2 Marks Questions

Question 512 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 522 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 532 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 542 Marks

If $A$ is a square matrix such that $|A| = 2,$ write the value of $\big|\text{A}\text{A}^{\text{T}}\big|.$

Answer

In a square matrix, $A = A^T.$
Since they are of same order, $AA^T = AA^T.$
Given, $A = 2$
$\Rightarrow AA^T= 2^2 = 4$

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Question 552 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 $C_2 \rightarrow C_2 - C_1]$
$=2\times\begin{vmatrix}4785&1\\4789&1\end{vmatrix}$
$=2\times(4785-4789)$
$=2\times(-4)=-8$
$\Rightarrow\begin{vmatrix}4785&4787\\4789&4791\end{vmatrix}=-8$

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Question 562 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 572 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 $R_2 \rightarrow R_2 - R_1]$
$\Rightarrow\begin{vmatrix}\text{a}&0&1\\-\text{a}&\text{b}&0\\1-\text{a}&1&0\end{vmatrix}=0 [$Applying $R_3 \rightarrow R_3 - R_1]$
$\Rightarrow\triangle=\begin{vmatrix}-\text{a}&\text{b}\\1-\text{a}&1\end{vmatrix}=0$
$\Rightarrow-\text{a}-\text{b}(1-\text{a})=0$
$\Rightarrow\text{a}+\text{b}=\text{ab}$

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Question 582 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 592 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 602 Marks

Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$

Answer

Let $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ we get
$\triangle=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta \end{vmatrix}$
$=\sin\theta\cos\theta$
$=\frac{\sin2\theta}{2}$
We know that $-1\leq\sin2\theta\leq1$
$\therefore$ Maximum value of $\triangle=\frac{1}{2}\times1=\frac{1}{2}$

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Question 612 Marks

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

Answer

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

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Question 622 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 632 Marks

Evaluate the following determinant:
$\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$

Answer

Let $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2,$ we get
$\Rightarrow\triangle=\begin{vmatrix}1&4&9-4\\4&9&16-9\\9&16&25-16 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&4&5\\4&9&7\\9&16&9\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&5&5\\4&13&7\\9&25&9 \end{vmatrix} [$Applying $C_2 \rightarrow C_1 + C_2]$
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix} [$Applying $C_2 \rightarrow 5C_1 - C_2$ and $C_3 \rightarrow 5C_1 - C_3]$
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$

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Question 642 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 652 Marks

Write rthe value of the determinant $\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}$

Answer

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

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Question 662 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 672 Marks

A matrix $A$ of order $3 \times 3$ has determinant $5. $What is the value of $|3A|$?

Answer

If $A = [a_{ij}]$ is a square matrix of order $n$ and $k$ is a constant, then
$|kA| = k^n|A|$
Here,
Number of rows $= n$
$k$ is a common factor from each row of $k$
$|3A| = 3^3|A| = 27 \times 5 = 135\ [$Given matrix is $3 \times 3$ such that $|A| = 5]$
Thus$, |3A| = 135$

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Question 682 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 692 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 702 Marks

If $|A| = 2,$ where $A$ is $2 \times 2$ matrix, find $|$adj $A|.$

Answer

For any square matrix $A$ of order $n, |$adj $A| = |A|^{n-1}$
Given, $|A| = 2$
Here, order is $2$
$\Rightarrow |$adj $A| = |2|^{2-1} = 2$

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MATHS 2 Marks Questions