MCQ 511 Mark
Which of the following is not a property of determinant:
AnswerCorrect option: A. The value of determinant changes if all of its rows and columns are interchanged
The value of determinant remains unchanged if all of its rows and columns are interchanged i.e. $|A| = |A\ ’|$
where $A$ is a square matrix and $A\ ’$ is the transpose of the matrix $A.$
View full question & answer→MCQ 521 Mark
If $A$ is a singular matrix, then $\text{adj A}$ is.
AnswerGiven $∣A∣ = 0$
We know $∣\text{adj A}∣ = ∣A∣ n - 1$
$\therefore\ ∣\text{adj A}∣ = 0$
Hence$, \text{adj A}$ is singular
View full question & answer→MCQ 531 Mark
Evaluate $\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$ is:
AnswerExpanding along $\text{R}_1,$ we get
$\triangle=\begin{bmatrix}3&-1&3\\6&-5&4\\3&-2&3\end{bmatrix}$
$\triangle=3\begin{bmatrix}-5&45\\-2&3\end{bmatrix}-(-1)\begin{bmatrix}6&4\\3&3\end{bmatrix}+3\begin{bmatrix}6&-5\\-3&-2\end{bmatrix}$
$\triangle=3(-15+90)+(18-12)+3(-12+15) $
$\triangle=3(75)+6+9=240. $
View full question & answer→MCQ 541 Mark
Choose the correct answer from given four options in each of the Exercise:
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
- A
$f(a) = 0$
- B
$f(b) = 0$
- ✓
$f(0) = 0$
- D
$f(1) = 0$
AnswerCorrect option: C. $f(0) = 0$
$\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
$\Rightarrow\ \text{f}(0)=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{vmatrix},$ Which is skew$-$symmetric determinant of order $3$
Hence $f(0) = 0.$
View full question & answer→MCQ 551 Mark
If $A$ and $B$ are square matrices such that $B=-A^{-1} B A,$ then $(A+B)^2=$
- A
$O$
- ✓
$A^2+B^2$
- C
$A^2+2 A B+B^2$
- D
$A + B$
AnswerCorrect option: B. $A^2+B^2$
$B=-A^{-1} B A$
$\Rightarrow A B=-A A^{-1} B A$
$\Rightarrow A b=-I B A$
$\Rightarrow A B=-B A$
$\Rightarrow A B+B A=0 \ldots . . .(i)$
Consider,
$(A+B)^2=A^2+A B+B A+B^2$
$(\because A B \neq B A)$
$(A+B)^2=A^2+O+B^2$
$(A+B)^2=A^2+B^2$
View full question & answer→MCQ 561 Mark
If $\triangle_1=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix},\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix},$ then:
- ✓
$\triangle_1+\triangle_2=0$
- B
$\triangle_1+2\triangle_2=0$
- C
$\triangle_1=\triangle_2$
- D
AnswerCorrect option: A. $\triangle_1+\triangle_2=0$
$\triangle_2=\begin{vmatrix}1&\text{bc}&\text{a}\\1&\text{ca}&\text{b}\\1&\text{ab}&\text{c}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}1&\text{abc}&\text{a}^2\\1&\text{bca}&\text{b}^2\\1&\text{cab}&\text{c}^2\end{vmatrix}\left[R_1, R_2, R_3\right.$ are multiplies by $a, b$ and $c$ respectively, therefore we divide by $abc]$
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}1&1&\text{a}^2\\1&1&\text{b}^2\\1&1&\text{c}^2\end{vmatrix} [$Taking $abc$ common from $\left.C _2\right]$
$=-\begin{vmatrix}1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\\1&\text{c}&\text{c}^2\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged. so,
$\triangle_2=-\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2\end{vmatrix} $
$=-\triangle_1$
$\triangle_1+\triangle_2=0$
View full question & answer→MCQ 571 Mark
If $\text{A}=\begin{bmatrix} 3 & 4 \\ 2 & 4 \end{bmatrix},\text{B}=\begin{bmatrix} -2 & -2 \\ 0 & -1 \end{bmatrix}$ then $(A+B)^{-1}=$
AnswerWe have
$(\text{A}+\text{B})=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
$\therefore|\text{A}+\text{B}| = -1\neq0$
Thus, $(A+B)^{-1}$ exists.
