MCQ 1011 Mark
If $A^5 = 0 $ Such that $\text{A}^{\text{n}}\neq\text{I for }1\leq\text{n}\leq4,\text{ then}(\text{I}-\text{A})^{-1}$ equals:
AnswerCorrect option: C. $I + A$
$A^5 = 0$
Using $a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$
$I - A^5 = (I - A)(I + A + A^2 + A^3 + A^4)$
$I = (I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)^{-1}I = (I - A)^{-1}(I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)-1 = I + A + A^2 + A^3 + A^4$
View full question & answer→MCQ 1021 Mark
Evaluate $\begin{bmatrix}8\text{x}+1&2\text{x}-2\\\text{x}^2-1&3\text{x}+5\end{bmatrix}$ is:
- A
$-2x^{3 }- 26x^{2 }+ 45x + 3$
- ✓
$-2x^{3 }+ 26x^{2 }+ 45x + 3$
- C
$-2x^{3 }+ 26x^{2 }+ 45x - 3$
- D
$-2x^{3 }- 26x^2- 45x + 3$
AnswerCorrect option: B. $-2x^{3 }+ 26x^{2 }+ 45x + 3$
Expanding along the first row, we get
$\triangle=8\text{x}+1(3\text{x}+5)-(2\text{x}-2)(\text{x}^2-1)$
$=(24\text{x}^2+43\text{x}+5)-(2\text{x}^3-2\text{x}^2-2\text{x}+2)$
$=-2\text{x}^3+26\text{x}^2+45\text{x}+3.$
View full question & answer→Question 1031 Mark
If $\text{A} = \begin{bmatrix}1&\text{amp; } \log_{\text{b}}\text{a}\\ \log_\text{a}\text{b}&\text{amp; } 1\end{bmatrix}$then $ |\text{A}|$ is equal to:
- $0$
- $\log_\text{a}\text{b}$
- $-1$
- $\log_\text{b}\text{a}$
Answer
- $0$
Solution:
On solving the given matrix,
$|\text{A}|=1-\log_\text{a}\text{b}.\log_\text{b} \text{a}=1-1=0$ View full question & answer→Question 1041 Mark
If $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is prthogonal, than x + y =
- 3
- 0
- -3
- 1
Answer
- $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is not an prthogonal, question is incorrect.
Solution:
$\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$
$\text{A}^\text{T}\text{A}=\text{I}$
$\frac{1}{3}\begin{bmatrix} 1 & 2 & \text{x} \\ 1 & 1 & 2 \\ 2 & -2 & \text{y} \end{bmatrix}\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 &\text{y} \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\frac{1}{9}\begin{bmatrix} 1+4+\text{x}^2 & 1+2+2\text{x} & \text{xy}-2 \\ 1+2+2\text{x} & 1+1+4 & 2-2+2\text{y} \\ 2-4+\text{xy} & 2-2+2\text{y} & 4+4+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 5+\text{x}^2 & 3+2\text{x} & \text{xy}-2 \\ 3+2\text{x} & 6 & 2\text{y} \\ -6+\text{xy} & 2\text{y} & 8+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Equality of two matrices does not hold Matrix A is not orthogonal.
Hence, the given question is incorrect. View full question & answer→Question 1051 Mark
If $\text{S}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},$ then adj A is:
- $\begin{bmatrix} -\text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- $\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- $\begin{bmatrix} \text{d} & \text{b} \\ \text{c} & \text{a} \end{bmatrix}$
- $\begin{bmatrix} \text{d} & \text{c} \\ \text{b} & \text{a} \end{bmatrix}$
Answer
- $\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Solution:
Adjoint of a square matrix of order 2 is obtained by interchancing the diagoinal elements and changing the signs of off-diagonal elements.
