Maths M.C.Q (1 Marks)

3. 10

Solution:

$\text{A(adj A)}\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$

By definition, we have

A(adj A) = |A|I = (adj A)A (Where I is the identity matrix)

⇒ |A|I = A(adj A)

$\Rightarrow|\text{A}|\text{I}=10\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\Rightarrow|\text{A}|=10$

4. None of these.

Solution:

$\triangle=\begin{vmatrix}\text{x}&2&\text{x}\\\text{x}^2&\text{x}&6\\\text{x}&\text{x}&6\end{vmatrix}$

$=\text{x}\begin{vmatrix}\text{x}&6\\\text{x}&6\end{vmatrix}-\text{x}^2\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}+\text{x}\begin{vmatrix}2&\text{x}\\\text{x}&6\end{vmatrix}$

= 0 - x²(12 - x²) + x(12 - x²)

= x⁴ - 12x² + 12x - x³

= ax⁴ + bx³ + cx² + dx + e

⇒ x⁴ - 12x² + 12x - x³ = ax⁴ + bx³ + cx² + dx + e

⇒ a = 1, b = -1, c = -12, d = 12, e = 0

Thus,

5a + 4b + 3c + 2d + e = 5 - 4 - 36 + 24 + 0 = -11

2. 9

Solution:

The minor of the element 2 can be obtained by deleting the first row and the first column

$\therefore\text{M}_{11}=9$

4. −13

Solution:

Given, determinant of the matrix $ \displaystyle \begin{vmatrix} 2 &\text{amp;} -4 \\ 9 &\text{amp; d}-3 \end{vmatrix}=4$

$\Rightarrow2(\text{d}−3)−(9)(−4)=4$

$\Rightarrow2\text{d}-6+36 = 4$

$\Rightarrow 2\text{d}=-26$

$\Rightarrow\text{d} = -13$

4. None of these

Solution:

We have,

$\begin{vmatrix}\text{a}-\text{b}&\text{b}+\text{c}&\text{a}\\\text{b}-\text{a}&\text{c}+\text{a}&\text{b}\\\text{c}-\text{a}&\text{a}+\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{a}+\text{c}&\text{b}+\text{c}+\text{a}&\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{b}\\\text{c}+\text{b}&\text{a}+\text{b}+\text{c}&\text{c}\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2\text{ and C}_2\rightarrow\text{C}_2+\text{C}_3\big]$

$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}+\text{c}&1&\text{a}\\\text{b}+\text{c}&1&\text{b}\\\text{c}+\text{b}&1&\text{c}\end{vmatrix}$ $\big[\text{Taking }(\text{a}+\text{b}+\text{c})\text{ common from C}_2\big]$

$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&0&\text{a}-\text{c}\\0&0&\text{b}-\text{c}\\\text{c}+\text{b}&1&\text{c}\end{vmatrix}$ $\big[\because\ \text{R}_2\rightarrow\text{R}_2-\text{R}_3\text{ and R}_1\rightarrow\text{R}_1-\text{R}_3\big]$

$=(\text{a}+\text{b}+\text{c})\big[-(\text{b}-\text{c}).(\text{a}-\text{b})\big]$ [expanding along R₂]

$=(\text{a}+\text{b}+\text{c})(\text{c}-\text{b})(\text{a}-\text{b})$

4. None of these.

Solution:

$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$

$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$

[on taking (b - a) common from C₁ and C₃ each]

$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{c}&\text{c}\\\text{a}-\text{b}&\text{a}-\text{b}&\text{b}\\\text{c}-\text{a}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_3\big]$

$=0$

[Since, two columns C₁ and C₂ are identical, so the value of determinant is zero]

3. $\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$

Solution:

Determinant of the matrix $\text{A}=\begin{bmatrix}5&8&9\\3&4&6\end{bmatrix}$is not possible as it is a rectangular matrix and not a square matrix.

