Maths M.C.Q (1 Marks)

2. 2

Solution:

Consider the element 2 in the determinant $\triangle=\begin{bmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{bmatrix}$

The minor of the element 2 is given by

$\therefore\text{M}_{22}=\begin{bmatrix}5&8\\5&8\end{bmatrix}=40-40=0$

$\Rightarrow\text{A}^{22}=(-1)^2+2 (0)=0.$

4. 0

Solution:

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

$\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}7\\1\\5\end{bmatrix}$

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

Here,

$\text{A}=\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7\\1\\5\end{bmatrix}$

Now,

$|\text{A}|=2(9-8)-1(-3-2)-1(4+3)$

$=2+5-7$

$=0$

Let c_ij be the co-factors of the elements a_ij in A = [a_ij]. Then,

$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}-3&2\\4&-3\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&2\\1&-3\end{vmatrix}=5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&-3\\1&4\end{vmatrix}=7$

$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&-1\\4&-3\end{vmatrix}=-1,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}2&-1\\1&-3\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}2&1\\1&4\end{vmatrix}=-7$

$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&-1\\-3&2\end{vmatrix}=5,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}2&-1\\1&2\end{vmatrix}=-5,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}2&1\\1&-3\end{vmatrix}=-7$

$\text{adj }\text{A}=\begin{bmatrix}1&5&7\\-1&-5&-7\\5&-5&-7\end{bmatrix}^\text{T}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}$

$\Rightarrow(\text{adj }\text{A})\text{B}=\begin{bmatrix}1&-1&5\\5&-5&-5\\7&-7&-7\end{bmatrix}\begin{bmatrix}7\\1\\5\end{bmatrix}$

$=\begin{bmatrix}7-1+25\\35-5-25\\49-7-35\end{bmatrix}=\begin{bmatrix}32\\5\\6\end{bmatrix}\neq0$

The given system of equations is inconsistent. Thus, it has no solution.

3. ±6

Solution:

We have, $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$

$\Rightarrow\ 2\text{x}^2-40=18+17$

$\Rightarrow\ 2\text{x}^2=32+40$

$\Rightarrow\ \text{x}^2=\frac{72}{2}=36$

$\Rightarrow\ \text{x}^2=36$

$\Rightarrow\ \text{x}=\pm6$

2. 2

Solution:

Let $\triangle=\begin{vmatrix}\text{cosec x}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosec x}\end{vmatrix}$

$=(\text{cosec x})^3\begin{vmatrix}1&\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&1&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}&1\end{vmatrix}$

$=(\text{cosec x})^3\begin{vmatrix}1&\tan\text{x}&\tan\text{x}\\\tan\text{x}&1&\tan\text{x}\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$

 $=(\text{cosec x})^3\begin{vmatrix}1-\tan\text{x}&\tan\text{x}-1&0\\0&1-\tan\text{x}&\tan\text{x}-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Applying R₁ → R₁ - R₂, R₂ → R₂ - R₃]

$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{vmatrix}1&-1&0\\0&1&-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$ [Taking out $(1-\tan\text{x})$ common from R₁ and R₂]

 $=(\text{cosec x})^3(1-\tan\text{x})^2\begin{Bmatrix}1\begin{vmatrix}1&-1\\\tan\text{x}&1\end{vmatrix}+\tan\text{x}\begin{vmatrix}-1&0\\1&-1\end{vmatrix}\end{Bmatrix}$ [Expanding along C₁]

 $=(\text{cosec x})^3(1-\tan\text{x})^2\{1+\tan\text{x}+\tan\text{x}\}$

$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+2\tan\text{x}\}$

$\triangle=0$

$=(\text{cosec x})^3(1-\tan\text{x})^2(1+2\tan\text{x})=0$

$(1-\tan\text{x})=0,(\text{coses x})^3=0$ and $(1+2\tan\text{x})=0$

Or $\tan\text{x}=1,\text{cosec x}=0$ and $\tan\text{x}=\frac{-1}{2}$

$\Rightarrow-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ $\Big[\tan\text{x}=1,\text{x}=\frac{-1}{2}$ are 2 real roots as $\text{cosec x}=0$ has no solution$\Big]$

Thus, these are 2 solutions.

2. λ ≠ 5

Solution:

x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ

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

= 2λ - 9 - λ + 3 + 1

= λ - 5

For unique solution determinant ≠ 0

⇒ λ ≠ 5

The right hand side is non zero what so ever be the value of µ.

