Question 12 Marks
If $\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0,$ find x.
Answer$\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0$
$\Rightarrow9(2\text{x}+5)-3(5\text{x}+2)=0$
$\Rightarrow18\text{x}+45-15\text{x}-6=0$
$\Rightarrow3\text{x}+39=0$
$\Rightarrow3\text{x}=-39$
$\Rightarrow\text{x}=\frac{-39}{3}$
$\Rightarrow\text{x}=-13$
View full question & answer→Question 22 Marks
Find the value of the determinant $\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
$\triangle=\begin{vmatrix}4200&1\\4205&1\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$]
$\triangle=4200 - 4202$
$\triangle=-2$
View full question & answer→Question 32 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&-1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\-1&0\end{bmatrix},$ find |AB|.
Answer$\text{A}=\begin{bmatrix}1&2\\3&-1\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$\text{AB}=\begin{bmatrix}1&2\\3&-1\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$=\begin{bmatrix}1-2&0+0\\3+1&0+0\end{bmatrix}=\begin{bmatrix}-1&0\\4&0\end{bmatrix}$
$|\text{AB}|=0-0=0$
View full question & answer→Question 42 Marks
If $A$ is a square matrix of order $3$ with determinant $4,$ then write the value of $|-A|.$
Answer$|A| = 4$
Here,
Order of the matrix $(n) = 3$
Using properties of matrices, we get
$|kA| = k^n|A| [$For a square matrix of order $n$ and constant $k]$
$\Rightarrow |-A| = (-1)^3 |A| = (-1) \times 4 = -4$
View full question & answer→Question 52 Marks
If $\begin{vmatrix}2\text{x}&\text{x}+3\\2(\text{x}+1)&\text{x}+1\end{vmatrix}=\begin{vmatrix}1&5\\3&3\end{vmatrix},$ then write the value of x.
Answer$\begin{vmatrix}2\text{x}&\text{x}+3\\2(\text{x}+1)&\text{x}+1\end{vmatrix}=\begin{vmatrix}1&5\\3&3\end{vmatrix}$
⇒ (2x)(x + 1) - 2(x + 1)(x + 3) = 13 - 15
⇒ (x + 1)(2x - 2x - 6) = -12
⇒ -6x - 6 = -12
⇒ -6x = -6
⇒ x = 1
Hence, the value of x is 1.
View full question & answer→Question 62 Marks
If $\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ find the value of $|\text{A}|+|\text{B}|.$
Answer$\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$$\Rightarrow|\text{A}|= 0-\text{i}^2$
$\Rightarrow|\text{A}|=-(-1)=1$
Also,
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow|\text{B}|=0-1=-1$
So,
$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$
View full question & answer→Question 72 Marks
If $\text{x}\in\text{N}$ and $\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8,$ then find the value of x.
Answer$\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8$
$\Rightarrow(\text{x}+3)2\text{x}-(-2)(-3\text{x})=8$
$\Rightarrow2\text{x}^2+6\text{x}-6\text{x}=8$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2-4=0$
$\Rightarrow\text{x}^2=4$
$\Rightarrow\text{x}=2$ $[\text{x}\neq-2\ \because\text{x}\in\text{N}]$
View full question & answer→Question 82 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&2&7\\12&3&5\\16&4&3 \end{vmatrix}$ [Applying $C_1 → C_1 - 4C_2$]
$\Rightarrow\triangle=0$
View full question & answer→Question 92 Marks
Let $A = [a_{ij}]$ be a square matrix of order $3 \times 3$ and $C_{ij}$ denote cofactor of $a_{ij}$ in $A.$ if $|A| = 5,$ write the value of $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33.}$
AnswerIf $A = a_{ij}$ is a square matrix of order $n$ and $C_{ij}$ is a cofactor of $a_{ij}$, then
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, $|A| = 5$ and matrix $A$ is of order $3 \times 3$
Since $a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$ represent expansion of A along third column, we get
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = |A| = 5$
$\Rightarrow a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} = 5$
View full question & answer→Question 102 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + 2y = 5,
3x + 6y = 15
AnswerUsing the equations, we get
$\text{D}=\begin{vmatrix}1&2\\3&6\end{vmatrix}=6-6=0$
$\text{D}_1=\begin{vmatrix}5&2\\15&6\end{vmatrix}=30-30=0$
$\text{D}_2=\begin{vmatrix}1&5\\3&15\end{vmatrix}=15-15=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2$
Hence, the system of linear equation has infinitely many solutions.
