2 Marks Questions - Page 2 - MATHS STD 12 Science Questions - Vidyadip

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2 Marks Questions

Question 512 Marks

Evaluate the following determinant:
$\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}102&18&36\\1&3&4\\17&3&6\end{vmatrix}$
Applying $R_3 → 17R_2 - R_3$, we get
$\triangle=\begin{vmatrix}102&18&36\\1&3&4\\0&48&62\end{vmatrix}$
Applying $R_2 → 102R_2 - R_1$, we get
$\triangle=\begin{vmatrix}102&18&36\\0&288&327\\0&48&62\end{vmatrix}$
Thus, $\triangle=102(288\times62-372\times48)$
$\triangle=0$

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Question 522 Marks

For what value of x, the following matrix is singular?
$\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}$

Answer

If a matrix A is singular, then |A| = 0
$\therefore\begin{vmatrix}5-\text{x}&\text{x}+1\\2&4\end{vmatrix}=0$
⇒ 4(5 - x) - 2(x + 1) = 0
⇒ 20 - 4x - 2x - 2
⇒ 18 - 6x = 0
⇒ 18 = 6x
⇒ x = 3

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Question 532 Marks

A matrix A of order $3 \times 3$ is such that $|A| = 4$. Find the value of $|2A|$.

Answer

$|KA| = k^n |A|$
Here, n is the order of A.
Given, $|A| = 4$
$\Rightarrow |2A| = 2^3 \times 4 = 32$

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Question 542 Marks

If A is a square matrix satisfying $A^T$ A = l, write the value of |A|.

Answer

Let $|\text{A}|=|\text{A}|^{\text{T}} $ [By property of determinants]
Given,
$\text{A}^{\text{T}}\text{A}=\text{I}$
$\Rightarrow|\text{A}^{\text{T}}\text{A}|=1$
Then,
$|\text{A}^{\text{T}}\text{A}|=|\text{A}^{\text{T}}||\text{A}|$ [Since the determinants are of the same order]
$\Rightarrow|\text{A}^{\text{T}}||\text{A}|=1$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}^{\text{T}}|}$
$\Rightarrow|\text{A}|=\frac{1}{|\text{A}|}$ $\big[\therefore|\text{A}|=|\text{A}^{\text{T}}|\big]$
$\Rightarrow|\text{A}|^2=1$
$\Rightarrow|\text{A}|=\pm1$

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Question 552 Marks

Write the cofactor of $a_{12}$ in the following matrix $\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$

Answer

Given,
$\begin{bmatrix}2&-3&5\\6&0&4\\1&5&-7\end{bmatrix}$
Here, $\text{a}_{12}=-3$
Cofactor of $\text{a}_{12}=(-1)^{1+2}\begin{vmatrix}6&4\\1&-7\end{vmatrix}$
$\text{a}_{12}=-(-42-4)=46$

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Question 562 Marks

What is the value of the determinant $\begin{vmatrix}0&2&0\\2&3&4\\4&5&6\end{vmatrix}?$

Answer

$\begin{vmatrix}0&2&0\\2&3&4\\4&5&6\end{vmatrix}$
$= 0(18 - 20) - 2(12 - 16) + 0(10 - 12)$
$ = 8$

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Question 572 Marks

Evaluate the following determinant:
$\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$

Answer

Let $\text{A}=\begin{vmatrix}\text{x}&-7\\\text{x}&5\text{x}+1 \end{vmatrix}$
$|\text{A}|=\text{x}(5\text{x}+1)+7\times\text{x}$
$=5\text{x}^2+\text{x}+7\text{x}$
$=5\text{x}^2+8\text{x}$
Hence, $|\text{A}|=5\text{x}^2+8\text{x}$

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Question 582 Marks

If $A = [A_{ij}]$ is a $3 \times 3$ diaginal matrix such that $a_{11} = 1, a_{22} = 2, a_{33} = 3,$ then find $|A|.$

Answer

If $A = [A_{ij}]$ is a diagonal matrix of order n, then $|A| = a_{11} \times a_{22} \times a_{33} \times ...... \times a_{mn}.$
Given, $a_{11} - 1, a_{22} - 2$ and $a_{33} - 3$
$\Rightarrow |A| = 1 \times 2 \times 3 = 6 [$Applying the above property$]$

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Question 592 Marks

Evaluate the following determinant:
$\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$

Answer

$\triangle=\begin{vmatrix}1&-3&2\\4&-1&2\\3&5&2\end{vmatrix}$
$=1\begin{vmatrix} -1&2\\5&2\end{vmatrix}-(-3)\begin{vmatrix}4&2\\3&2 \end{vmatrix}+2\begin{vmatrix}4&-1\\3&5 \end{vmatrix}$
$=1(-2-10)+3(8-6)+2(20+3)$
$=(-12)+6+46$
$=40$