Here,
$\text{A}=\begin{bmatrix}\text{a} & \text{bc} & \text{d} \end{bmatrix}$
$\Rightarrow\text{adj A}=\begin{bmatrix}\text{d} & -\text{b}-\text{c} & \text{a} \end{bmatrix}$ View full question & answer→MCQ 1061 Mark
Choose the correct answer from given four options in each of the Exercise:
If $\text{A}=\begin{vmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{vmatrix},$ then $A^{-1}$ exists, if:
- A
$\lambda=2$
- B
$\lambda\neq2$
- C
$\lambda\neq-2$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
We have, $\text{A}=\begin{vmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{vmatrix}$
Expanding along $R_1,$ we get
$\text{A}=2(6-5)-\lambda(-5)-3(-2)$
$=2+5\lambda+6$
$=5\lambda+8$
We know that, $A^{-1}$ exists, if $A$ is non-singular matrix i.e., $|\text{A}|\neq0.$
$\therefore\ 5\lambda+8\neq0$
$\Rightarrow\ 5\lambda\neq-8$
$\therefore\ \lambda\neq\frac{-8}{5}$
Thus, $A^{-1}$ exists for all values of $\lambda\text{ except }\frac{-8}{5}.$
View full question & answer→MCQ 1071 Mark
The value of $\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$ is:
- A
$5^2$
- ✓
$0$
- C
$5^{13}$
- D
$5^9$
Answer$\begin{vmatrix}5^2&5^3&5^4\\5^3&5^4&5^5\\5^4&5^5&5^6\end{vmatrix}$
$=5^2\times5^3\times5^4\begin{vmatrix}1&5&5^2\\1&5&5^2\\1&5&5^2\end{vmatrix} [$Taking out common factors from $R_1, R_2, R_3]$
$=5^2\times5^3\times5^4\times5\begin{vmatrix}1&1&5^2\\1&1&5^2\\1&1&5^2\end{vmatrix}$
$=5^2\times5^3\times5^4\times0$
$=0$
View full question & answer→Question 1081 Mark
If the points $(\text{k} + 1, 1), (2\text{k} + 1, 3)$ and $(2\text{k} + 2, 2\text{k})$ are collinear, then the value of $\text{k}$is:
- $2$
- $-2$
- $\frac{1}{2}$
- $1$
View full question & answer→Question 1091 Mark
For which of the following elements in the determinant $\triangle\begin{bmatrix}2&8\\4&7\end{bmatrix},$ the minor of the element is 2:
- 2
- 7
- 4
- 8
Answer
- 7
Solution:
Consider the element 7 in the determinant $\triangle\begin{bmatrix}2&8\\4&7\end{bmatrix}$
The minor of the element 7 can be obtained by deleting $\text{R}_2$ and $\text{C}_2$
$\therefore\text{M}_{22}=2$
Hence, the minor of the element 7 is 2. View full question & answer→Question 1101 Mark
If A is a singular matrix, then adj A is:
- Non-singular.
- Singular.
- Symmetric.
- Not defined.
Answer
- Singular.
Solution:
If A is singular matrix then adjoint of A is also singular.
View full question & answer→MCQ 1111 Mark
Choose the correct answer from given four options in each of the Exercise:
There are two values of a which makes determinant $\begin{bmatrix}1 & -2 & 5 \\2 & \text{a} & 1\\0 & 4 & 2\text{a} \end{bmatrix} = 86,$ then sum of these number is:
AnswerWe have,
$\begin{bmatrix}1 & -2 & 5 \\2 & \text{a} & 1\\0 & 4 & 2\text{a} \end{bmatrix} = 86,$
$\Rightarrow\ 1(2\text{a}^{2} + 4) - 2(-4\text{a} - 20) + 0 = 86 [$Expanding along $C_1]$
$\Rightarrow\ \text{a}^{2} + 4\text{a} - 21 = 0$
$\Rightarrow\ \text{(a + 7)} (\text{a} - 3) = 0$
$\Rightarrow\ \text{a} = -7 \text{ and } 3$
$\therefore$ Required sum $= -7 + 3 = -4$
View full question & answer→Question 1121 Mark
If A is a skew symmetric matrix, then ∣A∣ is:
- 11
- -1
- 0
- None
Answer
- 0
Solution:
Since 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 determinant of skew symmetric matrix is zero.