View full question & answer→Question 112 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
Answer$\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
$=\begin{vmatrix}2&3&4\\5&6&8\\2&3&4\end{vmatrix}$ [Taking 2x common from $R_3$]
$=0$
View full question & answer→Question 122 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
$=1\begin{vmatrix}6&10\\11&38\end{vmatrix}-3\begin{vmatrix}2&10\\31&38\end{vmatrix}+5\begin{vmatrix}2&6\\31&11\end{vmatrix}$
$=(228-110)-3(76-310)+5(22-186)$
$=1(118)-3(-234)+5(-164)$
$=118+702-820$
$=0$
View full question & answer→Question 132 Marks
If $A$ is a square matrix such that $|A| = 2$, write the value of $\big|\text{A}\text{A}^{\text{T}}\big|.$
AnswerIn 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$
View full question & answer→Question 142 Marks
A is a skew-symmetric of order 3, write the value of |A|.
AnswerWe know that if a skew symmetric matrix A is of odd order, then |A| = 0
Since the order of the given matrix is 3, |A| = 0.
View full question & answer→Question 152 Marks
If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.
AnswerSince A & B are square matrix of the same order, by the property of determinants we get
|AB| = |A| × |B|
|A| = 3, AB = I
⇒ |AB| = 1
⇒ |A| × |B| = 1
⇒ 3 × |B| = 1
$\Rightarrow|\text{B}|=\frac{1}{3}$
View full question & answer→Question 162 Marks
Write the value of $a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23}$.
AnswerWe know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column) is zero. Therefore,
$A = [a_{ij}]$ is a square matrix of order n.
$\Rightarrow\sum\limits_{\text{n}}^{\text{i}=1}\text{a}_{\text{ij}}\text{C}_\text{kj}=0$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ik}=0$
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
[Since the elements are of first row and the cofactors are of elements of second row]
$\Rightarrow a_{11}C_{21} + a_{12}C_{22} + a_{13}C_{23} = 0$
View full question & answer→Question 172 Marks
Find the value of x, if:
$\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$$\Rightarrow2\times5-3\times4=\text{x}\times5-3\times2\text{x}$
$\Rightarrow10-12=5\text{x}-6\text{x}$
$\Rightarrow-2=-\text{x}$
$\Rightarrow\text{x}=2$
View full question & answer→Question 182 Marks
If $A = [A_{ij}]$ is a $3 \times 3$ scalar matrix such that $a_{11} = 2$, then write the value of $|A|$.
AnswerA scalar matrix is a digonal matrix, in which all the diagonal elements are equal to a given scalar number.
Given, $A = [a_{ij}]$ is $3 \times 3$ matrix, where $a_{11} = 2$
$\Rightarrow\text{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$\Rightarrow\text{A}=\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}$
$\Rightarrow|\text{A}|=2\times\begin{vmatrix}2&0\\0&2\end{vmatrix}$ [Expanding along $C_1$]
$\Rightarrow|\text{A}|=2\times2\times2$
$\Rightarrow|\text{A}|=8$
View full question & answer→Question 192 Marks
Write rthe value of the determinant $\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}$
Answer$\begin{vmatrix}\text{p}&\text{p}+1\\\text{p}-1&\text{p}\ \end{vmatrix}=\text{p}^2-(\text{p}+1)(\text{p}-1)$
$=\text{p}^2-(\text{p}^2-1)$
$=\text{p}^2-\text{p}^2+1$
$=1$
View full question & answer→Question 202 Marks
Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1$ 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 $\triangle=\frac{1}{2}\times1=\frac{1}{2}$
View full question & answer→Question 212 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
$=\text{a}\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}-\text{h}\begin{vmatrix}\text{h}&\text{f}\\\text{g}&\text{c} \end{vmatrix}+\text{g}\begin{vmatrix}\text{h}&\text{b}\\\text{g}&\text{f} \end{vmatrix}$
$=\big(\text{bc}-\text{f}^2\big)-\text{h}\big(\text{hc}-\text{fg}\big)+\text{g}\big(\text{hf}-\text{gb}\big)$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{g}^2\text{b}$
$=\text{abc}+2\text{fgh}-\text{af}^2-\text{ch}^2-\text{bg}^2$
View full question & answer→Question 222 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix},$ find |AB|.