Now,
$(\text{A}+\text{B})^\text{T}=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
Here,
$(\text{A}+\text{B})^\text{T}\neq-(\text{A}+\text{B})$
Hence, it is not a akew symmetric matrix.
We also know that $A^{-1}+B^{-1}$ is not the same as $(A+B)^{-1}$.
View full question & answer→MCQ 581 Mark
Maximum value of a second order determinant whose every element is either $0, 1$ or $2$ only is:
AnswerSo, $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$
Given $a, b, c \ \&\ D$ can only be $0, 1, 2$
det $A = ad-bc$
So for max. value of $A,$
$a = 2$ and $d = 2$ and $b, c \in 0, 0$
So, Max value of det $\text{A}=\begin{bmatrix}2&0\\0&2\end{bmatrix}=4$
View full question & answer→MCQ 591 Mark
The equations in terms of $x$ and $y$ are:
- A
$x – y = 50, 2x – y = 550$
- ✓
$x – y = 50, 2x + y = 550$
- C
$x + y = 50, 2x + y = 550$
- D
$x + y = 50, 2x – y = 550$
AnswerCorrect option: B. $x – y = 50, 2x + y = 550$
View full question & answer→MCQ 601 Mark
If $a>0$ and discriminant of $a x^2+2 b x+c$ is negative, then $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$ is:
AnswerDiscriminant $D$ of $a x^2+2 b x+c=(2 b)^2-4 a c<0\ [$Given$]$
$\Rightarrow 4 b^2-4 a c < 0$
$\Rightarrow b^2-a c<0,$ where $a>0 \ldots . .(i)$
$\Rightarrow\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{ax}+\text{b}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}\text{ax}&\text{bx}&\text{ax}^2+\text{bx}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}\ [$Applying $\left.R _1 \rightarrow xR _1\right]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}\text{ax}+\text{b}&\text{bx}+\text{c}&\text{ax}^2+\text{bx}+\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}\ [$Applying $\left.R _1 \rightarrow R _1+ R _2\right]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{vmatrix}0&0&\text{ax}^2+2\text{bx}+\text{c}\\\text{b}&\text{c}&\text{bx}+\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}&0\end{vmatrix}\ [$Applying $\left.R_1 \rightarrow R_1-R_3\right]$
$\Rightarrow\triangle=\frac{1}{\text{x}}\begin{Bmatrix}\text{ax}^2+2\text{bx}+\text{c}\begin{vmatrix}\text{b}&\text{c}\\\text{ax}+\text{b}&\text{bx}+\text{c}\end{vmatrix}\end{Bmatrix}\ [$Expanding along $\left.R _1\right]$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{x}+\text{bc}-\text{acx}-\text{bc})$
$\Rightarrow\triangle=\frac{1}{\text{x}}(\text{ax}^2+2\text{bx}+\text{c})\text{ x }(\text{b}^2-\text{ac})$
$\Rightarrow\triangle=(\text{ax}^2+2\text{bx}+\text{c})(\text{b}^2\text{ac})<0\ [$From $\text{eq. (i)}]$
$\Rightarrow\triangle<0$
View full question & answer→MCQ 611 Mark
If $A$ is an invertible matrix of order $3$, then which of the following is not true:
- A
$|\text{adj A}|=|\text{A}|^2$
- B
$(\text{A}^{-1})^{-1}=\text{A}$
- ✓
If $B A=C A$, than $B \neq C$, where $B$ and $C$ are square matrices of order 3
- D
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1},$ where $\text{B}\neq\big[\text{b}_{\text{ij}}\big]_{3\times3}\text{ and |B|}\neq0$
AnswerCorrect option: C. If $B A=C A$, than $B \neq C$, where $B$ and $C$ are square matrices of order 3
c. If $B A=C A$, than $B \neq C$, where $B$ and $C$ are square matrices of order 3
Solution:
$BA = CA$
$\Rightarrow BAA ^{-1}= CAA ^{-1}$
$\Rightarrow BI = CI$
$\Rightarrow B = C$
Hence, $(c)$ is not correct.