View full question & answer→MCQ 1131 Mark
The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals:
Answer$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$
$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$
$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Taking $(b - a)$ common from $C_1$ and $C_3]$
$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix} [$Applying $C_{1 }\rightarrow C_1 - C_2 - C_3]$
$=0$
Hence, the correct option is $(d)$
View full question & answer→Question 1141 Mark
Evaluate $\begin{bmatrix}2&5\\-1&-1\end{bmatrix}$
- 3
- -7
- 5
- -2
Answer
- 3
Solution:
Expanding along $\text{R}_1,$ we get
$\triangle=2(-1)-5(-1)=2+5$
$=3$ View full question & answer→Question 1151 Mark
Find the cofactor of element -3 in the determinant $\triangle=\begin{bmatrix}1&4&4\\-3&5&9\\2&1&2\end{bmatrix}$ is:
- -4
- 4
- -5
- -3
Answer
- -4
Solution:
The minor of element -3 is given by
$\text{M}_{21}=\begin{bmatrix}4&4\\1&2\end{bmatrix}=4(2)-4=4$ (Obtained by eliminating $\text{R}_2$ and $\text{C}_1$)
$\therefore\text{A}_{21}=(-1)^{2+1} \text{M}_{21}=(-1)^3 4=-4.$ View full question & answer→MCQ 1161 Mark
Choose the correct answer from given four options in each of the Exercise: The area of a triangle with vertices $(-3, 0), (3, 0)$ and $(0, k)$ is $9$ sq. units. The value of k will be:
AnswerWe know that, area of a triangle with vertices$ (x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by
$\triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_2&\text{y}_2&1\\\text{x}_3&\text{y}_3&1\end{vmatrix}\end{vmatrix}$
$\therefore$ Area of triangle with vertices $(-3, 0), (3, 0)$ and $(0, k)$ is
$\therefore\ \ \triangle=\begin{vmatrix}\frac{1}{2} \begin{vmatrix}-3&0&1\\3&0&1\\0&\text{k}&1\end{vmatrix}\end{vmatrix}=9 \ ($given$)$
$\Rightarrow\ [-3(-\text{k)}-0+1(3\text{k})]=\pm18$
$\Rightarrow\ 6\text{k}=\pm18$
$\therefore \ \text{k}=\pm\frac{18}{6}=\pm3$
View full question & answer→Question 1171 Mark
If one of the roots of $\left |\begin{matrix} 3 &\text{amp; 5} &\text{amp; x}\\ 7 &\text{amp; x} &\text{amp; 7}\\ \text{x} &\text{amp; 5} &\text{amp; 3}\end{matrix}\right |=0$ is $-10,$ the other roots are:
- 3, 7
- 4, 7
- 3, 9
- 3, 4
Answer
- 3, 7
Solution:
Given, $\left |\begin{matrix} 3 &\text{amp; 5} &\text{amp; x}\\ 7 &\text{amp; x} &\text{amp; 7}\\ \text{x} &\text{amp; 5} &\text{amp; 3}\end{matrix}\right |=0$
$\Rightarrow3(3\text{x}-35)-5(21-7\text{x})+\text{x}(35-\text{x}^2)=0$
$\Rightarrow9\text{x}-105-105+35\text{x}+35\text{x}-\text{x}^3=0$
$\Rightarrow\text{x}^3-79\text{x}+210=0$
$\Rightarrow(\text{x}+10)(\text{x}-3)(\text{x}-7)=0$
$\Rightarrow\text{x}=10, 3, 7$ View full question & answer→MCQ 1181 Mark
Choose the correct answer If $a, b, c,$ are in $A.P,$ then the determinant:
$\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+2\text{a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$ is