Answer$\Rightarrow\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$
⇒ |A| = -1 - 6 = -7
$\Rightarrow\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
⇒ |B| = -2 + 12 = 10
If A and B are square matrix of the same order, then |AB| = |A| |B|.
⇒ |AB| = |A| |B|
⇒ |AB| = -7 × 10 = -70
View full question & answer→Question 232 Marks
Write the value of $\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
$=\text{a}^2-\text{iab}+\text{iab}-\text{i}^2\text{b}^2-(-\text{c}^2-\text{icd}+\text{icd}+\text{i}^2\text{d}^2)$
$=\text{a}^2-\text{i}^2\text{b}^2+\text{c}^2-\text{i}^2\text{d}^2$
Here, $\text{i}^2=-1$
$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
View full question & answer→Question 242 Marks
If w is an imaginary cube root of unity, find the value of $\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
Answer$\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
$=\begin{vmatrix}1+\text{w}+\text{w}^2&\text{w}&\text{w}^2\\\text{w}+\text{w}^2+1&\text{w}^2&1\\\text{w}^2+1+\text{w}&1&\text{w}\end{vmatrix}$ [Applying $C_1 → C_1 + C_2 + C_3$]
$=\begin{vmatrix}0&\text{w}&\text{w}^2\\0&\text{w}^2&1\\0&1&\text{w}\end{vmatrix}$ $[\because 1 + w + w^2 = 0$, w is the imaginary cube root of unity$]$
View full question & answer→Question 252 Marks
Find the value of x from the following: $\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
Answer$\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
$\Rightarrow2\text{x}^2-8=0$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2=\frac{8}{2}=4$
$\Rightarrow\text{x}=\sqrt{4}=\pm2$
View full question & answer→Question 262 Marks
Evaluate: $\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
Answer$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
$=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$=\cos(15^\circ+75^\circ)$ $\big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})\big]$
$=\cos90^\circ$
$=0$
$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}=0$
View full question & answer→Question 272 Marks
Find the value of x, if:
if $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10,$ find the value of x.
AnswerGiven, $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10$
$\Rightarrow12\text{x}-14=10$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=2$
View full question & answer→Question 282 Marks
If A is a singular matrix, then write the value of |A|.
AnswerSince A is a singular matrix
Thus, |A| = 0
View full question & answer→Question 292 Marks
If the matrix $\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$ is a singular, find the value of x.
AnswerA matrix is said to be singular if its determinant is zero. since the given matrix is singular, we get
$\text{A}=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=0$
$\Rightarrow5\text{x}+20=0$ [Expanding]
$\Rightarrow\text{x}=-\frac{20}{5}$
$\Rightarrow\text{x}=-4$
View full question & answer→Question 302 Marks
On expanding by first row, the value of the determinant of $3 \times 3$ square matrix $A = [a_{ij}]$ is $a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}$, where $[C_{ij}]$ is the cofactor of $a_{ij}$ in $A.$ Write the expression for its value on expanding by second column.
AnswerIf $A = a_{ij}$ is a square matrix of order n, then the sum of the products of elements of a row $($or a column$)$ with their cofactors is always equal to det (A). Therefore,
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, $|A| = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} [$Expanding along $R_1]$
Now,
$|A| = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32} [$Expanding along $R_2] [a_{12}, a_{22}$ and $a_{32}$ are elements of $C_3]$
View full question & answer→Question 312 Marks
Find the value of the determinant $\begin{vmatrix}243&156&300\\81&52&100\\-3&0&4\end{vmatrix}$
Answer[Applying $R_1 → R_1 - 3R_2$]
$=\begin{vmatrix}0&0&0\\81&52&100\\-3&0&4\end{vmatrix}$
$=0$
View full question & answer→Question 322 Marks
Find the value of the determinant $\begin{vmatrix}2^2&2^3&2^4\\2^3&2^4&2^5\\2^4&2^5&2^6\end{vmatrix}$
Answer$\begin{vmatrix}2^2&2^3&2^4\\2^3&2^4&2^5\\2^4&2^5&2^6\end{vmatrix}$
$=2^2\times2^3\times2^4\begin{vmatrix}1&2&2^2\\1&2&2^2\\1&2&2^2\end{vmatrix}$ [Taking out common factors from $R_1, R_2$ and $R_3$]
$=2^2\times2^3\times2^4\times2\begin{vmatrix}1&1&2^2\\1&1&2^2\\1&1&2^2\end{vmatrix}=0$ [Two rows being identical]
$\begin{vmatrix}2^2&2^3&2^4\\2^3&2^4&2^5\\2^4&2^5&2^6\end{vmatrix}=0$
View full question & answer→Question 332 Marks
State whether the matrix $\begin{vmatrix}2&3\\6&4\end{vmatrix}$ is singular or non-singular.