View full question & answer→MCQ 621 Mark
If $A$ is a square matrix such that $A ^2=1,$ then $A ^{-1}$ is equal to:
Answer$A^2=1$
$A^{-1} A^2=A^{-1} I$
$A=A^{-1}$
View full question & answer→MCQ 631 Mark
If $A$ is an invertible matrix, then which of the following is not true:
- ✓
$(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
- B
$|\text{A}^{-1}|=|\text{A}|^{-1}$
- C
$(\text{A}^\text{T})^\text{-1}=(\text{A}^{-1})^\text{T}$
- D
$|\text{A}|\neq0$
AnswerCorrect option: A. $(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
We have, $\left|A^{-1}\right|=|A|^{-1},(AT)^{-1}=\left(A^{-1}\right)^{\top}$ and $|\text{A}|\neq0$ all are the properties of the inverse of a matrix $A.$
View full question & answer→MCQ 641 Mark
If $a, b, c$ are distinct, then the value of $x$ satisfying $\begin{vmatrix}0&\text{x}^2-\text{a}&\text{x}^3-\text{b}\\\text{x}^2+\text{a}&0&\text{x}^2+\text{c}\\\text{x}^4+\text{b}&\text{x}-\text{c}&0\end{vmatrix}=0$ is:
AnswerWhen we put $x = 0$ in the given matrix, then it turns out to be the skew symmetric matrix of order $3$ and the determinant of the skew symmetric matrix of odd order is always $0.$
View full question & answer→MCQ 651 Mark
Choose the correct answer from given four options in each of the Exercise $:$ If $A$ and $B$ are invertible matrices, then which of the following is not correct?
- A
$\text{adj A} = |\text{A}|.\text{A}^{-1}$
- B
det $(A)^{-1}=[$det$(A)]^{-1}$
- C
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
- ✓
$(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
AnswerCorrect option: D. $(\text{A}+\text{B})^{-1}=\text{B}^{-1}+\text{A}^{-1}$
Since$, A$ and $B$ are invertible matrices,
So, we can say that
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}\ \dots(\text{i})$
Also$, \text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj A})$
$\Rightarrow\ \text{adj A}=|\text{A}|.\text{A}^{-1}\ \ \dots(\text{ii})$
Also$,$ det $(A)^{-1}=[$det $(A)]^{-1}$
$\Rightarrow$ det $A)^{-1}=\frac{1}{\big[\text{det (A)}\big]}$
$\Rightarrow$ det $(A).$det $(A)^{-1}=1\ \ \dots(\text{iii})$
Which is true.
Again, $(\text{A}+\text{B})^{-1}=\frac{1}{\big|(\text{A}+\text{B})\big|}\text{ adj }(\text{A}+\text{B})$
$\Rightarrow\ (\text{A}+\text{B})^{-1}\neq\text{B}^{-1}+\text{A}^{-1}\ \ \dots(\text{iv})$
So, only option $(d)$ is incorrect.
View full question & answer→MCQ 661 Mark
If $\text{A}=\begin{bmatrix}\alpha &\text{amp; 2} \\2 &\text{amp; }\alpha \end{bmatrix}$and $ |\text{A}^3|=125,$ then $\alpha$ is equal to:
- ✓
$\pm 3$
- B
$\pm 2$
- C
$\pm 5$
- D
$0$
AnswerCorrect option: A. $\pm 3$
Given, $ |\text{A}^3| =|\text{A}|^3 = 125$
$\Rightarrow|\text{A}|=5$
$\Rightarrow\alpha^2-4=5$
$\Rightarrow\alpha^2=9$
$\Rightarrow \alpha=\pm 3$
View full question & answer→MCQ 671 Mark
If $ \begin{vmatrix} 6\text{i} &\text{amp;} -3\text{i} &\text{amp;} 1\\ 4 &\text{amp; } 3\text{i} &\text{amp;} -1 \\ 20 &\text{amp; } 3 &\text{amp; i}\end{vmatrix}=\text{x}+\text{iy},$ then?