Determinants can be calculated only if the matrix is a square matrix.

1. $\text{x}=2,-\frac{1}{3}$

Solution:

Given that, $\begin{bmatrix}1&-1\\3&-5\end{bmatrix}=\begin{bmatrix}\text{x}&\text{x}^2\\3&-5\end{bmatrix} -5—(-3)=5\text{x}-3\text{x}^2 $

$-2=5\text{x}-3\text{x}^2$

$3\text{x}^2-5\text{x}-2=0$

Solving for x, we get

$\text{x}=2,-\frac{1}{3}$

4. -2

Solution:

Given that, $\text{A}=\begin{bmatrix}9&8\\7&6\end{bmatrix}$

$\Rightarrow\triangle=\begin{bmatrix}9&8\\7&6\end{bmatrix}$

$=9(6)-7(8)=54-56=-2$

4. 2⁶

Solution:

|A| = d

|adj A| = |A|^n-1

Here, n = 3, |A| = 8

|adj A| = 8²

$|\text{adj A}|={(2^3)}^2=2^6$

2. 12

Solution:

Given, $\left [\begin{matrix} 3&\text{amp; } 6\\ -1 &\text{amp; } 2\end {matrix} \right]$

Let determinent be $ \left| \text{d} \right|$ 

Value of $ \left| \text{d} \right|$ wil be

$\left| \text{d} \right|∣\text{d}∣=3\times 2-\left( 6\times -1 \right)$

$=6+6=12$

3. 1

Solution:

We have,

$\begin{vmatrix}\sin\text{x}&\cos\text{x}&\cos\text{x}\\\cos\text{x}&\sin\text{x}&\cos\text{x}\\\cos\text{x}&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$

$\text{Applying C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3,$

$\Rightarrow\ \begin{vmatrix}2\cos\text{x}+\sin\text{x}&\cos\text{x}&\cos\text{x}\\2\cos\text{x}+\sin\text{x}&\sin\text{x}&\cos\text{x}\\2\cos\text{x}+\sin\text{x}&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$

$\Rightarrow\ (2\cos\text{x}+\sin\text{x})\begin{vmatrix}1&\cos\text{x}&\cos\text{x}\\1&\sin\text{x}&\cos\text{x}\\1&\cos\text{x}&\sin\text{x}\end{vmatrix}=0$

$\big[\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-\text{R}_1\big]$

$\Rightarrow\ (2\cos\text{x}+\sin\text{x}) \begin{vmatrix}1&\cos\text{x}&\cos\text{x}\\0&\sin\text{x}-\cos\text{x}&0\\0&0&\sin\text{x}-\cos\text{x}\end{vmatrix}=0$

$\Rightarrow\ (2\cos\text{x}+\sin\text{x})[1\cdot(\sin\text{x}-\cos\text{x})^2]=0$ (expanding along C₁)

$\Rightarrow\ (2\cos\text{x}+\sin\text{x})(\sin\text{x}-\cos\text{x})^2=0$

$\Rightarrow\ 2\cos\text{x}=-\sin\text{x or }\sin\text{x}=\cos\text{x}$

$\Rightarrow\ \tan\text{x}=-2,$ which is not possible as for $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$

or $\tan\text{x}=1$

$\therefore\ \ \text{x}=\frac{\pi}{4}$

So, only me one real root exist.