Answer$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+2\text{b}\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}+2&\text{x}+3&\text{x}+\text{2a}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\\text{x}+4&\text{x}+5&\text{x}+2\text{c}\end{vmatrix} (2b = a + c$ as $a,b$ and $c$ are in $A.P.)$
Applying$ R_{1 }\rightarrow R_1 - R_2$ and $R_3 \rightarrow R_3 - R_2,$ we have:
$\triangle=\begin{vmatrix}-1&-1&\text{a}-\text{c}\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$
Applying $R_1 \rightarrow R_1 + R_3,$ we have:
$\triangle=\begin{vmatrix}0&0&0\\\text{x}+3&\text{x}+4&\text{x}+(\text{a}+\text{c})\\1&1&\text{c}-\text{a}\end{vmatrix}$
View full question & answer→MCQ 1191 Mark
If $\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix},$ then $aI + bA + 2 A^2$ equals:
Answer$\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1+\text{a} & \text{b} & 3 \\ \text{a} & \text{b} & 2 \\ 3\text{a} & 2\text{b} & \text{a}+\text{b}+4 \end{bmatrix}$
$\Rightarrow\text{aI}+\text{bA}+2\text{A}^2$
$=\begin{bmatrix} 3\text{a}+2+\text{b} & 2\text{b} & 6+\text{b} \\ 2\text{a} & \text{a}+2\text{b} & \text{b}+4 \\ \text{ab}6\text{a} & 6\text{b}+\text{b}^2 & 3\text{a}+4\text{b}+8 \end{bmatrix}$
View full question & answer→Question 1201 Mark
If x, y, z are non-zero real numbers, then the inverse, then the inverse of the matrix $\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$, is:
- $\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- $\text{xyz}\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- $\frac{1}{\text{xyz}}\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$
- $\frac{1}{\text{xyz}}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Answer
- $\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
Solution:
A = IA
$\Rightarrow\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\text{x}^1 & 0 & 0\\ 0 & \text{y}^1 & 0 \\ 0 & 0 & \text{z}^1\end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1=\frac{1}{\text{x}}\text{R}_1,\text{R}_2=\frac{1}{\text{y}}\text{R}_2\text{ and R}_3=\frac{1}{\text{z}}\text{R}_3\Big]$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$ View full question & answer→Question 1211 Mark
Find the value of the following determinant: $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$
- $\displaystyle \frac{15}{34}$
- $\displaystyle \frac{32}{45}$
- $\displaystyle \frac{25}{33}$
- $\displaystyle \frac{38}{35}$
Answer
- $\displaystyle \frac{38}{35}$
Solution:
The value of $\begin{vmatrix}\displaystyle \frac{-4}{7} &\text{amp; } \displaystyle \frac{-6}{35}\\ 5 &\text{amp; } \displaystyle \frac{-2}{5}\end{vmatrix}$ is $\bigg(\frac{-4}{7}\times\frac{-2}{5}\bigg)-\bigg(\frac{-4}{7}\times5\bigg)$
$=\frac{8}{25}+\frac{30}{35}=\frac{38}{35}$ View full question & answer→Question 1221 Mark
If A is a singular matrix, then adj A is.