2. $\frac{1}{\text{det(A)}}$

Solution:

We know that $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$

3. 8

Solution:

$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$

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

$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix}$ [Applying C₂ → C₂ - C₁ and C₃ → C₃ - C₁]

$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$

$=8\text{n}+12-8\text{n}-4$

$=8$

Hence, the correct option is (c)

1. -4

Solution:

The minor of element -3 is given by

$\text{M}_{21}=\begin{bmatrix}4&4\\1&2\end{bmatrix}=4(2)-4=4$  (Obtained by eliminating $\text{R}_2$ and $\text{C}_1$)

$\therefore\text{A}_{21}=(-1)^{2+1} \text{M}_{21}=(-1)^3 4=-4.$

3. a⁶

Solution:

$\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}$

$\therefore|\text{A}|=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}=\text{a}^3\neq0$

and

n = 3

Thus, we have

|adj A| = |A|^n-1 = (a³)² = a⁶.

1. -20

Solution:

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

$\Rightarrow2\text{x}-15=3\text{x}+5 $

$\Rightarrow\text{x}=-20$

3. $\pm6$

Solution:

$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$

$\Rightarrow2\text{x}^2-40=18+14$

$\Rightarrow2\text{x}^2-40=32$

$\Rightarrow2\text{x}^2=72$

$\Rightarrow\text{x}^2=36$

$\Rightarrow\text{x}=\pm6$

Hence, the correct option is (C)

4. 32

Solution:

$\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}=\begin{vmatrix}\text{p}&\text{a}&\text{a}\\\text{q}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{p}&\text{a}&\text{p}\\\text{q}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}\\+​​​​\begin{vmatrix}\text{p}&\text{x}&\text{p}\\\text{q}&\text{y}&\text{q}\\\text{r}&\text{z}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{a}\\\text{y}&\text{b}&\text{b}\\\text{r}&\text{c}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{r}&\text{c}&\text{r}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{a}\\\text{y}&\text{y}&\text{b}\\\text{z}&\text{z}&\text{c}\end{vmatrix}+​​​​\begin{vmatrix}\text{x}&\text{x}&\text{p}\\\text{y}&\text{y}&\text{q}\\\text{z}&\text{z}&\text{r}\end{vmatrix}$

$=0+0+\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+0+0+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}+0+0$

$=\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{a}&\text{p}\\\text{y}&\text{b}&\text{q}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$

$=2\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}$

$=2\times16=32$

Hence, the correct option is (b)

3. $\pm6$

1. 0

Solution:

$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

$=\begin{vmatrix}0&0&2(\text{a}+\text{c}-2\text{b})\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

[Applying R₁ → R₁ + R₃ - R₂, R₁ → R₁ - R₂]

$=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ $[\because$ a, b, c are in A.P.$]$

$=0$

3. -2n³

Solution:

$\text{A}_\text{r}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\sum\limits_{\text{r}=1}^\text{n}1&\sum\limits_{\text{r}=1}^\text{n}\text{r}&\sum\limits_{\text{r}=1}^\text{n}2\text{r}\\\sum\limits_{\text{r}=1}^\text{n}2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

As $\sum\limits_{\text{r}=1}^\text{r}1=1+1+1\ ......+1(\text{n times})=\text{n}$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{r}\text{r}=1+2+3+\ .....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}$

Let $\text{S}=\sum\limits_{\text{r}=1}^\text{r}2^\text{r}=2+2^2+2^3=\ .....+2^{\text{n}}$

$\Rightarrow2\text{S}=2^2+3^2=\ ....+2^{\text{n}}+2^{\text{n}+1}$

$\Rightarrow2\text{S}-\text{S}$

$\Rightarrow\text{S}=\sum\limits_{\text{r}=1}^\text{n}2^{\text{r}}=2^{\text{n+1}}-2$

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

[Applying R₁ → R₁ - R₂]

$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}-\text{n}&\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2-2^{\text{n}+1}\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

$=\begin{vmatrix}0&0&-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$

$=-2\times\begin{vmatrix}2\text{n}&\text{n}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}\end{vmatrix}$

$=-2\big[\text{n}^{3}+\text{n}^2-\text{n}^2\big]$

$=-2\text{n}^3$

1. 0

solution:

$\text{A}+\text{B}+\text{C}=\pi$

$\text{A}+\text{C}=\pi-\text{B},\text{A}+\text{B}=\pi-\text{C}$ and $\text{B}+\text{C}=\pi-\text{A}$

Thus the determinant becomes

$\begin{vmatrix}\sin\pi&\sin(\pi-\text{B})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\pi-\text{C})&\tan(\pi-\text{A})&0\end{vmatrix}$

