True False[1 Marks ]

Question 11 Mark

State True or False for the statements of the following Exercise:
Let $ \begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16,$ then $\Delta_1=\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b} +\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+ \text{z}&\text{c}+\text{r}\end{vmatrix}=32.$

Answer

True.
Solution:
Given $ \begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=16$
Now $\Delta_1=\begin{vmatrix}\text{p}+\text{x}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}+\text{y}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}+\text{z}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}=32$
$\big[\text{Applying C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\big]$
$=\begin{vmatrix}2(\text{p}+\text{x}+\text{a})&\text{a}+\text{x}&\text{a}+\text{p}\\2(\text{q}+\text{y}+\text{b})&\text{b} +\text{y}&\text{b}+\text{q}\\2(\text{r}+\text{z}+\text{c})&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}$
$\big[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_2\big]$
$=2\begin{vmatrix}\text{p}&\text{a}+\text{x}&\text{a}+\text{p}\\\text{q}&\text{b}+\text{y}&\text{b}+\text{q}\\\text{r}&\text{c}+\text{z}&\text{c}+\text{r}\end{vmatrix}$
$\big[\text{Applying C}_3\rightarrow\text{C}_3-\text{C}_1\big]$
$=2\begin{vmatrix}\text{p}&\text{a}+\text{x}&\text{a}\\\text{q}&\text{b}+\text{y}&\text{b}\\\text{r}&\text{c}+\text{z}&\text{c}\end{vmatrix}$
$\big[\text{Applying C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
$=2\begin{vmatrix}\text{p}&\text{x}&\text{a}\\\text{q}&\text{y}&\text{b}\\\text{r}&\text{z}&\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{p}&\text{x}\\\text{b}&\text{q}&\text{y}\\\text{c}&\text{r}&\text{z}\end{vmatrix}=2\times16=32$

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Question 21 Mark

State True or False for the statements of the following Exercise:
If the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ splits into exactly K determinants of order 3, each element of which contains only one term, then the value of K is 8.

Answer

True.
Solution:
Since, there are two terms in each element of the determinant $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}.$
Hence, we can break $\begin{vmatrix}\text{x}+\text{a}&\text{p}+\text{u}&\text{l}+\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ as $\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}.$
Again, we can break $\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}+\text{b}&\text{q}+\text{v}&\text{m}+\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$ as
$=\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{y}&\text{q}&\text{m}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{x}&\text{p}&\text{l}\\\text{b}&\text{v}&\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{y}&\text{q}&\text{m}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}+\begin{vmatrix}\text{a}&\text{u}&\text{f}\\\text{b}&\text{v}&\text{g}\\\text{z}+\text{c}&\text{r}+\text{w}&\text{n}+\text{h}\end{vmatrix}$
Similarly, we can split these 4 determinants in 8 determinants by splitting each one in two determinants further. So, given statement is true.

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Question 31 Mark

State True or False for the statements of the following Exercise:
$\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0,$ where a, b, c are in A.P.

Answer

True.
Solution:
Since, a, b and c are in AP, then 2b = a + c
Now $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+\text{R}_3-2\text{R}_2\big]$
$=\begin{vmatrix}0&0&\text{a}+\text{c}-2\text{b}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=\begin{vmatrix}0&0&0\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$

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Question 41 Mark

State True or False for the statements of the following Exercise:
The determinant $\begin{vmatrix}\sin\text{A}&\cos\text{A}&\sin\text{A}+\cos\text{B}\\\sin\text{B}&\cos\text{A}&\sin\text{B}+\cos\text{B}\\\sin\text{C}&\cos\text{A}&\sin\text{C}+\cos\text{B}\end{vmatrix}$ is equal to zero.