2. 0

Solution:

Let $\begin{vmatrix}\text{x}^2+3\text{x}&\text{x}-1&\text{x}+ 3\\\text{x}+1&-2\text{x}&\text{x}-4\\\text{x}-3&\text{x}+4&3\text{x}\end{vmatrix}$

$=(\text{x}^2+3\text{x})\begin{vmatrix}-2\text{x}&\text{x}-4\\\text{x}+4&3\text{x}\end{vmatrix}-(\text{x}-1)\begin{vmatrix}\text{x}+1&\text{x}-4\\\text{x}-3&3\text{x}\end{vmatrix}+(\text{x}+3)\begin{vmatrix}\text{x}+1&-2\text{x}\\\text{x}-3&\text{x}+4\end{vmatrix}$

= (x² + 3x)(-6x - x² + 16) - (x - 1)(3x² + 3x - x² + 7x - 12) + (x + 3)(x² + 5x + 4 + 2x² - 6x)

= -7x⁴ + 16x² + 48x + 21x³ + 8x² - 22x - 2x³ - 12 + 8x² + x + 3x³ + 12

= -7x⁴ + 22x³ + 32x² + 27x + 0

But x is a root of ax⁴ + bx³ + cx² + dx + e

e = 0

3. f(0) = 0

Solution:

Let $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$

Now, $\text{f(a)}=\begin{vmatrix}0&\text{a}-\text{a}&\text{a}-\text{b}\\\text{a}+\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&0&\text{a}-\text{b}\\2\text{a}&0&\text{a}-\text{c}\\\text{a}+\text{b}&\text{a}+\text{c}&0\end{vmatrix}$

$=(\text{a}-\text{b})(2\text{a}^2+2\text{ac})\neq0$

$\text{f(b)}=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}-\text{b}\\\text{b}+\text{a}&0&\text{b}-\text{c}\\\text{b}+\text{b}&\text{b}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&\text{b}-\text{a}&0\\\text{b}+\text{a}&0&\text{b}-\text{c}\\2\text{a}&\text{b}+\text{c}&0\end{vmatrix}$

$=(\text{b}-\text{a})(2\text{ab}-2\text{ac})\neq0$

$\text{f(0)}=\begin{vmatrix}0&\text{0}-\text{a}&\text{0}-\text{b}\\\text{0}+\text{a}&0&\text{0}-\text{c}\\\text{0}+\text{b}&\text{0}+\text{c}&0\end{vmatrix}$

$=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&\text{c}\\\text{b}&\text{c}&0\end{vmatrix}$

$=\text{a}(\text{bc})-\text{b}(\text{ac})=0$

Hence, the correct option is (c)

1. $\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$

Solution:

A = BX

B^-1A = B^-1BX

X = B^-1A

Using adjoint method of inverse

$\text{B}^{-1}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$

$\text{X}=\text{B}^{-1}\text{A}$

$\text{X}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$

$\text{x}=\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$ 

1. 0

Solution:

$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}$

$=\begin{vmatrix}1+\omega^{\text{n}}+\omega^{2\text{n}}&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}+1+\omega^{\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}+\omega^{2\text{n}}+1&\omega^{2\text{n}}&1\end{vmatrix}$ [Applying C₁ → C₁ + C₂ + C₃]

Now, $1+\omega+\omega^2=0$ $[\because$ is a complex cube root of unity$]$ 

$1+\omega^{\text{n}}+\omega^{2\text{n}}=0$ $[\because$ n is not a multiple of 3$]$

$\triangle=\begin{vmatrix}1&\omega^{\text{n}}&\omega^{2\text{n}}\\\omega^{2\text{n}}&1&\omega^{\text{n}}\\\omega^{\text{n}}&\omega^{2\text{n}}&1\end{vmatrix}=0$ 

1. 0

Solution:

We have, $ \begin{vmatrix}-1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&-1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&-1\end{vmatrix}$

$ =\begin{vmatrix}-\text{a}+\text{b}\cos\text{C}+\cos\text{B}&\cos\text{C}&\cos\text{B}\\\text{a}\cos\text{C}-\text{b}+\text{c}\cos\text{A}&-1&\cos\text{A}\\\text{a}\cos\text{B}+\text{b}\cos\text{A}-\text{C}&\cos\text{A}&-1\end{vmatrix}$ $\big[\text{C}_1\rightarrow\text{a C}_1+\text{b C}_2+\text{c C}_3\big]$