$=\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

$[\sin\pi=0,\sin(\pi-\text{B}),\cos(\pi-\text{C})=-\cos\text{C},\tan(\pi-\text{A})=-\tan\text{A}]$

It is a skew symmetric matrix of the odd order 3. Thus by property of determinants, we get

$|\triangle|=0$

$\begin{vmatrix}0&\sin\text{B}&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\-\cos\text{C}&-\tan\text{A}&0\end{vmatrix}$

4. $\text{Det (A)}\in[2,4]$

Solution:

$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$

$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$ [Applying C₃ → C₃ + C₁]

$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}$ [Expanding along C₃]

$=2(\sin^2\theta+1)$

Given, $0\leq\theta\leq2\pi$

$-1\leq\sin\theta\leq1$

$0\leq\sin^2\theta\leq1$

$|\text{A}|=2(\sin^2\theta+1)$

$|\text{A}|=2\times1=2$ $[\theta=0]$

$|\text{A}|=2\times2=4$ $[\theta=2\pi]$

$\text{Det (A)}\in[2,4]$

2. 0

Solution:

if A is skew symmetric matrix

then $\text{A} = \text{-A}^\text{T}$

$\therefore |\text{A}|=-|\text{A}^\text{T}|=-|\text{A}|$

$\Rightarrow 2|\text{A}|=0$

$\Rightarrow|\text{A}|=0$

3. $5\sqrt{3}+4$

Solution:

Evaluating along  $\text{R}_1$,we get

$\triangle5(\sqrt3)-(-4)^1=5\sqrt{3}+4$

1. sq.units

Solution:

Area of triangle $=\frac{1}{2} \begin{vmatrix} 0 &\text{amp; }0 &\text{amp; 1} \\ 0&\text{amp; 2} &\text{amp; 1} \\3 &\text{amp;1} &\text{amp; 1} \end{vmatrix}= \frac{1}{2}\times|-6|=3$

4. -1

Solution:

$\begin{vmatrix}1+\text{x}&1&1\\1&1+\text{y}&1\\1&1&1+\text{z}\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}&0&-\text{z}\\0&\text{y}&-\text{z}\\1&1&1+\text{z}\end{vmatrix}=0$ [Applying R₂ → R₂ - R₃ and R₁ → R₁ - R₃]

$\Rightarrow\text{x}\big[\text{y}(1+\text{z})+\text{z}\big]+1(\text{yz})=0$ [Expanding along first column]

$\Rightarrow\text{x}[\text{y}+\text{yz}+\text{z}]+\text{yz}=0$

$\Rightarrow\text{xy}+\text{xyz}+\text{xz}+\text{yz}=0$

$\Rightarrow\text{xy}+\text{yz}+\text{zx}=-\text{xyz}$

$\Rightarrow\frac{\text{xy}}{\text{xyz}}+\frac{\text{yz}}{\text{xyz}}+\frac{\text{zx}}{\text{xyz}}=-\frac{\text{xyz}}{\text{xyz}}$

$\Rightarrow\frac{1}{\text{z}}+\frac{1}{\text{x}}+\frac{1}{\text{y}}=-1$

$\Rightarrow\text{x}^{-1}+\text{y}^{-1}+\text{z}^{-1}=-1$

Hence, the correct option is (d)

3. -4

We have,

$\begin{bmatrix}1 & -2 & 5 \\2 & \text{a} & 1\\0 & 4 & 2\text{a} \end{bmatrix} = 86,$

$\Rightarrow\ 1(2\text{a}^{2} + 4) - 2(-4\text{a} - 20) + 0 = 86$ [Expanding along C₁]

$\Rightarrow\ \text{a}^{2} + 4\text{a} - 21 = 0$

$\Rightarrow\ \text{(a + 7)} (\text{a} - 3) = 0$

$\Rightarrow\ \text{a} = -7 \text{ and } 3$

$\therefore$ Required sum = -7 + 3 = -4

4. None of these

Solution:

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

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

$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Taking (b - a) common from C₁ and C₃]

$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Applying C₁ → C₁ - C₂ - C₃]

$=0$

Hence, the correct option is (d)

1. 3

Solution:

Expanding along $\text{R}_1,$ we get

$\triangle=2(-1)-5(-1)=2+5$

$=3$

1. -4

Solution:

The minor of element -3 is given by

$\text{M}_{21}=\begin{bmatrix}4&4\\1&2\end{bmatrix}=4(2)-4=4$  (Obtained by eliminating $\text{R}_2$ and $\text{C}_1$)