- non−singular
- singular
- symmetric
- not defined
Answer
- singular
Solution:
Given ∣A∣ = 0
We know ∣adjA∣ = ∣A∣ n - 1
∴ ∣adjA∣ = 0
Hence, adj A is singular
View full question & answer→Question 1231 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:
- f(a) = 0
- f(b) = 0
- f(0) = 0
- f(1) = 0
Answer
- f(0) = 0
Solution:
$\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b} \\\text{x}+\text{a} &0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix}$
$\Rightarrow\ \text{f}(0)=\begin{vmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{vmatrix},$ Which is skew-symmetric determinant of order 3
Hence f(0) = 0. View full question & answer→MCQ 1241 Mark
If $A$ and $B$ are square matrices such that $B = -A^{-1} BA,$ then $(A + B)^2 =$
- A
$O$
- ✓
$A^2 + B^2$
- C
$A^2 + 2AB + B^2$
- D
$A + B$
AnswerCorrect option: B. $A^2 + B^2$
$B = -A^{-1} BA$
$\Rightarrow AB = -AA^{-1}BA$
$\Rightarrow Ab = -IBA$
$\Rightarrow AB = -BA$
$\Rightarrow AB + BA = 0 .....(i)$
Consider,
$(A + B)^2 = A^2 + AB + BA + B^2$
$(\because\text{AB}\neq\text{BA})$
$(A + B)^2 = A^2 + O + B^2$
$(A + B)^2 = A^2 + B^2$
View full question & answer→MCQ 1251 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 1261 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 $BA = CA,$ than $\text{B}\neq\text{C},$ 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 $BA = CA,$ than $\text{B}\neq\text{C},$ where $B$ and $C$ are square matrices of order $3$
$BA = CA$
$\Rightarrow BAA^{-1} = CAA^{-1}$
$\Rightarrow BI = CI$
$\Rightarrow B = C$
Hence, $(c)$ is not correct.
View full question & answer→Question 1271 Mark
Maximum value of a second order determinant whose every element is either 0, 1 or 2 only is:
- 0
- 1
- 2
- 4
Answer
- 4
Solution:
So, $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$
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→Question 1281 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:
- c
- a
- b
- 0
Answer
- 0
Solution:
When 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 1291 Mark
If $A$ is a square matrix such that $A^2 = I,$ then $A^{-1}$ is equal to:
Answer$A^2 = I$
$A^{-1}A^2 = A^{-1}I$
$A = A^{-1}$
View full question & answer→Question 1301 Mark
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4
has a unique solution if
- k ≠ 0
- −1 < k < 1
- −2 < k < 2
- k = 0
Answer
- k ≠ 0
Solution:
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 ≠ 0
⇒ k ≠ 0 View full question & answer→Question 1311 Mark
The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
- 3
- 2
- 1
- 0
Answer
- 0
Solution:
From the given equation we get,
$\triangle=\begin{vmatrix}2&1&-1\\1&-3&2\\1&4&-3\end{vmatrix}$
⇒ 2(9 - 8) -1(-3 - 2) - 1(4 + 3)
⇒ 2(1) - 1(-5) - 1(7)
⇒ 2 + 5 - 7
⇒ 2 + 5 -7
⇒ 0
$\triangle_1=\begin{vmatrix}7&1&-1\\1&-3&2\\5&4&-3\end{vmatrix}$
⇒ 7(9 - 8) - 1(-3 - 10) - 1(4 + 15)
⇒ 7(1) - 1(-13) - 1(19)
⇒ 7 + 13 - 19
⇒ 20 - 19
$\Rightarrow1\neq0$
Hence the gvien system no solution. View full question & answer→Question 1321 Mark
Find the value of x if $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}.$
- $\text{x}=1,-\frac{1}{3}$
- $\text{x}=-1,-\frac{1}{3}$
- $\text{x}=1,\frac{1}{3}$
- $\text{x}=-1,\frac{1}{3}$
Answer
- $\text{x}=1,-\frac{1}{3}$
Solution:
Given that $\begin{bmatrix}3&\text{3}\\2&\text{x}^2\end{bmatrix}=\begin{bmatrix}5&3\\3&2\end{bmatrix}$
$\Rightarrow3\text{x}^2-2\text{x}=5(2)-3(3)$
$⇒3\text{x}^2-2\text{x}=1$
solving for x, we get
$\text{x}=1,-\frac{1}{3}$ View full question & answer→MCQ 1331 Mark
Consider the system of equations:
$a_1x + b_1y + c_1z = 0 , a_2x + b_2y + c_2z = 0 , a_3x + b_3y + c_3z = 0,$ 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 $(\text{adj} \ A)B = 0,$
then the system is consistent and has infinitely many solutions.
Clearly, it has more than two solutions.
View full question & answer→Question 1341 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:
- $\text{a}=1,\text{b}=1$
- $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
- $\text{a}=\sin2\theta,\text{b}=\cos2\theta$
- None of these.