We know that, $\text{a}=\text{b}\cos\text{C}+\text{c}\cos\text{B, b}=\text{c}\cos\text{A}+\text{a}\cos\text{C and c}=\text{a}\cos\text{B}+\text{b}\cos\text{A}$

Substituting these values we get

$\begin{bmatrix}-\text{a}+\text{a}&\cos\text{C}&\cos\text{B}\\\text{b}-\text{b}&-1&\cos\text{A}\\\text{c}-\text{c}&\cos\text{A}&-1\end{bmatrix}$

$\begin{bmatrix}0&\cos\text{C}&\cos\text{B}\\0&-1&\cos\text{A}\\0&\cos\text{A}&-1\end{bmatrix}$

$=0$

2. $\frac{1}{19}$

Solution:

$\text{adj A}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

$|\text{A}|=-19$

$\therefore\ \text{A}^{-1}=-\frac{1}{|\text{A}|}\text{ adj A}$

$\Rightarrow\ \text{A}^{-1}=-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

Now,

A^-1 = kA

$\Rightarrow-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}=\text{kA}$

$\Rightarrow\frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=\text{kA}$

$\Rightarrow\frac{1}{19}\text{A}=\text{kA}$

$\Rightarrow\text{k}=\frac{1}{19}$

1. $\pm \sqrt{3}$

Solution:

We have,

$\begin{vmatrix} 2 &\text{amp; } 4 \\ 5 &\text{amp; } 1 \end{vmatrix}​= \begin{vmatrix} 2\text{x} &\text{amp; }4 \\ 6 &\text{amp; x} \end{vmatrix}\Rightarrow2-20=2\text{x}^2-24\Rightarrow2\text{x}^2=6$

$\Rightarrow\pm\sqrt{3}$

1. There is only one solution.

Solution:

x + 2y + 3z = 1
2x + y + 3z = 2
5x + 5y + 9z = 4

The determinant of the coefficient matrix $\begin{bmatrix}1&2&3\\2&1&3\\5&5&9\end{bmatrix}$ is

$= -6 - 2(18 - 15) + 3(10 - 5)$

$= -6 - 6 + 15$

$=3\neq0$

The right hand side is also non zero.

The system has a unique solution.

3. -8

Solution:

Expanding along the first row, we get

$\triangle=5\begin{bmatrix}4&1\\6&1\end{bmatrix}-4\begin{bmatrix}3&1\\5&1\end{bmatrix}+3\begin{bmatrix}3&4\\5&6\end{bmatrix}$

$=5(4-6)-4(3-5)+3(18-20)$

$=5(-2)-4(-2)+3(-2)=-10+8-6=-8.$
 

1. A unique solution.

Solution:

The given system of equations can be written in matrix form as follows:

$\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$\text{AX}=\text{B}$

Here,

$\text{A}=\begin{bmatrix}1&1&1\\3&-1&2\\3&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\6\\-18\end{bmatrix}$

$|\text{A}|=1(-1-2)-1(3-6)+1(3+3)$

$=-3+3+6$

$=6\neq0$

So, the given system of equations has a unique solution.

1. n

Solution:

Let A = nx, B = (n - 1)x, C = (n + 2)x

⇒ C - B = x, B - A = x, C - A = 2x

Thus, the given determinant is

$\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{A}&\cos\text{B}&\cos\text{C}\\\sin\text{A}&\sin\text{B}&\sin\text{C}\end{vmatrix}$

$=\text{a}^2(\cos\text{B}\sin\text{C}-\cos\text{C}\sin\text{B})-\text{a}\times(\cos\text{A}\sin\text{C}-\cos\text{C}\sin\text{A})\\+1\times(\cos\text{A}\sin\text{B}-\sin\text{A}\cos\text{B})$

$=\text{a}^2\sin(\text{C}-\text{B})-\text{a}\sin(\text{C}-\text{A})+\sin(\text{B}-\text{A})$

