NCERT Exemplar Class 12 Maths Solutions Chapter 4 Determinants

Question:1

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

Answer:

$
\left|\begin{array}{cc}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{array}\right|
$
If $A=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$
The value of the determinant of A can then be found by:

$
\begin{aligned}
& |A|=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c \\
& =\left(x^2-x+1\right) \times(x+1)-(x+1) \times(x-1) \\
& =x\left(x^2-x+1\right)+1\left(x^2-x+1\right)-\left(x^2-1\right) \\
& \quad\left(\text { Since }(a-b)(a+b)=\left(a^2-b^2\right)\right) \\
& =x^3-x^2+x+x^2-x+1-x^2+1 \\
& =x^3-x^2+2
\end{aligned}
$

Question:2

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

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 evaluation:
$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$

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 evaluation:
$\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|$

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 - $C_1 \rightarrow C_1+C_2+C_3$

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

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

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

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

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

Now, expand the determinant along Column 1

$\begin{aligned} & =(x+y+z)[1 \times\{(3 y)(2 z+x)-(-3 z)(x-y)\}] \\ & =(x+y+z)[6 y z+3 y x+(3 z)(x-y)] \\ & =(x+y+z)[6 y z+3 y x+3 z x-3 z y] \\ & =(x+y+z)[3 y z+3 z x+3 y x] \\ & =3(x+y+z)(y z+z x+y x)\end{aligned}$

Question:5

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

Answer:

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

$\begin{aligned} & =(x+y+z)[1 \times\{(3 y)(2 z+x)-(-3 z)(x-y)\}] \\ & =(x+y+z)[6 y z+3 y x+(3 z)(x-y)] \\ & =(x+y+z)[6 y z+3 y x+3 z x-3 z y] \\ & =(x+y+z)[3 y z+3 z x+3 y x] \\ & =3(x+y+z)(y z+z x+y x)\end{aligned}$

Question:6

Using the properties of determinants in evaluation:
$\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$

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

$\begin{aligned} & =(a+b+c)[1 \times\{-(a+b+c) \times\{-(a+b+c)\}-0\} \\ & =(a+b+c)\left[(a+b+c)^2\right] \\ & =(a+b+c)^3\end{aligned}$

Question:7

Using the properties of determinants to 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 of 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 to 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,

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

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

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

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

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

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

$\begin{aligned} & =2 x y z[(1)\{(1)-0\}-(1)\{0-1\}+0\}] \\ & =2 x y z[1+1] \\ & =4 x y z \\ & =\text { RHS } \\ & \therefore \text { LHS }=\text { RHS }\end{aligned}$

Hence Proved

Question:9

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

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|
$
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|
$

$
=\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]
$

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|
$

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}\right|
$

$
\begin{aligned}
& =(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 $R_2$

$
=(a-1)^2\left|\begin{array}{ccc}
a+1 & 1 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right|
$

Expand along $\mathrm{C}_3$

$
\begin{aligned}
& =(a-1)^2[1\{(a+1)-2\}] \\
& =(a-1)^2[a+1-2] \\
& =(a-1)^3 \\
& =\text { RHS }
\end{aligned}
$
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:

Let the determinant be:

$
D=\left|\begin{array}{ccc}
1 & \cos C & \cos B \\
\cos C & 1 & \cos A \\
\cos B & \cos A & 1
\end{array}\right|
$
Expand the determinant:

$
D=1-\cos ^2 A-\cos ^2 B-\cos ^2 C+2 \cos A \cos B \cos C
$
Use identity:

$
\cos ^2 A+\cos ^2 B+\cos ^2 C=1+2 \cos A \cos B \cos C
$
So:

$
D=1-(1+2 \cos A \cos B \cos C)+2 \cos A \cos B \cos C=0
$
Hence, proved.