Answer
- $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
Solution:
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
On comparing, we get
$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}\text{ and b}=\frac{2\tan\theta}{\sec^2\theta}$
$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta\text{ and b}=2\sin\theta\cos\theta$
$\Rightarrow\text{a}=\cos2\theta\text{ and b}=\sin2\theta$ View full question & answer→MCQ 1351 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 $R_3 \rightarrow R_3 \ - \cos y R_1\ + \sin y R_2]$
$=(\sin\text{y}-\cos\text{y})(\cos^2\text{x}+\sin^2\text{x})$
$=\sin\text{y}-\cos\text{y}$
$=\sqrt{2}\Big(\frac{1}{\sqrt{2}}\sin\text{y}-\frac{1}{\sqrt{2}}\cos\text{y}\Big)$
$=\sqrt{2}\Big(\cos\frac{\pi}{4}\sin\text{y}-\sin\frac{\pi}{4}\cos\text{y}\Big)$
$=\sqrt{2}\sin\Big(\text{y}-\frac{\pi}{4}\Big)$
Therefore, $-\sqrt{2}\leq\triangle\leq\sqrt{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 1361 Mark
Let $\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix},\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$. If $AX = B,$ then $X$ is equal to:
- ✓
$\begin{bmatrix}1\\2\\3\end{bmatrix}$
- B
$\begin{bmatrix}-1\\-2\\-3\end{bmatrix}$
- C
$\begin{bmatrix}-1\\2\\3\end{bmatrix}$
- D
$\begin{bmatrix}0\\2\\1\end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}1\\2\\3\end{bmatrix}$
Here,
$\text{A}=\begin{bmatrix}1&-1&2\\2&0&1\\3&2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}_1\\\text{x}_2\\\text{x}_3\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}3\\1\\4\end{bmatrix}$
$|\text{A}|=1(0-2)+1(2-3)+2(4-0)$
$=-2-1+8$
$=5$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}].$ 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→Question 1371 Mark
The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on
- µ only.
- λ only.
- λ and µ both.
- neither λ nor µ.
Answer
- $\mu$ only.
Solution:
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 1381 Mark
If the determinant $\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}=0,$ then:
- A
$a, b, c$ are in $H.P.$
- ✓
$\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $G.P.$
- C
$a, b, c$ are in $G.P.$ only.
- D
$a, b, c$ are in $A.P.$
AnswerCorrect option: B. $\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$ or $a, b, c$ are in $G.P.$
Let $\triangle=\begin{vmatrix}\text{a}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}&\text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+3\text{b}&2\text{b}\alpha+3\text{c}&0\end{vmatrix}$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2\text{a}\alpha+ 3\text{b}-2\text{b}\alpha-3&2\text{b}\alpha+3\text{c}&0\end{vmatrix} [$Applying $C_1 \rightarrow C_1- C_2]$
$=\begin{vmatrix}\text{a}-\text{b}&\text{b}&2\text{a}\alpha+3\text{b}\\\text{b}-\text{c}& \text{c}&2\text{b}\alpha+3\text{c}\\2(\text{a}-\text{b})\alpha+3(\text{b}-\text{c})&2\text{b}\alpha+3\text{b}&0\end{vmatrix}$
$=2\alpha(2\text{a}\alpha+3\text{b})-3(2\text{b}\alpha+3\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}\\\text{b}-\text{c}&\text{c}\end{vmatrix} [$Expanding along $R_3]$
$=-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)$
But $\triangle=0 [$Given$]$
$\Rightarrow-(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})(\text{ac}-\text{b}^2)=0$
$\Rightarrow(4\text{a}\alpha^2+12\text{b}\alpha+9\text{c})=0$
Or $(\text{ac}-\text{b}^2)=0$
$\Rightarrow\alpha$ is a root of $4ax^2 + 12bx + 9c = 0$
Or $ac = b^2,$ i.e. $a, b, c$ are in $G.P.$
View full question & answer→Question 1391 Mark
If $\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$ then $|\text{A}|$
- 0
