|=32

Answer:

\\ We \: \: have\\ \Delta=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=16$ \\We need to prove\\ $\Delta_{1}=\left|\begin{array}{ccc}p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r\end{array}\right|=32 .$C_{1} \rightarrow C_{1}+C_{2}+C_{3} \left|\begin{array}{ccc} 2(p+x+a) & a+x & a+p \\ 2(q+y+b) & b+y & b+q \\ 2(r+z+c) & c+z & c+r \end{array}\right| =32
2 can be taken common from Column 1
2\left|\begin{array}{ccc} (p+x+a) & a+x & a+p \\ (q+y+b) & b+y & b+q \\ (r+z+c) & c+z & c+r \end{array}\right| =32
After that apply C1 → C1 – C2 and C2→ C2 – C3
\\ \begin{aligned} &\left|\begin{array}{lll} p & x-p & a+p \\ q & y-q & b+q \\ r & z-r & c+r \end{array}\right|=16\\ &\left|\begin{array}{lll} p & x & a+p \\ q & y & b+q \\ r & z & c+r \end{array}\right|-\left|\begin{array}{lll} p & p & a+p \\ q & q & b+q \\ r & r & c+r \end{array}\right|=16\\ &\text { Second determinants of column } 2 \text { and } 3 \text { are identical }\\ &\left|\begin{array}{lll} p & x & a+p \\ q & y & b+q \\ r & z & c+r \end{array}\right|-0=16 \end{aligned}
\\ \left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|+\left|\begin{array}{lll} p & x & p \\ q & y & q \\ r & z & r \end{array}\right|=16
Again, second determinant of column 1 and 3 are identical
\\ \left|\begin{array}{lll} p & x & a \\ q & y & b \\ r & z & c \end{array}\right|=16
Hence the statement given in question is true

Question:58

State True or False for the statements

The maximum value of \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right| is 1/2.

Answer:

\\ \begin{array}{l} \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cos \theta \end{array}\right| \\ \text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\ \Rightarrow \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ 0 & 0 & \cos \theta \end{array}\right| \end{array}
\\ \vspace{\baselineskip}= cos $ \theta $ . sin $ \theta $ \\ \\ \vspace{\baselineskip}\text{Multiply and divide by 2},\\ \vspace{\baselineskip} = 1/2 (2sin $ \theta $ cos $ \theta $ )\\ \vspace{\baselineskip} \text{We know}, 2 sin $ \theta $ cos $ \theta $ = sin 2$ \theta $ \\ \\ \vspace{\baselineskip}= 1/2 (sin 2$ \theta $ )\\ \\Since the maximum value of sin 2$ \theta $ is 1, $ \theta $ = 45$ ^{\circ} $ .\\

\\ \\ \vspace{\baselineskip}$ \therefore $ $ \Delta $ = 1/2 (sin 2(45$ ^{\circ} $ ))\\ \\ \vspace{\baselineskip}= 1/2 sin 90$ ^{\circ} $ \\ \\ \vspace{\baselineskip}= 1/2 (1)\\ \\ \vspace{\baselineskip}$ \therefore $ $ \Delta $ = 1/2\\ \vspace{\baselineskip} Thus the given statement is true.

More About NCERT Exemplar Solutions for Class 12 Maths Chapter 4

Determinants are related to the matrices that are solved in chapter 3 of 12 Class NCERT Maths book. Studying determinants is not just about passing exams, but is about being prepared for higher education in any field of maths, science, economics, etc. In Class 12 Maths NCERT exemplar solutions chapter 4, the students will learn about determinants, their elements, and how to calculate determinants of various square matrices.

NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants Sub-topics covered

The sub-topics that are covered in this chapter of NCERT Class 12 solution are:

  • Introduction
  • Determinant
  • Determinant of a matrix of order one
  • Determinant of a matrix of order 2
  • Determinant of a matrix of order 3x3
  • Properties of determinants
  • Area of triangle
  • Minors and co-factors
  • Adjoint and inverse of a matrix
  • Adjoint of a matrix
  • Applications of matrices and determinants
  • Solution of a system of linear equations using the inverse of matrices

What will Students Learn in NCERT Exemplar Class 12 Maths Solutions Chapter 4?

  • The determinants are a value of scalar nature derived from the square matrix-like 2x2, 3x3, etc. This determinant then helps in understanding the linear equation defined by the matrix.
  • Determinants derived help in determining whether the solution of the linear system is unique or not.
  • There are many other related topics covered in this Class 12 Maths NCERT exemplar solutions chapter 4, like the area of triangle, co-factors, minors, inverse of a matrix, finding the equation of line, etc.
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NCERT Exemplar Class 12 Maths Solutions

Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 4

  • We will help in finding the NCERT exemplar Class 12 Maths chapter 4 solutions to the questions that are given in the NCERT book in detail. We don't skip steps or try to solve the questions in a trick or the easy way; instead, our expert guides will help in solving every question in the minutest detail kept in mind.
  • The language used for solving the questions is simple and easy to understand. Not only the questions are solved exhaustively; we also keep a marking scheme in mind to help our students score better.
  • NCERT Exemplar Class 12 Maths solutions chapter 4 will help in gaining confidence in solving more difficult questions. Detailed steps in easy language will help the students to understand the topics and its intricacies.
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Frequently Asked Question (FAQs)

1. Can I download the solutions for this chapter?

Yes, you download NCERT exemplar Class 12 Maths solutions chapter 4 pdf by using the webpage to pdf tool available online.

2. What are the important topics of this chapter?

The Properties of Determinants, Adjoint and Inverse of a Matrix and Application of Determinants and Matrices are the more important topics among others as per their weightage.

3. How to study well for boards?

Practice, Practice and Practice. Once you have read the chapters well and made notes, you must practice being fast and precise with answers.

4. How many questions are there in this chapter?

The NCERT exemplar solutions for Class 12 Maths chapter 4 has one exercise with 58 questions for practice.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

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Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

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2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

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12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

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\; K\;
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zero\;

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K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

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increase two fold
Option 3)

remain unchanged

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be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

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Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
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half that in 8 g He

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558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

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more than 9
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