$
D=0
$

Question:11

If the coordinates 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:

The 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

$\Rightarrow\left(\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3}\end{array}\right|\right)^{2}=\left(\frac{\sqrt{3}}{4} \mathrm{a}^{2}\right)^{2}$

$\\\begin{aligned} &\Rightarrow\left|\begin{array}{lll} \mathrm{x}_{1} & \mathrm{x}_{2} & \mathrm{x}_{3} \\ \mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} \\ \mathrm{z}_{1} & \mathrm{z}_{2} & \mathrm{z}_{3} \end{array}\right|^{2}=\frac{3}{4} \mathrm{a}^{4}\\ &\text { Hence Proved } \end{aligned}$

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{aligned}
& |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{ll}
\mathrm{a}_{22} & \mathrm{a}_{23} \\
\mathrm{a}_{32} & a_{33}
\end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{ll}
\mathrm{a}_{21} & a_{23} \\
\mathrm{a}_{31} & 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}{ccc}
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|
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow(1)\{-6-\{(-7) \cos 2 \theta\}\}-1\{8-7 \cos 2 \theta\}+\sin 3 \theta\{28-21\}=0 \\
& \Rightarrow-6+7 \cos 2 \theta-8+7 \cos 2 \theta+7 \sin 3 \theta=0 \\
& \Rightarrow 14 \cos 2 \theta+7 \sin 3 \theta-14=0 \\
& \Rightarrow 2 \cos 2 \theta+\sin 3 \theta-2=0
\end{aligned}
$
We know,

$
\begin{aligned}
& \cos 2 \theta=1-2 \sin ^2 \theta \\
& \sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta \\
& \Rightarrow 2\left(1-2 \sin ^2 \theta\right)+\left(3 \sin \theta-4 \sin ^3 \theta\right)-2=0 \\
& \Rightarrow 2-4 \sin ^2 \theta+3 \sin \theta-4 \sin ^3 \theta-2=0 \\
& \Rightarrow-2+4 \sin ^2 \theta-3 \sin \theta+4 \sin ^3 \theta+2=0 \\
& \Rightarrow \sin \theta\left(4 \sin \theta-3+4 \sin ^2 \theta\right)=0
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \sin \theta\left(4 \sin ^2 \theta-6 \sin \theta+2 \sin \theta-3\right)=0 \\
& \Rightarrow \sin \theta[2 \sin \theta(2 \sin \theta-3)+1(2 \sin \theta-3)]=0 \\
& \Rightarrow \sin \theta(2 \sin \theta+1)(2 \sin \theta-3)=0 \\
& \Rightarrow \sin \theta=0 \text { or } 2 \sin \theta+1=0 \text { or } 2 \sin \theta-3=0 \\
& \Rightarrow \theta=n \pi \text { or } 2 \sin \theta=-1 \text { or } 2 \sin \theta=3 \\
& \Rightarrow \text { 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{aligned}
$
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

$
\Rightarrow(12+x)[(1)\{0-(2 x)(-2 x)\}]=0
$
$
\Rightarrow(12+x)\left(4 x^2\right)=0
$
$
\Rightarrow 12+x=0 \text { or } 4 x^2=0
$
$
\Rightarrow x=-12 \text { or } x=0
$
Hence, the value of $\mathrm{x}=-12$ 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:

$a_1, a_2, \ldots, a_r$ are in G.P
We know that, $a_{r+1}=A R^{(r+1)-1}=A R^r \ldots(i)$
[ $\because a_n=a r^{n-1}$, where $a=$ first term and $r=$ common ratio]
A is the first term of G.P
R is the common ratio of G.P.

$
\therefore\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|=\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[\text { from (i) }]
$
Taking $A R^r, A R^{r+6}$ and $A R^{r+10}$ common from $R_1, R_2$ and $R_3$

$
=\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 of 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}{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|=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 the 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 and 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

$
\begin{aligned}
& \Rightarrow(\cos B-\cos A)(\cos C-\cos A)[(1)\{\cos C+\cos A+1-(\cos B+\cos A+1)\}]=0 \\
& \Rightarrow(\cos B-\cos A)(\cos C-\cos A)[\cos C+\cos A+1-\cos B-\cos A-1]=0 \\
& \Rightarrow(\cos B-\cos A)(\cos C-\cos A)(\cos C-\cos B)=0 \\
& \Rightarrow \cos B-\cos A=0 \backslash \text { or } \cos C-\cos A=0 \backslash \text { or } \cos C-\cos B=0 \\
& \Rightarrow \cos B=\cos A \text { or } \cos C=\cos A \text { or } \cos C=\cos B \\
& \Rightarrow B=A \text { or } C=A \text { or } C=B
\end{aligned}
$
$
\text { Hence, } \triangle A B C \text { 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:

To find $\operatorname{adj} \mathrm{A}$

$
\begin{aligned}
& a_{11}=\left|\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right|=0-1=-1 \\
& a_{12}=-\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|=-(0-1)=1 \\
& a_{13}=\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|=1-0=1 \\
& a_{21}=-\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|=-(0-1)=1 \\
& a_{22}=\left|\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right|=0-1=-1 \\
& a_{23}=-\left|\begin{array}{ll}
0 & 1 \\
1 & 1
\end{array}\right|=-(0-1)=1 \\
& \mathrm{a}_{31}=\left|\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right|=1-0=1 \\
& \mathrm{a}_{32}=-\left|\begin{array}{ll}
0 & 1 \\
1 & 1
\end{array}\right|=-(0-1)=1 \\
& \therefore \text { adjA }=\left[\begin{array}{lll}
\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{aligned}
$
$
\therefore A^{-1}=\frac{\operatorname{adj} A}{|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}
$

Now, we need to prove that.t

$
\begin{aligned}
& 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\} \\
& =\frac{1}{2}\left\{\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]-\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\right\} \\
& 2-3 \\
& =\frac{1}{2}\left\{\left[\begin{array}{cc}
1 \\
1 & 2-3 \\
1 & 1 \\
1 & 1
\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{aligned}
$
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:

Find IAI Expand IAI along Column 1

$
\begin{aligned}
& |\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}{cc}
\mathrm{a}_{12} & \mathrm{a}_{13} \\
a_{32} & a_{33}
\end{array}\right|+\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc}
\mathrm{a}_{12} & \mathrm{a}_{13} \\
\mathrm{a}_{22} & a_{23}
\end{array}\right| \\
& \left.|\mathrm{A}|=(1)\left|\begin{array}{cc}
0 & 0 \\
-1 & -2 \\
-1 & 1
\end{array}\right|-(-2)|+0| \begin{array}{cc}
2 & 0 \\
-1 & -2
\end{array} \right\rvert\, \\
& =(-1+2)+2(0)+0 \\
& =1
\end{aligned}
$
To find adj $A$

$
\begin{aligned}
& 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 \\
& 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}{cc}
1 & 2 \\
-2 & -1
\end{array}\right|=-1+4=3
\end{aligned}
$

$
\begin{aligned}
& \therefore \operatorname{adj} 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]=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right]=\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{aligned}
$
According to the linear equation:

$
x-2 y=10
$
$
2 x-y-z=8
$
$
-2 y+z=7
$
We know that, $\mathrm{AX}=\mathrm{B}$

$
A=\left[\begin{array}{ccc}
1 & -2 & 0 \\
2 & -1 & -1 \\
0 & -2 & 1
\end{array}\right]
$
Here,
So, transpose of $A^{-1}$

$
\begin{aligned}
& A^{-1}=\left[\begin{array}{lll}
-3 & 2 & 2 \\
-2 & 1 & 1 \\
-4 & 2 & 3
\end{array}\right] \\
& \Rightarrow X=A^{-1} B
\end{aligned}
$
$\begin{aligned} & \Rightarrow\left[\begin{array}{l}x \\ y \\ z \\ x \\ y \\ y\end{array}\right]=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]\left[\begin{array}{l}10 \\ 8 \\ -30+16+14 \\ -20+8+7 \\ -40+16+21\end{array}\right] \\ & \therefore x=0, y=-5 \text { and } z=-3\end{aligned}$

Question:19

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

Answer:

Given system:

$
\begin{array}{r}
3 x+2 y-2 z=3 \\
x+2 y+3 z=6 \\
2 x-y+z=2
\end{array}
$
Matrix form:

$
\left[\begin{array}{ccc}
3 & 2 & -2 \\
1 & 2 & 3 \\
2 & -1 & 1
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
3 \\
6 \\
2
\end{array}\right]
$
The inverse of the coefficient matrix:

$
A^{-1}=\left[\begin{array}{ccc}
\frac{7}{17} & \frac{8}{17} & -\frac{4}{17} \\
\frac{1}{17} & \frac{5}{17} & \frac{6}{17} \\
\frac{5}{17} & -\frac{2}{17} & \frac{3}{17}
\end{array}\right]
$

Multiply:

$
X=A^{-1} B=\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right]
$
Final Answer:

$
x=1, \quad y=2, \quad 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{aligned}
& 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] \\
& =\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] \\
& B A=6 I \ldots(i)
\end{aligned}
$
Now, the given system of equations is:

$
y+2 z=7,
$
$
x-y=3
$
$
2 x+3 y+4 z=17
$

So,

$
\left[\begin{array}{ccc}
0 & 1 & 2 \\
1 & -1 & 0 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
7 \\
3 \\
17
\end{array}\right]
$
Apply, $\mathrm{R}_1 \leftrightarrow \mathrm{R}_2$

$
\begin{aligned}
& \mathrm{R}_2 \leftrightarrow \mathrm{R}_3 \\
& {\left[\begin{array}{lll}
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]}
\end{aligned}
$
So, $B A=6 I[$ from eg(i) $]$

$
\begin{aligned}
& {\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]}
\end{aligned}
$

$\therefore x=2, y=-1$ and $z=4$

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|
$
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|
$
Take $(a+b+c)$ common from Column 1

$
=(a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|
$
Expand along Column 1

$
\begin{aligned}
& =(a+b+c)\left[(1)\left(b c-a^2\right)-(1)\left(b^2-a c\right)+(1)\left(b a-c^2\right)\right] \\
& =(a+b+c)\left[b c-a^2-b^2+a c+a b-c^2\right] \\
& =(a+b+c)\left[-\left(a^2+b^2+c^2-a b-b c-a c\right)\right] \\
& =-\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]}
\end{aligned}
$

Given that $\Delta=0$

$
\Rightarrow-\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0 $

$\Rightarrow(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0$

$\text {Either }(a+b+c)=0 \text { or }(a-b)^2+(b-c)^2+(c-a)^2=0$

$\text {but it is given that }(a+b+c) \neq 0 \therefore(a-b)^2+(b-c)^2+(c-a)^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:

$\left|\begin{array}{lll}b c-a^2 & c a-b^2 & a b-c^2 \\ c a-b^2 & a b-c^2 & b c-a^2 \\ a b-c^2 & b c-a^2 & c a-b^2\end{array}\right|$ is given.
Apply, $R_1 \rightarrow R_1-R_2$,

$
\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 \\
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 \\
=\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{array}\right|
\end{aligned}
$
Take (a+b+c) common from Column 1

$
=(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|
$

Apply $R_2 \rightarrow R_2-R_3$

$
\begin{aligned}
& =(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

$
=(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|
$

Expand along Column 1

$
\begin{aligned}
& =(a+b+c)^2\left[a b+b c+c a-\left(a^2+b^2+c^2\right)\right]^2 \\
& =(a+b+c)(a+b+c)\left[a b+b c+c a-\left(a^2+b^2+c^2\right)\right]^2
\end{aligned}
$
The determinant is divisible by $(\mathrm{a}+\mathrm{b}+\mathrm{c})$ and the quotient is $(a+b+c)\left[a b+b c+c a-\left(a^2+b^2+c^2\right)\right]^2$

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}{ccc}
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

$
\begin{aligned}
& =x a\{(z a)(y a)-(x c)(x b)\}-(y b)\{(y c)(y a)-(z b)(x b)\}+(z c)\{(y c)(x c)- \\
& (z b)(z a)\} \\
& =x a\left\{a^2 y z-x^2 b c\right\}-y b\left\{y^2 a c-b^2 x z\right\}+z c\left\{c^2 x y-z^2 a b\right\} \\
& =a^3 x y z-x^3 a b c-y^3 a b c+b^3 x y z+c^3 x y z-z^3 a b c \\
& =x y z\left(a^3+b^3+c^3\right)-a b c\left(x^3+y^3+z^3\right)
\end{aligned}
$
Given $x+y+z=0$

$
\begin{aligned}
& \Rightarrow x^3+y^3+z^3=3 x y z=x y z\left(a^3+b^3+c^3\right)-a b c(3 x y z)=x y z\left(a^3+b^3+c^3-3 a b c\right) \\
& =x y z\left|\begin{array}{ccc}
a & b & c \\
c & a & b \\
b & c & a
\end{array}\right|
\end{aligned}
$

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:

$
\begin{aligned}
& \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| \\
& A=\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|
\end{aligned}
$
Then the determinant of A is