- 10
- 12
- 60
Answer
- 0
Solution:
$\text{A}=\begin{vmatrix} 10 & 2 \\ 30 & 6 \end{vmatrix}$
$=10\times6-30\times2=60-60=0$ View full question & answer→MCQ 1401 Mark
Let $\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix},$ then $\lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}$ is equal to:
Answer$\text{f(x)}=\begin{vmatrix}\cos\text{x}&\text{x}&1\\2\sin\text{x}&\text{x}&2\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix}$$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}&\text{x}&\text{x}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_3]$
$=\begin{vmatrix}\cos\text{x}&\text{x}&1\\\sin\text{x}&0&\text{x}\\\sin\text{x}-\cos\text{x}&0&\text{x}-1\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=-\text{x}[\text{x}\sin\text{x}-\sin\text{x}-\text{x}\sin\text{x}+\text{x}\cos\text{x}]$
$=-\text{x}(\text{x}\cos\text{x}-\sin\text{x})$
$\therefore\ \lim_\limits{\text{x}\rightarrow0}\frac{\text{f(x)}}{\text{x}^2}=\lim_\limits{\text{x}\rightarrow0}\frac{\text{x}(\sin\text{x}-\text{x}\cos\text{x})}{\text{x}^2}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}^2}-\lim_\limits{\text{x}\rightarrow0}\cos\text{x}$
$=1-1=0$
Hence, the correct option is $(a)$
View full question & answer→Question 1411 Mark
The value of the determinant $\begin{vmatrix} 5 &\text{amp; } 1 \\ 3 &\text{amp; } 2 \end{vmatrix}$
- 4
- 5
- 6
- 7
Answer
- 7
Solution:
5 × 2 - 1 × 3 = 7
View full question & answer→Question 1421 Mark
The matrix $\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is a singular matrix, if the value of b is:
- -3
- 3
- 0
- Now-existent
Answer
- Non-existent.
Solution:
$\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is singular matrix.
$\Rightarrow\begin{vmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{vmatrix}=0$
⇒ 5(-4b + 12) - 10(-2b + 6) + 3(4 - 4) = 0
⇒ -20b + 60 + 20b - 60 = 0
b does not exist. View full question & answer→MCQ 1431 Mark
If $\text{A}=\begin{vmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}\end{vmatrix}$ and $C_{ij}$ is cofactor of $a_{ij}$ in a, then value of $|A|$ is given by:
- A
$a_{11}C_{31} + a_{12}C_{32} + a_{13}C_{33}$
- B
$a_{11}C_{11} + a_{12}C_{21} + a_{13}C_{31}$
- C
$a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13}$
- ✓
$a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
AnswerCorrect option: D. $a_{11}C_{11} + a_{21}C_{21} + a_{13}C_{31}$
Properties of determinants state that if $a$ is a square matrix of the order $n,$ then Det $(A)$ is the sum of products of elements of a row $($or a column$)$ with the corresponding cofactor of that element.
View full question & answer→Question 1441 Mark
If $\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix},$ then $\text{A}^{-1}$ exists if:
- $\lambda=2$
- $\lambda\neq2$
- $\lambda\neq-2$
- $\text{None of these}$
Answer
- $\text{None of these}$
Solution:
$\text{A}=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}$
The inverse of a matrix exists if its determinant is not equal to 0.
Consider,
$|\text{A}|=\begin{bmatrix}2&\lambda&-3\\0&2&5\\1&1&3\end{bmatrix}\neq0$
$\Rightarrow|\text{A}| = 2 (6 – 5) – \lambda (0 – 5) + (-3) (0 – 2)\neq0$
$\Rightarrow2 + 5\lambda + 6 \neq 0$
$\Rightarrow5\lambda + 8 \neq 0$
$\Rightarrow5\lambda \neq -8$
$\Rightarrow\lambda\neq\frac{-8}{5}$ View full question & answer→MCQ 1451 Mark
If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if:
- A
$\lambda\neq1$
- B
$\lambda\neq2$
- C
$\lambda\neq-1$
- ✓
$\lambda\neq0$
AnswerCorrect option: D. $\lambda\neq0$
A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus, $A^{-1}$ exists if $\lambda\neq0$.