$=\text{a}^2\sin{\text{x}}-\text{a}\sin2\text{x}+\sin\text{x}$ [Independent of n] 

2. 0

Solution:

$\triangle=\begin{bmatrix}5&0&5\\1&4&3\\0&8&6\end{bmatrix}$

Expanding along $\text{R}_1,$ we get:

$\triangle=5\begin{bmatrix}4&3\\8&6\end{bmatrix}-0\begin{bmatrix}1&3\\0&6\end{bmatrix}+5\begin{bmatrix}1&4\\0&8\end{bmatrix}$

$\triangle=5(24-24)-0+5(8-0)$

$\triangle=0-0+40=40.$
 

1. AB is non-singular

Solution:

A and B are non-singular matrices of order n × n.

$\therefore|\text{A}|\neq0\text{ and }|\text{B}|\neq0\ .....(\text{i})$

A and B are of the same order, so AB is defined and is of the same order.

Thus,

|AB| = |A||B|

$\Rightarrow|\text{AB}|\neq0\ \big[\text{Using (1)}\big]$

Thus, Ab is non-singular.

4. A + B = 0

Solution:

Let $\text{A}=[\text{a}_{\text{ij}}]$ and $\text{B}=[\text{b}_{\text{ij}}]$ be a square matrix of order 2

As their orders are same, A + B is defined as

$\text{A}+\text{B}=[\text{a}_{\text{ij}}+\text{b}_\text{ij}]$

$\Rightarrow|\text{A}+\text{B}|=|\text{a}_{\text{ij}}+\text{b}_\text{ij}|$

Now,

$|\text{A}+\text{B}|=0$

$\Rightarrow|\text{a}_{\text{ij}}+\text{b}_\text{ij}|=0$

$\Rightarrow[\text{a}_{\text{ij}}+\text{b}_\text{ij}]=0$

[corrsponding term is 0]

$\Rightarrow\text{A}+\text{B}=0$

1. 4

Solution:

$\triangle=\begin{vmatrix}-2\text{a}&\text{a}+\text{b}&\text{a}+\text{c}\\\text{b}+\text{a}&-2\text{b}&\text{b}+\text{c}\\\text{c}+\text{a}&\text{c}+\text{b}&-2\text{c}\end{vmatrix}$

Let a + b = 2C, b + c = 2A and c + a = 2B

⇒ a + b + b + c + c + a = 2A + 2B + 2C

⇒ 2(a + b + c) = (A + B + C)

Also, a = (a + b + c) - (b + c) = (A + B + C) - 2A = B + C - A

Similarly, b = C + A - B, c = A + B - C

Hence, 4 is the order factor of the determinant.

1. $\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if n is an even natural number.

Solution:

$\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$

$\text{A}^2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=1$

If n is an natural number.

1. $\frac{1}{2}$

Solution:

Since,

$\Delta=\begin{vmatrix}1 & 1&1 \\1 &1+\sin\theta&1\\1+\cos\theta &1&1\end{vmatrix}$

$=\begin{vmatrix}0 &0&0 \\0 &\sin\theta&1\\\cos\theta &0&1\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_2-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_3\big]$

$=-(\sin\theta.\cos\theta)$

$=\frac{1}{2}.2\sin\cos\theta=\frac{1}{2}\sin2\theta$

Since, the maximum value of $\sin2\theta$ is 1. So, for maximum value of $\theta$ should be 45°

$\therefore\ \Delta-\frac{1}{2}\sin2.45^\circ$

$=\frac{1}{2}\sin90^\circ=\frac{1}{2}.1=\frac{1}{2}$

3. 240

Solution:

Expanding along $\text{R}_1,$ we get

$\triangle=\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$

$\triangle=3\begin{bmatrix}-5&45\\-2&3\end{bmatrix}-(-1)\begin{bmatrix}6&4\\3&3\end{bmatrix}+3\begin{bmatrix}6&-5\\-3&-2\end{bmatrix}$