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Question 602 Marks

Evaluate $\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4785&2\\4789&2\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$]
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$

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Question 612 Marks

Write the value of $\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}\sin20^{\circ}&-\cos20^{\circ}\\\sin70^{\circ}&\cos70^{\circ}\end{vmatrix}$
$=\sin20^{\circ}\cos70^{\circ}+\cos20^{\circ}\sin70^{\circ}$
$=\sin(20^{\circ}+70^{\circ})$ [trignometric identity]
$=\sin90^{\circ}$
$=1$

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Question 622 Marks

Using determinants prove that the points $(a, b), (a', b)$ and $(a - a', b - b')$ are collinear if $ab' = a'b$.

Answer

$\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'&\text{b}'&1\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\\text{a}-\text{a}'&\text{b}-\text{b}'&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\begin{vmatrix}\text{a}&\text{b}&1\\\text{a}'-\text{a}&\text{b}'-\text{b}&0\\-\text{a}'&-\text{b}'&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}\text{a}'-\text{a}&\text{b}'-\text{b}\\-\text{a}'&-\text{b}'\end{vmatrix}$
$=-\text{b}'(\text{a}'-\text{a})+\text{a}'(\text{b}'-\text{b})$
$=-\text{b}'\text{a}'+\text{b}'\text{a}+\text{a}'\text{b}'-\text{a}'\text{b}$
$=\text{b}'\text{a}-\text{a}'\text{b}$
If the points are collinear then $\triangle=0$
$\text{a}\text{b}'-\text{a}'\text{b}=0$
Thus, $\text{a}\text{b}'=\text{a}'\text{b}$

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Question 632 Marks

If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.

Answer

Let A & B be non-singular matrices of order n.
A ≠ 0 and B ≠ 0 By definition.
Since they are of same order, AB = AB, AB = 0 if either A = 0 or B = 0 But it is not the case here. Thus, AB is non-zero and AB is non-singular matrix.

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Question 642 Marks

Find the value of the determinant $\begin{vmatrix}243&156&300\\81&52&100\\-3&0&4\end{vmatrix}$

Answer

[Applying $R_1 → R_1 - 3R_2$]
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$

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Question 652 Marks

If $I_3$ denotes identity matrix of order $3 \times 3$, write the value of its determinant.

Answer

In an identity matrix, all the diagonal elements are $1$ and rest of the elements are 0.
Here,
$\text{I}_3=\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}$
$\text{I}_3=1\times\begin{vmatrix}1&0\\0&1\end{vmatrix}$ [Expanding along $C_1$]
$\text{I}_3=1$
$\text{I}_3=1$

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Question 662 Marks

If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.

Answer

$|\text{A}|=\lambda$ [Order of A is n]
$\Rightarrow|-\text{A}|=(-1)^{\text{n}}|\text{A}|=(-1)^{\text{n}}\lambda$

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Question 672 Marks

Evaluate the following determinant:
$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{vmatrix}$

Answer

$\triangle=\cos^2\theta-(-\sin^2\theta)$
$\triangle=\cos^2\theta+\sin^2\theta=1$

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Question 682 Marks

Show that the following systems of linear equations is inconsistent:

3x + y = 5,

-6x - 2y = 9

Answer

$\text{D}=\begin{vmatrix}3&1\\-6&-2\end{vmatrix}=-6+6=0$
$\text{D}_1=\begin{vmatrix}5&1\\9&-2\end{vmatrix}=-10-9=-19\neq0$
Since D = 0 but $\text{D}_1\neq0$
Hence the given system of equations is inconsistent.

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Question 692 Marks

If $\text{A}=\begin{bmatrix}5&3&8\\2&0&1\\1&2&3\end{bmatrix}.$ Write the cofactor of element $a_{32}$.

Answer

Minor of $\text{a}_{32}=\text{M}_{32}=\begin{vmatrix}5&8\\2&1\end{vmatrix}=5-16=-11$
Cofactor of $\text{a}_{\text{n}}=\text{A}_{32}=(-1)^{3+2}\text{M}_{32}=11$
Hence, the cofactor of the elements $a_n$ is $11$.

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Question 702 Marks

If the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, prove that $a + b = ab$.

Answer

If the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear, then
$\begin{vmatrix}\text{a}&0&1\\0&\text{b}&1\\1&1&1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1&1&1\end{vmatrix}=0$ [Applying $R_2 → R_2 - R_1$]
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0$ [Applying $R_3 → R_3 - R_1$]
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$

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