AnswerLet $\triangle=\begin{vmatrix}2&3\\6&4\end{vmatrix}$
= 2 × 4 - 6 × 3
= 18 - 18 = -10
A matrix is said to be singular if its determinant is equal to zero. Since $\triangle=-10\neq0,$ the given matrix is non-singular.
View full question & answer→Question 342 Marks
Find the value of x, if:
$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&5\\8&3\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&5\\8&3\end{vmatrix}$
$\Rightarrow2\text{x}^2-40=18-40$
$\Rightarrow2\text{x}^2=18$
$\Rightarrow\text{x}^2=9$
$\Rightarrow\text{x}=\pm3$
View full question & answer→Question 352 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
AnswerLet $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying $R_3 → R_3 - R_2$, 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 $C_2 → C_1 + C_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix}$ [Applying $C_2 → 5C_1 - C_2$ and $C_3 → 5C_1 - C_3$]
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$
View full question & answer→Question 362 Marks
A matrix $A$ of order $3 \times 3$ has determinant $5$. What is the value of $|3A|?$
AnswerIf $A = [a_{ij}]$ 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, $|3A| = 135$
View full question & answer→Question 372 Marks
If $|A| = 2,$ where $A$ is $2 × 2$ matrix, find $|adj\ A|.$
AnswerFor any square matrix $A$ of order $n, |adj\ A| = |A|^{n-1}$
Given, $|A| = 2$
Here, order is $2$
$\Rightarrow |adj\ A| = |2|^{2-1} = 2$
View full question & answer→Question 382 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}2&3&4\\2\text{x}&3\text{x}&4\text{x}\\5&6&8\end{vmatrix}$
$=\text{x}\begin{vmatrix}2&3&4\\2&3&4\\5&6&8\end{vmatrix}$ [Taking out x common from $R_2$]
$=0$
View full question & answer→Question 392 Marks
write the value of the determinant $\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}2&-3&5\\4&-6&10\\6&-9&15\end{vmatrix}$
$=\begin{vmatrix}2&-3&5\\4-4&-6+6&10-10\\6&-9&15\end{vmatrix}$ [Applying $R_2 → R_2 - 2R_1$]
$=\begin{vmatrix}2&-3&5\\0&0&0\\6&-9&15\end{vmatrix}$
$=0$
View full question & answer→Question 402 Marks
Find the value of x, if:
$\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}3&\text{x}\\\text{x}&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}$$\Rightarrow3-\text{x}^2=3-8$
$\Rightarrow-\text{x}^2=-8$
$\Rightarrow\text{x}^2=8$
$\Rightarrow\text{x}=\pm2\sqrt{2}$
View full question & answer→Question 412 Marks
Find the value of x, if:
$\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
AnswerGiven, $\begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2\text{x}&4\\6&\text{x}\end{vmatrix}$
$\Rightarrow2-20=2\text{x}^2-24$
$\Rightarrow-18=2\text{x}^2-24$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}^2=3$
$\Rightarrow\text{x}=\pm\sqrt{3}$
View full question & answer→Question 422 Marks
If $\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix},$ then write the value of $x$.
Answer$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow (x + 1)(x + 2) - (x - 1)(x - 3) = 12 + 1$
$\Rightarrow x^2 + 3x + 2 - x^2 + 4x - 3 = 13$
$\Rightarrow 7x - 1 = 13$
$\Rightarrow 7x = 14$
$\Rightarrow x = 2$
Hence, the value of $x$ is $2$
View full question & answer→Question 432 Marks
Write the value of the determinant $\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&1&\text{b}+\text{c}\\\text{b}&1&\text{c}+\text{a}\\\text{c}&1&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{b}+\text{c}&1&\text{b}+\text{c}\\\text{b}+\text{c}+\text{a}&1&\text{c}+\text{a}\\\text{c}+\text{a}+\text{b}&1&\text{a}+\text{b}\end{vmatrix}$ [Applying $C_1 → C_1 + C_3$]
$=\text{a}+\text{b}+\text{c}\begin{vmatrix}1&1&\text{b}+\text{c}\\1&1&\text{c}+\text{a}\\1&1&\text{a}+\text{b}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})\times0$
$=0$
View full question & answer→Question 442 Marks
If $\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix},$ write the value of x.