$\triangle=3(-15+90)+(18-12)+3(-12+15) $

$\triangle=3(75)+6+9=240. $

1. $\triangle_1+\triangle_2=0$

Solution:

$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$

$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}$ [R₁, R₂, R₃ are multiplies by a, b and c respectively, therefore we divide by abc]

$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix}$ [Taking abc common from C₂]

$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$

We know that the value of a determinant remains unchanged if its rows and columns are interchanged. so,

$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $

$=-\triangle_1$

$\triangle_1+\triangle_2=0$

3. Negative

Solution:

Discriminant D of ax² + 2bx + c = (2b)² - 4ac < 0 [Given]

⇒ 4b² - 4ac < 0

⇒ b² - ac < 0, where a > 0 .....(i)

$\Rightarrow\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$

$\Rightarrow\triangle=\begin{vmatrix}\text{ax}&\text{bx}&\text{ax}^2+\text{bx}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → xR₁]

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}\text{ax}+\text{b}&\text{bx}+\text{c}&\text{ax}^2+\text{bx}+\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → R₁ + R₂] 

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}0&0&\text{ax}^2+2\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ [Applying R₁ → R₁ - R₃]

$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{Bmatrix}\text{ax}^2+2\text{bx}+\text{c}\begin{vmatrix}\text{b}&\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}\end{vmatrix}\end{Bmatrix}$ [Expanding along R₁]

$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{x}+\text{bc}-\text{acx}-\text{bc})$

$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})\text{ x }(\text{b}^2-\text{ac})$

$\Rightarrow\triangle=(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{ac})<0$ [From eq. (i)]

$\Rightarrow\triangle<0$

1. $(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$

Solution:

We have, |A^-1| = |A|^-1, (AT)^-1 = (A^-1)^T and $|\text{A}|\neq0$ all are the properties of the inverse of a matrix A.

4. $(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$

Solution:

Since, A and B are invertible matrices, So, we can say that

$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}\ \dots(\text{i})$

Also, $\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj A})$

$\Rightarrow\ \text{adj A}=|\text{A}|.\text{A}^{-1}\ \ \dots(\text{ii})$

Also, $\text{det (A)}^{-1}=[\text{det (A)}]^{-1}$

$\Rightarrow\ \text{det (A)}^{-1}=\frac{1}{\big[\text{det (A)}\big]}$

$\Rightarrow\ \text{det (A)}.\text{det (A)}^{-1}=1\ \ \dots(\text{iii})$

Which is true.

Again, $(\text{A}+\text{B})^{-1}=\frac{1}{\big|(\text{A}+\text{B})\big|}\text{ adj }(\text{A}+\text{B})$

$\Rightarrow\ (\text{A}+\text{B})^{-1}\neq\text{B}^{-1}+\text{A}^{-1}\ \ \dots(\text{iv})$

So, only option (d) is incorrect.

1. $\pm 3$

Solution:

Given, $ |\text{A}^3| =|\text{A}|^3 = 125$

$\Rightarrow|\text{A}|=5\Rightarrow\alpha^2-4=5\Rightarrow\alpha^2=9$

$\Rightarrow \alpha=\pm 3$

4. x = 0, y = 0

Solution:

$ \begin{vmatrix} 6\text{i} &\text{amp;} -3\text{i} &\text{amp;} 1\\ 4 &\text{amp; } 3\text{i} &\text{amp;} -1 \\ 20 &\text{amp; } 3 &\text{amp; i}\end{vmatrix}=\text{x}+\text{iy},$

$\Rightarrow6\text{i}(3\text{i}^2+3)+3\text{i}(4\text{i}+20)+1(12-60\text{i})=\text{x}+\text{iy}\Rightarrow0=\text{x}+\text{iy}$

$\therefore \text{x}=\text{y}=0$