Answer

True.
Solution:
Since, $\begin{vmatrix}\sin\text{A}&\cos\text{A}&\sin\text{A}+\cos\text{B}\\\sin\text{B}&\cos\text{A}&\sin\text{B}+\cos\text{B}\\\sin\text{C}&\cos\text{A}&\sin\text{C}+\cos\text{B}\end{vmatrix}$
$=\begin{vmatrix}\sin\text{A}&\cos\text{A}&\sin\text{A}\\\sin\text{B}&\cos\text{A}&\sin\text{B}\\\sin\text{C}&\cos\text{A}&\sin\text{C}\end{vmatrix}+\begin{vmatrix}\sin\text{A}&\cos\text{A}&\cos\text{B}\\\sin\text{B}&\cos\text{A}&\cos\text{B}\\\sin\text{C}&\cos\text{A}&\cos\text{B}\end{vmatrix}$
$=0+\begin{vmatrix}\sin\text{A}&\cos\text{A}&\cos\text{B}\\\sin\text{B}&\cos\text{A}&\cos\text{B}\\\sin\text{C}&\cos\text{A}&\cos\text{B}\end{vmatrix}$
[Since, in first determinant $C_1$ and $C_3$ are identicals]
$=\cos\text{A}.\cos\text{B}\begin{vmatrix}\sin\text{A}&1&1\\\sin\text{B}&1&1\\\sin\text{C}&1&1\end{vmatrix}$
[Taking cos A common from $C_2$ and cos B common from $C_3$]
= 0 [since, $C_2$ and $C_3$ are identicals]

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Question 51 Mark

State True or False for the statements of the following Exercise:
$\big(\text{A}^3\big)^{-1}=(\text{A}^{-1})^3,$ where A is square matrx and |A| ≠ 0.

Answer

True.Solution:
$(\text{A}^3)^{-1}=(\text{A.A.A})^{-1}=\text{A}^{-1}.\text{A}^{-1}.\text{A}^{-1}=\big(\text{A}^{-1})^3$

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Question 61 Mark

State True or False for the statements of the following Exercise:
$\big(\text{aA})^{-1}-\frac{1}{\text{a}}\text{A}^{-1},$ where a is any real number and A is a square matrix.

Answer

False.
Solution:
We know that $\big(\text{aA})^{-1}-\frac{1}{\text{a}}\text{A}^{-1}$ where a is any non-zero scalar (i.e., a ≠ 0)
Since, it is given that a is any real number.
$\therefore\ (\text{aA})^{-1}\neq\frac{1}{\text{a}}\text{A}^{-1}$

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Question 71 Mark

State True or False for the statements of the following Exercise:
The maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta\end{vmatrix}$ is $\frac{1}{2}.$

Answer

True.
Solution:
$\Delta=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta\end{vmatrix}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-\text{R}_1\big]$
$=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta\end{vmatrix}=\cos\theta.\sin\theta=\frac{1}{2}\sin2\theta$
Therefore, maximum value of $\Delta\text{ is }\frac{1}{2},$ when $\sin2\theta=1.$

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Question 81 Mark

State True or False for the statements of the following Exercise:
If the value of a third order determinant is 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.

Answer

True.
Solution:
Let A is the determinant
$\therefore |A| = 12$
Also, we know that, if A is a square matrix of order n, then $ |adj\ A| = |A|^{n-1}$
For $n = 3, |adj\ A| = |A|^{3-1} = |A|^2$
$= (12)2 = 144$

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Question 91 Mark

State True or False for the statements of the following Exercise:
If A and B are matrices of order 3 and |A| = 5, |B| = 3, then |3AB| = 27 × 5 × 3 = 405.

Answer

True.
Solution:
We know that, |AB| = |A|.|B|
$\therefore$ |3AB| = 27 |AB|
= 27 |A|.|B|
= 27 × 5 × 3 = 405

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Question 101 Mark

State True or False for the statements of the following Exercise:
$|adj.A| = |A|^2$, where A is a square matrix of order two.

Answer

False.
Solution:
We know that, $|adj\ A| = |A|^{n-1}$ where A is a square of matrix of order n.
Substituting $n = 2$, we get
$|adj\ A| = |A|^{2-1}$
$= |A|$

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Question 111 Mark

State True or False for the statements of the following Exercise:
$|A^{-1}| \neq |A|^{-1},$ where A is non-singular matrix.

Answer

False.
Solution:
We know that, $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$ where A is a non-singular matrix

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