- A
$x = 3, y = 1$
- B
$x = 1, y = 3$
- C
$x = 0, y = 3$
- ✓
$x = 0, y = 0$
AnswerCorrect option: D. $x = 0, y = 0$
$ \begin{vmatrix} 6\text{i} &\text{amp;} -3\text{i} &\text{amp;} 1\\ 4 &\text{amp; } 3\text{i} &\text{amp;} -1 \\ 20 &\text{amp; } 3 &\text{amp; i}\end{vmatrix}=\text{x}+\text{iy},$
$\Rightarrow6\text{i}(3\text{i}^2+3)+3\text{i}(4\text{i}+20)+1(12-60\text{i})=\text{x}+\text{iy}$
$\Rightarrow0=\text{x}+\text{iy}$
$\therefore \text{x}=\text{y}=0$
View full question & answer→MCQ 681 Mark
A determinant of second order is made with the elements $0$ and $1.$ The number of determinants with non $-$ negative values is:
AnswerThere are only three determinants of second order with negative value,
$\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}, \begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }1\end{bmatrix}, \begin{bmatrix}1&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$
Number of possible determinants with elements $0$ and $1$ are ${ 2 }^{ 4 }=16$
therefore, number of determinants with non $-$ negative values is $13.$
View full question & answer→MCQ 691 Mark
If $A$ is a $3 \times 3$ matrix and det$(3A) = k($det $A),$ then $k =$
View full question & answer→MCQ 701 Mark
The system of linear equations : $x + y + z = 2 , 2x + y − z = 3 , 3x + 2y + kz = 4$ has a unique solution if
- A
$k \neq 0$
- B
$−1 < k < 1$
- C
$−2 < k < 2$
- ✓
$k = 0$
AnswerCorrect option: D. $k = 0$
$x + y + z = 2$
$2x + y − z = 3$
$3x + 2y + kz = 4$
The determination of the coefficient matrix $\begin{bmatrix}1&1&1\\2&1&-1\\3&2&\text{k}\end{bmatrix}$ is
$= k + 2 -2k - 3 + 1$
$=-k$
To have a unique solution the determinant $\neq 0$
$\Rightarrow k \neq 0$
View full question & answer→MCQ 711 Mark
If $A$ is a skew symmetric matrix, then $∣A∣$ is
AnswerSince the Skew Symmetric Matrix Consist Of Elements of Opposite Sign at Opposite Side of Matrix Diagonal with all the Diagonal Elements as Zero Therefore the Determination of skew Symmetric matrix is Zero
View full question & answer→MCQ 721 Mark
For which of the following element in the determinant $\triangle=\begin{bmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{bmatrix},$ the minor and the cofactor both are zero.
AnswerConsider 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
$\therefore\text{M}_{22}=\begin{bmatrix}5&8\\5&8\end{bmatrix}=40-40=0$
$\Rightarrow\text{A}^{22}=(-1)^2+2 (0)=0.$
View full question & answer→MCQ 731 Mark
If $A$ is a singular matrix, then $A (\text{adj A})$ is a
AnswerGiven $A$ is a singular matrix.
$\Rightarrow ∣A∣ = 0$
$A \text{(adj A)} = \|A\| = 0I = O$
$\therefore A \text{(adj A)}$ is a zero matrix.
View full question & answer→MCQ 741 Mark
The number of solutions of the system of equations : $2 x+y-z=7, x-3 y+2 z=1 , x+4 y-3 z=5$
AnswerThe 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,
$\text{A}=\begin{bmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$ 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 $c _\text{ij}$ be the co $-$ factors of the elements $a _\text{ij}$ in $A =\left[ a _\text{ij }\right]$. 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.
View full question & answer→MCQ 751 Mark
The number of solutions of the system of equations $2x + y − z = 7, x − 3y + 2z = 1, x + 4y − 3z = 5$
AnswerFrom the given equation we get,
$\triangle=\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix}$
$\Rightarrow 2(9 - 8) -1(-3 - 2) - 1(4 + 3)$
$\Rightarrow 2(1) - 1(-5) - 1(7)$
$\Rightarrow 2 + 5 - 7$
$\Rightarrow 2 + 5 -7$
$\Rightarrow 0$
$\triangle_1=\begin{vmatrix}7&1&-1\\1&-3&2\\5&4&-3\end{vmatrix}$
$\Rightarrow 7(9 - 8) - 1(-3 - 10) - 1(4 + 15)$
$\Rightarrow 7(1) - 1(-13) - 1(19)$
$\Rightarrow 7 + 13 - 19$
$\Rightarrow 20 - 19$
$\Rightarrow1\neq0$
Hence the gvien system no solution.