$
\begin{aligned}
& |\mathrm{A}|=\left|\begin{array}{ll}
\mathrm{a} & \mathrm{~b} \\
\mathrm{c} & \mathrm{~d}
\end{array}\right|=\mathrm{ad}-\mathrm{bc} \\
& \Rightarrow(2 x)(x)-(5)(8)=(6)(3)-(7)(-2) \Rightarrow 2 x^2-40=18-(-14) \Rightarrow 2 x^2-40=18+14 \Rightarrow x^2-20=9+7 \Rightarrow x^2-20=16 \Rightarrow x^2=16+20 \Rightarrow x^2=36 \Rightarrow x=\sqrt{36} \Rightarrow x= \pm 6
\end{aligned}
$

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}{ccc}
a-b & b+c & a \\
b-c & c+a & b \\
c-a & a+b & c
\end{array}\right|
$

Apply C2 $\rightarrow \mathrm{C} 2+\mathrm{C} 3$

$
=\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|
$
Take $(a+b+c)$ 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|
$
Apply $\mathrm{C}_1 \rightarrow \mathrm{C}_1-\mathrm{C}_3$

$
\begin{aligned}
& =(a+b+c)\left|\begin{array}{ccc}
a-b-a & 1 & a \\
b-c-b & 1 & b \\
c-a-c & 1 & c
\end{array}\right| \\
& =(a+b+c)\left|\begin{array}{lll}
-b & 1 & a \\
-c & 1 & b \\
-a & 1 & c
\end{array}\right|
\end{aligned}
$
Expand along Row 1

$
\begin{aligned}
& =(a+b+c)\left[(-b)\{c-b\}-(1)\left\{-c^2-(-a b)\right\}+a\{-c-(-a)\}\right] \\
& =(a+b+c)\left(-b c+b^2+c^2-a b-a c+a^2\right)
\end{aligned}
$
$\begin{aligned} & =a\left(-b c+b^2+c^2-a b-a c+a^2\right)+b\left(-b c+b^2+c^2-a b-a c+a^2\right)+c(-b c+ \\ & \left.b^2+c^2-a b-a c+a^2\right) \\ & =-a b c+a b^2+a c^2-a^2 b-a^2 c+a^3-b^2 c+b^3+b c^2-a b^2-a b c+a^2 b-b c^2+ \\ & b^2 c+c^3-a b c-a c^2+a^2 c \\ & =a^3+b^3+c^3-3 a b c\end{aligned}$

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

$
\Rightarrow-(k)\{-3-3\}=18
$
$
\Rightarrow-k(-6)=18
$
$
\Rightarrow 6 k=18
$
$
\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 Columns 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 Columns 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

$
\begin{aligned}
& (2 \cos X+\sin X)[(1)\{(\sin X-\cos X)(\sin X-\cos X)\}] \\
& \Rightarrow(2 \cos X+\sin X)(\sin X-\cos X)^2=0 \\
& \Rightarrow 2 \cos X=-\sin X \operatorname{or}(\sin X-\cos X)^2=0 \\
& \Rightarrow 2=-\frac{\sin \mathrm{x}}{\cos \mathrm{x}} \text { or } \sin \mathrm{x}=\cos \mathrm{x} \\
& \Rightarrow \tan x=-2 \text { ortan } x=1\left[\because \tan x=\frac{\sin x}{\cos x}\right] \\
& \text { buttan } x=-2 \text { isnotpossibleasfor }-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}
\end{aligned}
$
$
\text { 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:

$
\left.\left|\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}\right| \right\rvert\,
$
Expand along Column 1

$
\begin{aligned}
& |\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}{ll}
\mathrm{a}_{12} & \mathrm{a}_{13} \\
\mathrm{a}_{22} & \mathrm{a}_{23}
\end{array}\right| \\
& \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}
\cos C & \cos B \\
-1 & \cos A
\end{array}\right| \\
& =\left[(-1)\left\{1-\cos ^2 A\right\}-\cos C\{-\cos C-\cos A \cos B\}+\cos B\{\cos A \cos C+\right. \\
& \cos B\}]
\end{aligned}
$

$
=-1+\cos ^2 A+\cos ^2 C+\cos A \cos B \cos C+\cos A \cos B \cos C+\cos ^2 B
$
$
=-1+\cos ^2 A+\cos ^2 B+\cos ^2 C+2 \cos A \cos B \cos C
$
Using the formula

