Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks:
If A is a matrix of order $3 × 3$, then $(A^2)^{-1} =$ ________.

Answer

If A is a matrix of order 3 \times 3, then $(A^2)^{-1} = (A^{-1})^2.$
Solution:
We know that, $(A^n)^{-1} = (A^{-1})^n,$ where n \in N.
Substituting $n = 2$ on both sides, we get
$(A^2)^{-1} = (A^{-1})^2$

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Question 21 Mark

Fill in the blanks:
If A is a matrix of order 3 × 3, then number of minors in determinant of A are ________.

Answer

If A is a matrix of order 3 × 3, then number of minors in determinant of A are 9.
Solution:
Since, the order of the matrix A is 3 × 3. It means there are 9 elements in the matrix A.
Hence, the number of minors in determinant of A is 9.

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Question 31 Mark

Fill in the blanks:
If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=$ _________.

Answer

If $\cos2\theta=0,$ then $\begin{vmatrix}0&\cos\theta&\sin\theta\\\cos\theta&\sin\theta&0\\\sin\theta&0&\cos\theta\end{vmatrix}^2=\frac{1}{2}.$
Solution:
Since, $\cos2\theta=0$
$\Rightarrow\ \cos2\theta=\cos\frac{\pi}{2}$ $\Rightarrow\ 2\theta=\frac{\pi}{2}$
$\Rightarrow\ \theta=\frac{\pi}{4}$
$\therefore\ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$ and $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\therefore\ \begin{vmatrix}0&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\end{vmatrix}$
Expanding along $R_1,$
$=\bigg[-\frac{1}{\sqrt{2}}\Big(\frac{1}{2}\Big)+\frac{1}{\sqrt{2}}\Big(-\frac{1}{2}\Big)\bigg]^2$ $=\bigg[\frac{-2}{2\sqrt{2}}\bigg]^2=\bigg(\frac{-1}{\sqrt{2}}\bigg)^2=\frac{1}{2}$

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Question 41 Mark

Fill in the blanks:
$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=$ ________.

Answer

$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}$ $=(\text{z}-\text{x})(\text{y}-\text{z})(\text{y}-\text{x}+\text{xyz})$
Solution:
We have, $\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=\begin{vmatrix}\text{z}-\text{x}&\text{xyz}&\text{x}-\text{z}\\\text{z}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_3\big]$
$=(\text{z}-\text{x})\begin{vmatrix}1&\text{xyz}&\text{x}-\text{z}\\1&0&\text{y}-\text{z}\\1&\text{z}-\text{y}&0\end{vmatrix}$
[Taking (z - x) common from column 1]
Expanding along $R_1,$
$= (\text{z} - \text{x})\Big[1.\big\{-1({\text{y} - \text{z})(\text{z} - \text{y})}\big\} -\text{xyz}(\text{z} - \text{y}) + (\text{x} - \text{z})(\text{z} - \text{y})\Big]$
$=(\text{z}-\text{x})(\text{z}-\text{y})(-\text{y}+\text{z}-\text{xyz}+\text{x}-\text{z})$
$=(\text{z}-\text{x})(\text{z}-\text{y})(\text{x}-\text{y}-\text{xyz})$
$=(\text{z}-\text{x})(\text{y}-\text{z})(\text{y}-\text{x}+\text{xyz})$

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Question 51 Mark

Fill in the blanks:
If A is a matrix of order 3 × 3, then |3A| = _______.

Answer

If A is a matrix of order 3 × 3, then |3A| = 27|A|
Solution:
If A is a matrix of order 3 × 3, then |3A| = 3 × 3 × 3|A| = 27|A|

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Question 61 Mark

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

Answer

If A is invertible matrix of order $3 × 3,$ then $|\text{A}^{-1}|=\frac{1}{|\text{A}|}$
Solution:
We know that, $|A|.|A^{-1}| = 1$
$\Rightarrow\ |\text{A}^{-1}|=\frac{1}{|\text{A}|}$
Therefore, if A is invertible matrix of order 3 × 3, then $|\text{A}^{-1}|=\frac{1}{|\text{A}|}.$

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Question 71 Mark

Fill in the blanks:
If x = -9 is a root of $\begin{vmatrix}\text{x}&3&7\\2&\text{x}&2\\7&6&\text{x}\end{vmatrix}=0,$ then other two roots are __________.