View full question & answer→MCQ 1461 Mark
If $\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix},$ then det $(adj\ (adj\ A))$ is:
- ✓
$14^4$
- B
$14^3$
- C
$14^2$
- D
$14$
AnswerCorrect option: A. $14^4$
$\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}$
$|\text{A}|=14$
$\text{det(adj(adj A))}=|\text{A}|^{{\text{n}-1}^{2}}$
$\text{det(adj(adj A))}=|14|^{{3-1}^{2}}=14^4$
View full question & answer→MCQ 1471 Mark
Choose the correct answer from given four options in each of the Exercise:
The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$is:
- A
$9x^2(x + y)$
- ✓
$9y^2(x + y)$
- C
$3y^2(x + y)$
- D
$7x^2(x + y)$
AnswerCorrect option: B. $9y^2(x + y)$
We have, $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}3(\text{x}+\text{y})&\text{x}+\text{y}&\text{y}\\3(\text{x}+\text{y})&\text{x}&\text{y}\\3(\text{x}+\text{y})&\text{x}+2\text{y}&-2\text{y}\end{vmatrix}$
$\big[\because\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\text{ and C}_3\rightarrow\text{C}_3 -\text{C}_2\big]$
$=3(\text{x}+\text{y})\begin{vmatrix}1&(\text{x}+\text{y})&\text{y}\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix}$
$ [$Taking $3(x + y)$ common from first column$]$
$=3(\text{x}+\text{y})\begin{vmatrix}0&\text{y}&0\\1&\text{x}&\text{y}\\1&(\text{x}+2\text{y})&-2\text{y}\end{vmatrix}$
$[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2]$
Expanding along $R_1,$
$=3(\text{x}+\text{y})\big[-\text{y}(-2\text{y})-\text{y}\big]$
$=3\text{y}^2.3(\text{x}+\text{y})=9\text{y}^2(\text{x}+\text{y})$
View full question & answer→Question 1481 Mark
The matrix $\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$ is:
- non-singular
- singular
- skew-symmetric
- symmetric
Answer
- singular
Solution:
Given $\text{A}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$
$\text{|A|}=\begin{bmatrix}1 &\text{amp; } 0 &\text{amp; 1} \\ 2 &\text{amp; 1}&\text{amp; 0} \\ 3 &\text{amp; 1} &\text{amp; 1}\end{bmatrix}$
$\Rightarrow∣\text{A}∣=1−1=0$
Hence, A is singular. View full question & answer→Question 1491 Mark
If $\text{A}=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix},$ find $|\text{A}|$
- 352
- 356
- 325
- 532
Answer
- 352
Solution:
$\Rightarrow|\text{A}|=\begin{bmatrix}2&5&9\\6&1&3\\4&8&2\end{bmatrix}$Evaluating along the first row, we get
$\triangle=2\begin{bmatrix}1&3\\8&2\end{bmatrix}-5\begin{bmatrix}6&3\\4&2\end{bmatrix}+9\begin{bmatrix}6&1\\4&8\end{bmatrix}$
$\triangle=2(2-24)-5(12-12)+9(48-4)$
$\triangle=2(-22)-0+9(44) $
$\triangle=-44+9(44)=44(-1+9)=352 $ View full question & answer→Question 1501 Mark
If A is a singular matrix, then A (adj A) is a.
- scalar matrix
- zero matrix
- identity matrix
- orthogonal matrix
Answer
- zero matrix
Solution:
Given A is a singular matrix.
⇒ ∣A∣ = 0
A(adjA) = ∣A∣ I = 0I = O
∴ A(adjA) is a zero matrix.
View full question & answer→