Answer$\begin{vmatrix}2\text{x}&5\\8&\text{x}\end{vmatrix}=\begin{vmatrix}6&-2\\7&3\end{vmatrix}$
$\Rightarrow2\text{x}^2-40=18+4$
$\Rightarrow2\text{x}^2-40=32$
$\Rightarrow2\text{x}^2=72$
$\Rightarrow\text{x}^2=36$
$\Rightarrow\text{x}^2=\pm6$
Hence, the value of x is $\pm6$
View full question & answer→Question 452 Marks
If $\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix},$ find the value of x.
Answer$\begin{vmatrix}3\text{x}&7\\-2&4\end{vmatrix}=\begin{vmatrix}8&7\\6&4\end{vmatrix}$
⇒ 12x + 14 = 32 - 42
⇒ 12x + 14 = -10
⇒ 12x = -24
⇒ x = -2
$\therefore$ x = -2
View full question & answer→Question 462 Marks
Find the value of x, if:
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
AnswerGiven,
$\begin{vmatrix}\text{x}+1&\text{x}-1\\\text{x}-3&\text{x}+2\end{vmatrix}=\begin{vmatrix}4&-1\\1&3\end{vmatrix}$
$\Rightarrow(\text{x}+1)(\text{x}+2)-(\text{x}-3)(\text{x}-1)=12+1$
$\Rightarrow\text{x}^2+3\text{x}+2-\text{x}^2+4\text{x}-3=13$
$\Rightarrow7\text{x}-1=13$
$\Rightarrow7\text{x}=14$
$\Rightarrow\text{x}=2$
View full question & answer→Question 472 Marks
Evaluate the following determinant:
$\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
AnswerConsider the determinant
$\triangle=\begin{vmatrix}67&19&21\\39&13&14\\81&24&26 \end{vmatrix}$
Applying $C_1 → C_1- 4C_3$, We get,
$\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}4&19&21\\-3&13&14\\-3&24&26 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\-3&13&14\\0&11&12\end{vmatrix}$ [Applying $R_3 → R_3 - R_2$ and $R_1 → R_1 + R_2$]
$\Rightarrow\triangle=\begin{vmatrix}1&32&35\\0&109&119\\0&11&12\end{vmatrix}$ [Applying $R_2 → 3R_1 + R_2$]
$\Rightarrow\triangle=1(109\times12-119\times11)$
$\Rightarrow\triangle=-1$
View full question & answer→Question 482 Marks
Show that the following systems of linear equations is inconsistent:
$2x - y = 5,$
$4x - 2y = 7$
AnswerConsider,
$2x - y = 5$
$4x - 2y = 7$
$\text{D}=\begin{vmatrix}2&-1\\4&-2\end{vmatrix}=-4+4=0$
$\text{D}_1=\begin{vmatrix}5&-1\\7&-2\end{vmatrix}=-10+7=-3$
$\text{D}_2=\begin{vmatrix}2&5\\4&7\end{vmatrix}=14-20=-6$
Hence, $D_1$ and $D_2$ are non zero. Thus the given system is inconsistent.
View full question & answer→Question 492 Marks
A matrix of order $3 \times 3$ has determinant $2$. What is the value of $|A(3I)|,$ where I is the identity matrix of order $3 \times 3.$
AnswerLet A be the given matrix. Then,
$|A| = 2 [$Order $= n = 3]$
$|I| = 1 [I$ is an identity matrix$]$
$3(I) = 3$
$|A^3(I)| = |3A| = 3^3|A| [A$ being of order $3]$
$= 27 \times 2 = 54$
$|A^3(I)| = 54$
View full question & answer→Question 502 Marks
Find the value of $\lambda$ so that the points $(1, - 5), (-4, 5)$ and $(\lambda,7)$ are collinear.
AnswerIf the points are collinear, then the area of the triangle must be zero.
Hence,
$\begin{vmatrix}1&-5&1\\-4&5&1\\\lambda&7&1\end{vmatrix}=0$
Expanding along $R_1$
$1(-2)+5(-4-\lambda)+1(-28-5\lambda)=0$
$-2-20-5\lambda-28-5\lambda=0$
$-50-10\lambda=0$
$\lambda=5$
Hence, $\lambda=5$
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