$
\begin{aligned}
& 1+\cos 2 A=2 \cos ^2 A \\
& =-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}
\end{aligned}
$
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}
$

$
\begin{aligned}
& =\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 \\
& \text { so, } \cos 2 A+\cos 2 B=2 \cos (A+B) \cos (A-B) \\
& =\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 \text { (i) }
\end{aligned}
$

We know that $A, B, C$ are angles of triangle

$
\begin{aligned}
& \Rightarrow A+B+C=\pi \\
& \Rightarrow A+B=\pi-C \\
& =\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}
\end{aligned}
$
$=-\cos C\{\cos (A-B)-\cos C\}+2 \cos A \cos B \cos C=-\cos C[\cos (A-B)-\cos \{\pi-(A+B)\}]+2 \cos A \cos B \cos C=-\cos C[\cos (A-B)+\cos (A+B)]+2 \cos A \cos B \cos C=-\cos C[2 \cos A \cos B]+2 \cos A \cos B \cos C=0$

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:

Given:

$
f(t)=\left|\begin{array}{ccc}
\cos t & t & 1 \\
2 \sin t & t & 2 t \\
\sin t & t & t
\end{array}\right|
$
Divide $\mathrm{R}_2$ and $\mathrm{R}_3$ 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{\sin t}{t} & \frac{t}{t} & \frac{t}{t}
\end{array}\right|
$
$
\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

$
\begin{aligned}
& =(1)(1-2)+(1)(2-1) \\
& =-1+1
\end{aligned}
$
$=0$

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)

We have:

$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1+\cos \theta & 1 & 1
\end{array}\right|
$
Apply, $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3$ 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|
$

$=0-0+1(\sin \theta \cdot \cos \theta)$ Multiply and divide by $2,=\frac{1}{2}(2 \sin \theta \cos \theta)$ We already know that $\underline{2 \sin \theta \cos \theta=\sin 2 \theta}=\frac{1}{2}(\sin 2 \theta)$
The maximum value of $: \sin 2 \theta$ is $1, \theta=45^{\circ}$.

$
\begin{aligned}
& \therefore \Delta=\frac{1}{2}\left(\sin 2\left(45^{\circ}\right)\right)=\frac{1}{2} \sin 90^{\circ}=\frac{1}{2}(1) \\
& \therefore \Delta=\frac{1}{2}
\end{aligned}
$

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|
$

If we put $x=a$

$
\begin{aligned}
& 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| \\
& =0-0+(a-b)[2 a(a+c)-0(a+b)]=(a-b)\left[2 a^2+2 a c-0\right]=(a-b)\left(2 a^2+2 a c\right) \neq 0 \text { If } x=b \\
& 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| \\
& =0-(b-a)[(b+a)(0)-(b-c)(2 b)]+0=-(b-a)\left[0-2 b^2+2 b c\right]=(a-b)\left(2 b^2-2 b c\right) \neq 0
\end{aligned}
$
If $x=0$ according to the given question:

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

$
\begin{aligned}
& =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| \\
& =0+a[a(0)-(-b c)]-b[a c-b(0)]
\end{aligned}
$
$
=a[b c]-b[a c]
$
$
=a b c-a b c=0
$
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
$A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right]$

$
\begin{aligned}
& \Rightarrow|A|=2(6-5)-\lambda(0-5)+(-3)(0-2) \\
& =2+5 \lambda+6 \\
& =5 \lambda+8
\end{aligned}
$

ithe inverse of A exists only if A is non-singular.e. $|A| \neq 0$.

$
\begin{aligned}
& .5 \lambda+8 \neq 0 \\
& \Rightarrow 5 \lambda \neq-8
\end{aligned}
$

$\therefore \lambda \neq-\frac{8}{5}$
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, that A and B are invertible matrices

$
\begin{aligned}
& \text { Consider }(A B) B^{-1} A^{-1} \Rightarrow(A B) B^{-1} A^{-1}=A\left(B B^{-1}\right) A^{-1}=A I A^{-1}=(A I) A^{-1}=A A^{-1}=I \Rightarrow(A B)^{-1}=B^{-1} A^{-1} \ldots \text { option }(C) \\
& \text { Also } A A^{-1}=I \Rightarrow\left|A A^{-1}\right|=|I| \Rightarrow|A|\left|A^{-1}\right|=1 \\
& \Rightarrow|\mathrm{~A}|^{-1}=\frac{1}{|\mathrm{~A}|} \\
& \therefore \operatorname{det}(A)^{-1}=[\operatorname{det}(A)]^{-1} \ldots(B)
\end{aligned}
$
We know that $\frac{|\mathrm{A}|^{-1}}{\operatorname{adj} \mathrm{~A}} \frac{|\vec{A}|}{\text { We }}$
$\Rightarrow \operatorname{adj} A=|A| \cdot A^{-1} \ldots$ option $(A)$