Answer

If x = -9 is a root of $\begin{vmatrix}\text{x}&3&7\\2&\text{x}&2\\7&6&\text{x}\end{vmatrix}=0,$ then other two roots are x = 7.
Solution:
Since, $\begin{vmatrix}\text{x}&3&7\\2&\text{x}&2\\7&6&\text{x}\end{vmatrix}=0$
Expanding along $R_1,$
$x(x^2 - 12) -3(2x - 14) + 7(12 - 7x) = 0$
$\Rightarrow x^3 - 12x - 6x + 42 + 84 - 49x = 0$
$\Rightarrow x^3 - 67x + 126 = 0 ...(i)$
Here, $126 \times 1 = 9 \times 2 \times 7$
For $x = 2, 2^3 - 67 \times 2 + 126 = 134 - 134 = 0$
Hence,$ x = 2$ is a roots.
For $x = 7, 7^3 - 67 \times 7 + 126 = 469 - 469 = 0$
Hence, $x = 7$ is also a root.

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Question 81 Mark

Fill in the blanks: If $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}=\text{A}+\text{Bx}+\text{Cx}^2+\ \dots,$ then A = ________.

Answer

If $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}=\text{A}+\text{Bx}+\text{Cx}^2+\ \dots,$ then A = 0.
Solution: We have, $\text{f(x)}=\begin{vmatrix}(1+\text{x})^{17}&(1+\text{x})^{19}&(1+\text{x})^{23}\\(1+\text{x})^{23}&(1+\text{x})^{29}&(1+\text{x})^{34}\\(1+\text{x})^{41}&(1+\text{x})^{43}&(1+\text{x})^{47}\end{vmatrix}$ [Taking $(1 + x)^{17}, (1 + x)^{23}, (1 + x)^{41}$ common from $R_1, R_2, R_3$ respectively]
$=(1+\text{x})^{17}(1+\text{x})^{23},(1+\text{x})^{41}\begin{vmatrix}1&(1+\text{x})^{2}&(1+\text{x})^{6}\\1&(1+\text{x})^{6}&(1+\text{x})^{11}\\1&(1+\text{x})^{2}&(1+\text{x})^{6}\end{vmatrix}=0$ $\big[\because\ \text{R}_1\text{ and R}_3\text{ are identical}\big]$ $\therefore\ \text{A}=0$

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Question 91 Mark

Fill in the blanks: If x, y, z ∈ R, then the value of determinant $\begin{vmatrix}(2^\text{x}+2^{-\text{x}})^2&(2^\text{x}-2^{-\text{x}})^2&1\\(3^\text{x}+3^{-\text{x}})^2&(3^\text{x}-3^{-\text{x}})^2&\\1(4^\text{x}+4^{-\text{x}})&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}$ is:

Answer

If x, y, z ∈ R, then the value of determinant $\begin{vmatrix}(2^\text{x}+2^{-\text{x}})^2&(2^\text{x}-2^{-\text{x}})^2&1\\(3^\text{x}+3^{-\text{x}})^2&(3^\text{x}-3^{-\text{x}})^2&1\\(4^\text{x}+4^{-\text{x}})&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}=0$ is:
Solution:
We have,
$ \begin{vmatrix}(2^\text{x}+2^{-\text{x}})^2&(2^\text{x}-2^{-\text{x}})^2&1\\(3^\text{x}+3^{-\text{x}})^2&(3^\text{x}-3^{-\text{x}})^2&1\\(4^\text{x}+4^{-\text{x}})&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}$
$[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_2]$
$ =\begin{vmatrix}4.2^\text{x}.2^{-\text{x}}&(2^\text{x}-2^{-\text{x}})^2&1\\4.3^\text{x}.3^{-\text{x}}&(3^\text{x}-3^{-\text{x}})^2&1\\4.4^\text{x}.4^{-\text{x}}&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}$ $\big[\because\ (\text{a}+\text{b})^2-(\text{a}-\text{b})^2=4\text{ab}\big]$
$ =\begin{vmatrix}4&(2^\text{x}-2^{-\text{x}})^2&1\\4&(3^\text{x}-3^{-\text{x}})^2&1\\4&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}=4$ $ =\begin{vmatrix}1&(2^\text{x}-2^{-\text{x}})^2&1\\1&(3^\text{x}-3^{-\text{x}})^2&1\\1&(4^\text{x}-4^{-\text{x}})^2&1\end{vmatrix}=0$

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Question 101 Mark

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

The sum of the products of elements of any row with the co-factors of corresponding elements is equal to value of the determinant.
Solution:
The sum of the products of elements of any row with the co-factors of corresponding elements is equal to value of the determinant.
Let $\Delta=\begin{vmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}\end{vmatrix}$
Expanding along $R_1,$
$\Delta=\text{a}_{11}\text{A}_{12}+\text{a}_{12}\text{A}_{12}+\text{a}_{13}\text{A}_{13}$
= Sum of products of elements of $R_1$ with their corresponding cofactors.

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Maths Fill In The Blanks[1 Marks ]

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