View full question & answer→MCQ 761 Mark
Choose the correct answer from given four options in each of the Exercise : If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ then value of $x$ is:
AnswerCorrect option: C. $ \pm6$
We have, $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow\ 2\text{x}^2-40=18+17$
$\Rightarrow\ 2\text{x}^2=32+40$
$\Rightarrow\ \text{x}^2=\frac{72}{2}=36$
$\Rightarrow\ \text{x}^2=36$
$\Rightarrow\ \text{x}=\pm6$
View full question & answer→MCQ 771 Mark
The number of distinct real roots of $\begin{vmatrix}\text{cosec}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosecx}\end{vmatrix}=0$ lies in the interval $-\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{4}$ is:
AnswerLet $\triangle=\begin{vmatrix}\text{cosec x}&\sec\text{x}&\sec\text{x}\\\sec\text{x}&\text{cosec}\text{x}&\sec\text{x}\\\sec\text{x}&\sec\text{x}&\text{cosec x}\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&1&\frac{\sec\text{x}}{\text{cosecx}}\\\frac{\sec\text{x}}{\text{cosecx}}&\frac{\sec\text{x}}{\text{cosecx}}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1&\tan\text{x}&\tan\text{x}\\\tan\text{x}&1&\tan\text{x}\\\tan\text{x}&\tan\text{x}&1\end{vmatrix}$
$=(\text{cosec x})^3\begin{vmatrix}1-\tan\text{x}&\tan\text{x}-1&0\\0&1-\tan\text{x}&\tan\text{x}-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix} [$Applying $\left.R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3\right]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{vmatrix}1&-1&0\\0&1&-1\\\tan\text{x}&\tan\text{x}&1\end{vmatrix} [$Taking out $(1-\tan\text{x})$ common from $R _1$ and $R _2]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\begin{Bmatrix}1\begin{vmatrix}1&-1\\\tan\text{x}&1\end{vmatrix}+\tan\text{x}\begin{vmatrix}-1&0\\1&-1\end{vmatrix}\end{Bmatrix} [$Expanding along $\left.C_1\right]$
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+\tan\text{x}+\tan\text{x}\}$
$=(\text{cosec x})^3(1-\tan\text{x})^2\{1+2\tan\text{x}\}$
$\triangle=0$
$=(\text{cosec x})^3(1-\tan\text{x})^2(1+2\tan\text{x})=0$
$(1-\tan\text{x})=0,(\text{coses x})^3=0$ 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.
View full question & answer→MCQ 781 Mark
Which of the following is correct?
- A
Determinant is a square matrix
- B
Determinant is a number associated to a matrix
- ✓
Determinant is a number associated to a square matrix
- D
AnswerCorrect option: C. Determinant is a number associated to a square matrix
Determinant is defined only for a square matrix.
and its denotes the value of that square matrix.
View full question & answer→MCQ 791 Mark
The system of equations : $x + y + z = 5, x + 2y + 3z = 9, x + 3y + \lambda z = \mu$ has a unique solution, if
AnswerCorrect option: B. $\lambda \neq 5$
$x + y + z = 5$
$x + 2y + 3z = 9$
$x + 3y + \lambda z = µ$
The determinant of the coefficient matrix $\begin{bmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{bmatrix}$ is
$= 2\lambda - 9 - \lambda + 3 + 1$
$= \lambda - 5$
For unique solution determinant $\neq 0$
$\Rightarrow \lambda \neq 5$
The right hand side is non zero what so ever be the value of $\mu.$
View full question & answer→MCQ 801 Mark
If $A$ is a square matrix of order $3$ and $|A| = 5,$ then the value of $|2A\ '|$ is :
AnswerAccording to the property of transpose of a matrix,
$(kA\ ') = kA\ '$
Also, from the property of determinant of a matrix,
$|A\ '| = |A|$
Thus $, |2A\ '| = 2|A|$
$= 2 \times 5$
$= 10$
View full question & answer→MCQ 811 Mark
If $A$ is an invertible matrix, then $\operatorname{det}\left(A^{-1}\right)$ is equal to :
AnswerCorrect option: B. $\frac{1}{\text{det(A)}}$
We know that $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$
View full question & answer→MCQ 821 Mark
If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
Answer$\therefore\ \text{Given: Area of triangle}=\text{Modulus of}\ \frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=35$
$\Rightarrow\ \text{Modulus of}\ \frac{1}{2}\begin{vmatrix}2&-6&1\\5&4&1\\k&4&1\end{vmatrix}=35$
$\Rightarrow\ \bigg|\frac{1}{2}\left[2(4-4)-(-6)(5-k)+1(20-4k)\right]\bigg|=35$
$\Rightarrow\ \begin{vmatrix}\frac{1}{2}\left[0+30-6k+20-4k\right]\end{vmatrix}=35\ \Rightarrow\ \begin{vmatrix}\frac{1}{2}\left[50-10k\right]\end{vmatrix}=35$
$\Rightarrow\ \begin{vmatrix}25-5k\end{vmatrix}=35\ \Rightarrow\ \ 25-5k=\pm35$
Taking positive sign, 25 - 5k = 35 $\ \ \Rightarrow k=-2$
Taking negative sign, 25 - 5k = -35 $\Rightarrow k=12$
Therefore, option (d) is correct.