$\therefore\text{A}_{21}=(-1)^{2+1} \text{M}_{21}=(-1)^3 4=-4.$

1. 0

Solution:

We have, $\text{f}(\text{t})=\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $

$\Rightarrow\ \frac{\text{f(x)}}{\text{t}^2}=\frac{1}{\text{t}^2}​​​​\begin{vmatrix}\cos\text{t}&\text{t}&1\\2\sin\text{t}&\text{t}&2\text{t}\\\sin\text{t}&\text{t}&\text{t}\end{vmatrix} $

$=\begin{vmatrix}\cos\text{t}&\text{t}&1\\\frac{2\sin\text{t}}{\text{t}}&1&2\\\frac{\sin\text{t}}{\text{t}}&1&1\end{vmatrix} $ $\big[\text{Dividing R}_2\text{ and R}_3\text{ by }'\text{t}'\big]$

$\Rightarrow\ \lim\limits_{\text{t}\rightarrow0}\frac{\text{f(t)}}{\text{t}^2}=\begin{vmatrix} \lim\limits_{\text{t}\rightarrow0}\text{t}\cos\text{t}&\lim\limits_{\text{t}\rightarrow0}\text{t}&\lim\limits_{\text{t}\rightarrow0}1\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{2\sin}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}2\\\lim\limits_{\text{t}\rightarrow0}\text{t}\frac{\sin\text{t}}{\text{t}}&\lim\limits_{\text{t}\rightarrow0}1&\lim\limits_{\text{t}\rightarrow0}1\end{vmatrix}$

$=\begin{vmatrix}1&0&1\\2&1&2\\1&1&1\end{vmatrix}$ $\bigg(\because\lim\limits_{\text{t}\rightarrow 0}\frac{\sin\text{t}}{\text{t}}=1\bigg)$

$=1(1-2)-0+1(2-1)$

$=0$

2. 63

Solution:

$\text{ax}^3+\text{bx}^2+\text{cx}+\text{d}=\begin{bmatrix}\text{x}+1&\text{amp;}2\text{x}&\text{amp; 3}\text{x}\\2\text{x}+3&\text{amp;}\text{x+1}&\text{amp;}\text{x}\\2-\text{x}&\text{amp;}3\text{x}+4&\text{amp;}5\text{x}-1\end{bmatrix}$ if 

$\text{x}=1\text{a}+\text{b}+\text{c}+\text{d}=\begin{bmatrix}2&\text{amp;}2&\text{amp;3}\\5&\text{amp;}2&\text{amp;1}\\1&\text{amp;}7&\text{amp;4} \end{bmatrix}$

$\text{a}+\text{b}+\text{c}+\text{d}=2-38+99=101-38=63$

2. Unique solution

Solution:

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

$\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$

Here,

$\text{A}=\begin{bmatrix}\frac{1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{-1}{\text{c}^2}\\\frac{1}{\text{a}^2}&\frac{-1}{\text{b}^2}&\frac{1}{\text{c}^2}\\\frac{-1}{\text{a}^2}&\frac{1}{\text{b}^2}&\frac{1}{\text{c}^2}\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1\\1\\1\end{bmatrix}$

Now,

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

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

$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times1(-1-1)-1(1+1)-1(1-1)$

$=\frac{1}{\text{a}^2\text{b}^2\text{c}^2}\times(-2-2)$

$=\frac{-4}{\text{a}^2\text{b}^2\text{c}^2}$

$\Rightarrow|\text{A}|\neq0$

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

3. -4

Solution:

$\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}=86$

⇒ 1(2a² + 4) - 2(-4a - 20) = 86

⇒ 2a² + 4 + 8a + 40 = 86

⇒ 2a² + 8a - 42 = 0

⇒ a² + 4a - 21 = 0

⇒ a² + 7a - 3a - 21 = 0

⇒ a(a + 7) - 3(a + 7) = 0

⇒ a = -7, 3

Sum of the two values of a = -7 + 3 = -4

Hence, the correct option is (c)

1. $\dfrac{-1}{2}$

Solution:

Here, $|\text{A}| =1\times 1-2\times 2 = -3$

$\therefore\text{f}(|\text{A}|)=\cfrac{1+(-3)}{1+3}=-\cfrac{1}{2}$

2. $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is not an prthogonal, question is incorrect.