$
\Rightarrow(\mathrm{A}+\mathrm{B})^{-1}=\frac{1}{|\mathrm{~A}+\mathrm{B}|} \operatorname{adj}(\mathrm{A}+\mathrm{B})
$

But $\mathrm{B}^{-1}+\mathrm{A}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}+\frac{1}{|\mathrm{~A}|}$ adj A

$
\therefore(A+B)^{-1} \neq B^{-1}+A^{-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

$
\Rightarrow x[y(1+z)+z]-0+1(y z)=0
$

$
x y+x y z+x z+y z=0
$

Divide both sides 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)

Matrix given:

$
\begin{aligned}
& \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| \\
& =x\left[x^2-(x+y)(x+2 y)\right]-(x+y)\left[(x+2 y)(x)-(x+y)^2\right]+(x+2 y)[(x+ \\
& \left.2 y)^2-x(x+y)\right] \\
& =x\left[x^2-x^2-3 x y-2 y^2\right]-(x+y)\left[x^2+2 x y-x^2-2 x y-y^2\right]+(x+2 y)\left[x^2+\right. \\
& \left.4 x y+4 y^2-x^2-x y\right] \\
& =x\left[-3 x y-2 y^2\right]-(x+y)\left[-y^2\right]+(x+2 y)\left[3 x y+4 y^2\right] \\
& =-3 x^2 y-2 x y^2+x y^2+y^3+3 x^2 y+4 x y^2+6 x y^2+8 y^3 \\
& =9 y^3+9 x y^2 \\
& =9 y^2(x+y)
\end{aligned}
$

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)

We have:

$
\begin{aligned}
& \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 \\
& 1\left[2 a^2-(-4)\right]+2[4 a-0]+5[8-0]=86
\end{aligned}
$

$
1\left[2 a^2+4\right]+2[4 a]+5[8]=86
$

$
2 a^2+4+8 a+40=86
$

$
2 a^2+8 a+44=86
$

$
2 a^2+8 a=42
$

$
2\left(a^2+4 a\right)=42
$

$
\left(a^2+4 a\right)=21 \Rightarrow a^2+4 a-21=0 \Rightarrow(a+7)(a-3)=0 \therefore a=-7 \text { or } 3
$

The 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 $|3 A|=27|A|$.
We know:

$
\begin{aligned}
& \text { if } A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|k \cdot A|=k^3|A| \\
& \therefore|3 A|=3^3|A|=27|A|
\end{aligned}
$

Question:39

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

Answer:

If A is an invertible matrix of order $3 \times 3$, then $\left|A^{-1}\right|=|A|^{-1}$
Given
$\mathrm{A}=$ invertible matrix $3 \times 3$

$
\begin{aligned}
& A A^{-1}=I \Rightarrow|A|\left|A^{-1}\right|=1 \\
& \therefore\left|A^{-1}\right|=\frac{1}{|A|}
\end{aligned}
$

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

$
\cos 2 \theta=0 \Rightarrow \cos 2 \theta=\cos \pi / 2 \Rightarrow 2 \theta=\pi / 2 \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 the number of minors in the determinant of A is

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

We know that the determinant is equal to the 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:

We know that

$
\begin{aligned}
& 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}
$
$\Rightarrow x\left[x^2-12\right]-3[2 x-14]+7[12-7 x]=0 \Rightarrow x^3-12 x-6 x+42+84-49 x=0 \Rightarrow x^3-67 x+126=0(x+9)(2-x)(7-x)=0 H$ ere, $126 \times 1=9 \times 2 \times 7$ Forx $=2, \Rightarrow 2^3-67 \cdot 2+126=134-134=0 \therefore x=2$ is one root. For $=7, \Rightarrow 7^3-677-7+126=469-469=0; x=7$ 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:

$
\begin{aligned}
& =(z-x)[1[0-(y-z)(z-y)]-(x y z)[0-(y-z)]+(x-z)[(z-y)-0]] \\
& =(z-x)(z-y)(-y+z-x y z+x-z) \\
& =(z-x)(z-y)(x-y-x y z) \\
& =(z-x)(y-z)(y-x+x y z)
\end{aligned}
$

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 $f(x)=\left|\begin{array}{ccc}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right| \Rightarrow \mathrm{f}(x)=(1+x)^{17}(1+x)^{23}(1+x)^{41}\left|\begin{array}{ccc}1 & (1+x)^2 & (1+x)^6 \\ 1 & (1+x)^6 & (1+x)^{11} \\ 1 & (1+x)^2 & (1+x)^6\end{array}\right|$ Wecanseethatrow1 androw3areidentical $\therefore \mathrm{f}(x)=(1+x)^{17}(1+x)^{23}$

$
\therefore 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:

$\left(A^3\right)^{-1}=\left(A^{-1}\right)^3$ Because, $\left(A^n\right)^{-1}=\left(A^{-1}\right)^n$, wheren $\in \mathbb{N}$.

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 the question is 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: $A A^{-1}=I$.

$
\Rightarrow|A|\left|A^{-1}\right|=1 \therefore\left|A^{-1}\right|=1 /|A|=|A|^{-1}
$
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:

$
\begin{aligned}
& |A B|=|A| \cdot|B| \text { andif } A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|k . A|=k^3|A| . \\
& \therefore|3 A|=27|A B|=27|A||B|=27 \cdot 5 \cdot 3=405
\end{aligned}
$
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 with its co-factor will be 144.

Answer:

Given $\therefore|A|=12$
For any square matrix of order $\mathrm{n}, \operatorname{adj} A\left|=|A|^{n-1}\right.$

$
\text { Forn }=3,|\operatorname{adj} A|=|A|^{3-1}=|A|^2=12^2=144
$
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 $\mathrm{n}, \operatorname{adj} A\left|=|A|^{n-1}\right.$
Here $\mathrm{n}=2$,

$
\Rightarrow|\operatorname{adj} A|=|A|^{n-1}=|A|
$
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 the right-hand side

Thus Determinant = 0

The 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 $\left|\begin{array}{ccc}x+a & p+u & 1+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$ Splitrow $1 \Rightarrow\left|\begin{array}{ccc}x+a & p+u & 1+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|=\left|\begin{array}{ccc}x & p & 1 \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|+\left|\begin{array}{ccc}a & u & f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right|$
Split row 2

$
\left|\begin{array}{ccc}
x & p & 1 \\
y & q & m \\
z+c & r+w & n+h
\end{array}\right|+\left|\begin{array}{ccc}
a & u & f \\
y & q & m \\
z+c & r+w & n+h
\end{array}\right|+\left|\begin{array}{ccc}
x & p & l \\
b & v & g \\
z+c & r+w & n+h
\end{array}\right|
$
$
+\left|\begin{array}{ccc}
a & u & f \\
b & v & g \\
z+c & r+w & n+h
\end{array}\right|
$
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:

$
\begin{aligned}
& \text { Wehave } \Delta=\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|=16 \text { Weneedtoprove } \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 \text {. } \\
& \mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3 \\
& \left|\begin{array}{lll}
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
\end{aligned}
$
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 $\mathrm{C} 1 \rightarrow \mathrm{C} 1-\mathrm{C} 2$ and $\mathrm{C} 2 \rightarrow \mathrm{C} 2-\mathrm{C} 3$

$
\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}{ccc}
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
\end{aligned}
$
The second determinant of columns 2 and 3 are identical.l

$
\left|\begin{array}{lll}
p & x & a+p \\
q & y & b+q \\
r & z & c+r
\end{array}\right|-0=16
$

$
\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, the second determinant of columns 1 and 3 is identic. al

$
\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:

$
\Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|
$
Apply, $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ 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|
$

$=\cos \theta \cdot \sin \theta$ Multiply and divide by $2,=1 / 2(2 \sin \theta \cos \theta)$ We know, $2 \sin \theta \cos \theta=\sin 2 \theta=1 / 2(\sin 2 \theta)$

Since the maximum value of $\sin 2 \theta$ is $1, \theta=45^{\circ}$.

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