View full question & answer→MCQ 831 Mark
If $\text{A}_{\text{r}}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix},$ then the value of $\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}$ is :
AnswerCorrect option: C. $-2 n^3$
$\text{A}_\text{r}=\begin{vmatrix}1&\text{r}&2^{\text{r}}\\2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\sum\limits_{\text{r}=1}^\text{n}1&\sum\limits_{\text{r}=1}^\text{n}\text{r}&\sum\limits_{\text{r}=1}^\text{n}2\text{r}\\\sum\limits_{\text{r}=1}^\text{n}2&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
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 $\left.R _1 \rightarrow R _1- R _2\right]$
$\Rightarrow\sum\limits_{\text{r}=1}^\text{n}\text{A}_\text{r}=\begin{vmatrix}\text{n}-\text{n}&\frac{\text{n}(\text{n}+1)}{2}-\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1}-2-2^{\text{n}+1}\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=\begin{vmatrix}0&0&-2\\2\text{n}&\text{n}&\text{n}^{2}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}&2^{\text{n}+1} \end{vmatrix}$
$=-2\times\begin{vmatrix}2\text{n}&\text{n}\\\text{n}&\frac{\text{n}(\text{n}+1)}{2}\end{vmatrix}$
$=-2\big[\text{n}^{3}+\text{n}^2-\text{n}^2\big]$
$=-2\text{n}^3$
View full question & answer→MCQ 841 Mark
The value of $\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$ is :
Answer$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\\text{n}&\text{n}+2&\text{n}+3\\\frac{\text{n}(\text{n}-1)}{2}&\frac{(\text{n}+2)(\text{n}+1)}{2}&\frac{(\text{n}+4)(\text{n}+3)}{2}\end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix}\ [$Applying $C _2 \rightarrow C _2- C _1$ and $\left.C _3 \rightarrow C _3- C _1\right]$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$
$=8\text{n}+12-8\text{n}-4$
$=8$
Hence, the correct option is $(c)$
View full question & answer→MCQ 851 Mark
If $\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix},$ then the value of $|\text{adj A}|$ is:
- A
$a^{27}$
- B
$a^9$
- ✓
$a^6$
- D
$a^2$
Answer$\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}$
$\therefore|\text{A}|=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}=\text{a}^3\neq0$
and
$n = 3$
Thus, we have
$|\operatorname{adj} A|=|A|^{n-1}=\left(a^3\right)^2=a^6$
View full question & answer→MCQ 861 Mark
If $\text{x},\text{ y}\in\text{R},$ then the determinant $\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$ lies in the interval :
AnswerCorrect option: A. $\Big[-\sqrt{2},\sqrt{2}\Big]$
$\triangle=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&0\end{vmatrix}$
$=\begin{vmatrix}\cos\text{x}&-\sin\text{x}&1\\\sin\text{x}&\cos\text{x}&1\\0&0&\sin\text{y}-\cos\text{y}\end{vmatrix}\ [$Applying $\left.R_3 \rightarrow R_3-\cos y R_1+\sin y R_2\right]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 871 Mark
If $A^5 = 0$ Such that $\text{A}^{\text{n}}\neq\text{I}$ for $1\leq\text{n}\leq4,$ then $(\text{I}-\text{A})^{-1}$ equals :
Answer$A^5=0$
Using $a^5-b^5=(a-b)\left(a^4+a^3 b+a^2 b^2+a b^3+b^4\right)$
$I-A^5=(I-A)\left(I+A+A^2+A^3+A^4\right)$
$I=(I-A)\left(I+A+A^2+A^3+A^4\right)$
$(I-A)^{-1} I=(I-A)^{-1}(I-A)\left(I+A+A^2+A^3+A^4\right)$
$(I-A)-1=I+A+A^2+A^3+A^4$
View full question & answer→MCQ 881 Mark
If two rows of a determinant are identical, then what is the value of the determinant ?