Solution:

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

$\text{A}^\text{T}\text{A}=\text{I}$

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

$\frac{1}{9}\begin{bmatrix} 1+4+\text{x}^2 & 1+2+2\text{x} & \text{xy}-2 \\ 1+2+2\text{x} & 1+1+4 & 2-2+2\text{y} \\ 2-4+\text{xy} & 2-2+2\text{y} & 4+4+\text{y}^2 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix} 5+\text{x}^2 & 3+2\text{x} & \text{xy}-2 \\ 3+2\text{x} & 6 & 2\text{y} \\ -6+\text{xy} & 2\text{y} & 8+\text{y}^2 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Equality of two matrices does not hold Matrix A is not orthogonal.

Hence, the given question is incorrect.

2. $\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$

Solution:

Adjoint of a square matrix of order 2 is obtained by interchancing the diagoinal elements and changing the signs of off-diagonal elements.

Here,

$\text{A}=\begin{bmatrix}\text{a} & \text{bc} & \text{d} \end{bmatrix}$

$\Rightarrow\text{adj A}=\begin{bmatrix}\text{d} & -\text{b}-\text{c} & \text{a} \end{bmatrix}$

1. $0$

Solution:

On solving the given matrix,

$|\text{A}|=1-\log_\text{a}\text{b}.\log_\text{b} \text{a}=1-1=0$

2. -9

Solution:

We know that, area of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by

$\triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}\end{vmatrix}$

$\therefore$ Area of triangle with vertices (-3, 0), (3, 0) and (0, k) is

$\therefore\ \ \triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}-3&0&1\\3&0&1\\0&\text{k}&1\end{vmatrix}\end{vmatrix}=9$ (given)

$\Rightarrow\ [-3(-\text{k)}-0+1(3\text{k})]=\pm18$

$\Rightarrow\ 6\text{k}=\pm18$

$\therefore\ \ \text{k}=\pm\frac{18}{6}=\pm3$

1. 3, 7

Solution:

Given, $\left |\begin{matrix} 3 &\text{amp; 5} &\text{amp; x}\\ 7 &\text{amp; x} &\text{amp; 7}\\ \text{x} &\text{amp; 5} &\text{amp; 3}\end{matrix}\right |=0$

$\Rightarrow3(3\text{x}-35)-5(21-7\text{x})+\text{x}(35-\text{x}^2)=0$

$\Rightarrow9\text{x}-105-105+35\text{x}+35\text{x}-\text{x}^3=0$

$\Rightarrow\text{x}^3-79\text{x}+210=0$

$\Rightarrow(\text{x}+10)(\text{x}-3)(\text{x}-7)=0$

$\Rightarrow\text{x}=10, 3, 7$

1. 0

$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$

$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ (2b = a + c as a,b and c are in A.P.)

Applying R₁ → R₁ - R₂ and R₃ → R₃ - R₂, we have:

$\triangle=\begin{vmatrix}-1&-1&\text{a}-\text{c}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$

Applying R₁ → R₁ + R₃, we have:

$\triangle=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$

4. None of these

Solution:

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

$\text{A}^2=\begin{bmatrix} 1+\text{a} & \text{b} & 3 \\ \text{a} & \text{b} & 2 \\ 3\text{a} & 2\text{b} & \text{a}+\text{b}+4 \end{bmatrix}$

$\Rightarrow\text{aI}+\text{bA}+2\text{A}^2$

$=\begin{bmatrix} 3\text{a}+2+\text{b} & 2\text{b} & 6+\text{b} \\ 2\text{a} & \text{a}+2\text{b} & \text{b}+4 \\ \text{ab}6\text{a} & 6\text{b}+\text{b}^2 & 3\text{a}+4\text{b}+8 \end{bmatrix}$

1. $\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$

Solution:

A = IA

$\Rightarrow\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$

$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\text{x}^1 & 0 & 0\\ 0 & \text{y}^1 & 0 \\ 0 & 0 & \text{z}^1\end{bmatrix}\text{A}$

$\Big[\text{Applying R}_1=\frac{1}{\text{x}}\text{R}_1,\text{R}_2=\frac{1}{\text{y}}\text{R}_2\text{ and R}_3=\frac{1}{\text{z}}\text{R}_3\Big]$

$\Rightarrow\text{A}^{-1}=\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$

4. $\displaystyle \frac{38}{35}$

Solution:

The value of $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$ is $\bigg(\frac{-4}{7}\times\frac{-2}{5}\bigg)-\bigg(\frac{-4}{7}\times5\bigg)$

$=\frac{8}{25}+\frac{30}{35}=\frac{38}{35}$