AnswerLet determinant of this matrix is $x,$ if we interchange the two identical rows of the matrix then by property the determinant of the new matrix is $- x,$ but overall the matrix will be same as we have interchanged only the two identical rows.
So, $x = -x,$ we have $x = 0.$
Hence, the determinant is zero.
View full question & answer→MCQ 891 Mark
$A$ and $B$ are two points and $C$ is any point collinear with $A$ and $B.$ If $AB=10, BC=5,$ then $AC$ is equal to :
- ✓
either $15$ or $5$
- B
necessarily $5$
- C
necessarily $16$
- D
AnswerCorrect option: A. either $15$ or $5$
Since $C$ is collinear with $A $ and $B,C$ lies either
$(i)$ to the left of point $B$ or
$(ii)$ to the right of point $B$
$\therefore $ In case $(i)\ AC = AB - BC = 10 - 5 = 5$
In case $(ii)\ AC = AB + BC = 10 + 5 = 15$
View full question & answer→MCQ 901 Mark
Evaluate $\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$ is :
Answer$\triangle=\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$
Taking $4, 6$ and $7$ from $\text{R}_1, \text{ R}_2,\text{ R}_3$ respectively
$\triangle=4\times6\times7\begin{bmatrix}4&8&12\\6&12&18\\7&14&21\end{bmatrix}$
Since the elements of all rows are identical, the determinant is zero.
View full question & answer→MCQ 911 Mark
Find the value of $x$ if $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}.$
- ✓
$\text{x}=1,-\frac{1}{3}$
- B
$\text{x}=-1,-\frac{1}{3}$
- C
$\text{x}=1,\frac{1}{3}$
- D
$\text{x}=-1,\frac{1}{3}$
AnswerCorrect option: A. $\text{x}=1,-\frac{1}{3}$
Given that $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}$
$\Rightarrow3\text{x}^2-2\text{x}=5(2)-3(3)$
$\Rightarrow 3\text{x}^2-2\text{x}=1$
solving for $x,$ we get
$\text{x}=1,-\frac{1}{3}$
View full question & answer→MCQ 921 Mark
Consider the system of equations : $a_1 x+b_1 y+c_1 z=0 , a_2 x+b_2 y+c_2 z=0, a_3 x+b_3 y+c_3 z=0,$ if $\begin{vmatrix}\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\\\text{a}_3&\text{b}_3&\text{c}_3\end{vmatrix}=0,$ then the system has
- ✓
- B
One trivial and one non $-$ trivial solutions.
- C
- D
Only trivial solution $(0, 0, 0).$
AnswerHere, $|A|=0$ and $B=0\ ($Given$) .$
If $|A|=0$ and $(\operatorname{adj} A) B=0,$ then the system is consistent and has infinitely many solutions.
Clearly, it has more than two solutions.
View full question & answer→MCQ 931 Mark
Find the value of $x,$ if $\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$ is:
Answer$\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$
$\Rightarrow2\text{x}-15=3\text{x}+5 $
$\Rightarrow\text{x}=-20$
View full question & answer→MCQ 941 Mark
If $\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}-1=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix},$ then :
- A
$\text{a}=1,\text{b}=1$
- ✓
$\text{a}=\cos2\theta,\text{b}=\sin2\theta$
- C
$\text{a}=\sin2\theta,\text{b}=\cos2\theta$
- D
AnswerCorrect option: B. $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
On comparing, we get
$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}$ and ${b}=\frac{2\tan\theta}{\sec^2\theta}$
$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta$ and ${b}=2\sin\theta\cos\theta$
$\Rightarrow\text{a}=\cos2\theta$ and ${b}=\sin2\theta$
View full question & answer→MCQ 951 Mark
Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to :
- A
$\begin{bmatrix}1\\2\\3\end{bmatrix}$
- B
$\begin{bmatrix}-1\\-2\\-3\end{bmatrix}$
- ✓
$\begin{bmatrix}-1\\2\\3\end{bmatrix}$
- D
$\begin{bmatrix}0\\2\\1\end{bmatrix}$
AnswerCorrect option: C. $\begin{bmatrix}-1\\2\\3\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$
$|\text{A}|=1(0-2)+1(2-3)+2(4-0)$
$=-2-1+8$
$=5$
Let $C _\text{ij }$ be the co $-$ factors of the elements $a_\text{ij }$ in $A =\left[ a _\text{ij }\right]$.
Then, $\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}0&1\\2&1\end{vmatrix}=-2,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}2&1\\3&1\end{vmatrix}=1,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}2&0\\3&2\end{vmatrix}=4$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}-1&2\\2&1\end{vmatrix}=5,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&2\\3&1\end{vmatrix}=-5,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&-1\\3&2\end{vmatrix}=-5$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}-1&2\\0&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&2\\2&1\end{vmatrix}=3,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&-1\\2&0\end{vmatrix}=2$
$\text{adj }\text{A}=\begin{bmatrix}-2&5&-1\\5&-5&-5\\-1&2&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-2&5&-1\\1&-5&3\\4&-5&2\end{bmatrix}\begin{bmatrix}3\\1\\4\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{5}\begin{bmatrix}-6+5-4\\3-5+12\\12-5+8\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\frac{1}{5}\begin{bmatrix}-5\\10\\15\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}=\begin{bmatrix}-1\\2\\3\end{bmatrix}$
View full question & answer→MCQ 961 Mark
Evaluate $\begin{bmatrix}\text{x}^2&\text{x}^3&\text{x}^4\\\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{x}^3&\text{x}^4\end{bmatrix}$ is:
- ✓
$0$
- B
$1$
- C
$xyz$
- D
$x^2 y z^3$
Answer$\begin{bmatrix}\text{x}^2&\text{x}^3&\text{x}^4\\\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{x}^3&\text{x}^4\end{bmatrix}$
If the elements of any two rows or columns are identical, then the value of determinant is zero.
Here, the elements of row $1$ and row $3$ are identical.
Hence, its determinant is $0$.
View full question & answer→MCQ 971 Mark
The existence of the unique solution of the system of equations : $x + y + z = \lambda , 5x − y + µz = 10, 2x + 3y − z = 6$ depends on
AnswerCorrect option: A. $\mu$ only.
For a unique solution, $|\text{A}|\neq0$
$\Rightarrow\begin{vmatrix}1&1&1\\5&-1&\mu\\2&3&-1\end{vmatrix}\neq0$
$\Rightarrow1(1-3\mu)-1(-5-2\mu)+1(15+2)\neq0$
$\Rightarrow1-3\mu+5+2\mu+17\neq0$
$\Rightarrow-\mu+23\neq0$
$\Rightarrow\mu\neq23$
So, existence of a unique solution depends only on $\mu$.
View full question & answer→MCQ 981 Mark
If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ then $x =$
AnswerCorrect option: C. $\pm6$
$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow2\text{x}^2-40=18+14$
$\Rightarrow2\text{x}^2-40=32$
$\Rightarrow2\text{x}^2=72$
$\Rightarrow\text{x}^2=36$
$\Rightarrow\text{x}=\pm6$
Hence, the correct option is $(C)$
View full question & answer→MCQ 991 Mark
The value of $y ($breadth of rectangular field$)$ is:
- ✓
$150m$
- B
$200m$
- C
$430m$
- D
$350m$
AnswerCorrect option: A. $150m$
View full question & answer→MCQ 1001 Mark
$\begin{bmatrix}2\text{x}&5\\8&\text{x}\end{bmatrix}=\begin{bmatrix}6\text{x}&-2\\7&3\end{bmatrix},$ then the value of $\text{x}$ is:
AnswerCorrect option: C. $\pm6$
View full question & answer→