Question 11 Mark
If $\begin{vmatrix} 3 \text{x}& 7 \$0.3em] -2 & 4\$0.3em] \end{vmatrix} = \begin{vmatrix} 8 & 7 \$0.3em] 6 & 4\$0.3em] \end{vmatrix}, $find the value of x.
View full question & answer→Question 21 Mark
If $\begin{bmatrix} 3 \text{x}& 7 \$0.3em] -2 & 4\$0.3em] \end{bmatrix} = \begin{bmatrix} 8 & 7 \$0.3em] 6 & 4\$0.3em] \end{bmatrix}, $find the value of x.
View full question & answer→Question 31 Mark
If $\begin{bmatrix} \text{x - y }& \text{z} \$0.3em] 2\text{x - y }& \text{w} \$0.3em] \end{bmatrix} = \begin{bmatrix} -1& 4 \$0.3em] 0 & 5\$0.3em] \end{bmatrix},$find the value of x + y.
View full question & answer→Question 41 Mark
$\text{If }x \in \text{N and} \begin{bmatrix} \text{x + 3} & -2 \\ \text{-3x} & \text{2x} \\ \end{bmatrix} = 8, $ then find the value of $x.$
Answer$\text{(x + 3)2x - (-2) (-3x) = 8}$
$\text{x} = 2$
View full question & answer→Question 51 Mark
What positive value of x makes the following pair of determinants equal? .
$\begin{vmatrix}\text{2x}&3\\5&\text{x} \end{vmatrix}, \begin{vmatrix}\text{16}&3\\5&\text{2} \end{vmatrix}$
View full question & answer→Question 61 Mark
Find the value of x from the following:
$ \begin{vmatrix} \text{x} & \text{4} \\ \text{2} & \text{2x} \\ \end{vmatrix}=0.$
View full question & answer→Question 71 Mark
Write the value of the following determinant:
$ \begin{vmatrix} \text{a - b} & \text{b - c} & \text{c - a} \\ \text{b - c} & \text{c - a} & \text{a - b} \\ \text{c - a} & \text{a - b} & \text{b - c} \end{vmatrix}.$
View full question & answer→Question 81 Mark
Evaluate:
$\begin{vmatrix} \sin30^{o} & \cos30^{o} \\ -\sin60^{o} & \cos60^{o} \\ \end{vmatrix}$.
View full question & answer→Question 91 Mark
If $ \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} = 8,$ write the value of x.
Answer$ \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix} = 8$
Expanding along $R_1$, we get
$x (-x^{2} - 1) - \sin \theta (-x \sin \theta - \cos \theta) + \cos \theta (-\sin \theta + x \cos \theta) = 8$
$\Rightarrow - x^{3} - x + x \sin^{2} \theta + \sin \theta \cos \theta - \sin \theta \cos\theta + x \cos^{2} \theta = 8$
$\Rightarrow - x^{3} - x + x \big( \sin^{2} \theta + \cos^{2} \theta\big) = 8$
$\Rightarrow - x^{3} - x + x = 8$
$\Rightarrow x^{3} + 8 = 0$
$\Rightarrow (x + 2) (x^{2} - 2x + 4) = 0$
$\Rightarrow x + 2 = 0 \text{ }\text{ }\text{ }\text{ }\text{ } [\because x^{2} - 2x + 4 > 0 \text{ }\forall x]$
$\Rightarrow x = -2$
View full question & answer→Question 101 Mark
Find the maximum value of $ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos\theta \end{bmatrix} $
Answer$\triangle = \begin{vmatrix} 1 & 0 & 0 \\ 1 & \sin\theta & 0 \\ 1 & 0 & \cos\theta \end{vmatrix} = \sin\theta\cos\theta $
$= \frac{1}{2}\sin 2\theta \therefore \text{Max value}= \frac{1}{2}$
View full question & answer→Question 111 Mark
If $\ \begin{bmatrix}2\text{x} & 5\\8 & \text{x}\\\end{bmatrix} \ = \ \begin{bmatrix}6 & -2\\7 & 3\\\end{bmatrix}, \ $ , Write the value of x.
Answer$\text{x} = \pm 6.$
View full question & answer→Question 121 Mark
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$\text{x} = \pm 6.$
View full question & answer→Question 131 Mark
Find the value of a if $\begin{bmatrix} \text{a} - \text{b}& 2\text{a} + \text{c}\\ 2\text{a} - \text{b} & 3\text{c} + \text{d} \\ \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 &13 \\ \end{bmatrix}$
AnswerGiven: $\begin{bmatrix} \text{a} - \text{b}& 2\text{a} + \text{c}\\ 2\text{a} - \text{b} & 3\text{c} + \text{d} \\ \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 &13 \\ \end{bmatrix}$
$\Rightarrow\text{a} - \text{b} = - 1 $ - - - - - (1)
2a + c = 5 - - - - - -(2)
2a – b = 0 - - - - - - -(3)
3c + d =13 - - - - - - - (4)
$\text{ From (iii)} 2\text{ a } = \text{b}\Rightarrow\text{a} = \frac{\text{b}}{2}$
Putting in (i) we get $\frac{\text{b}}{2} - \text{b} = -1$
$\Rightarrow\frac{\text{b}}{2} = 1 \Rightarrow\text{b} = 2 $
$\therefore\text{a} = 1 $
(ii) $\Rightarrow$c =5 - 2 x 1 =5 - 2 = 3
(iv) $\Rightarrow$d =13 – 3 x (3) =13 - 9 = 4
i.e. a = 1, b = 2, c = 3, d = 4.
View full question & answer→Question 141 Mark
If$\begin{bmatrix} \text{x} +1& \text{x} - 1\\ \text{x} - 3 & \text{x} + 2 \\ \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 1 &3 \\ \end{bmatrix}$, then write the value of x.
AnswerGiven $\begin{bmatrix} \text{x} +1& \text{x} - 1\\ \text{x} - 3 & \text{x} + 2 \\ \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 1 &3 \\ \end{bmatrix}$
$\Rightarrow(\text{x} + 1 ) (\text{x} + 2) - (\text{x} - 1 )(\text{x} - 3 ) = 12+ 1 $
$\Rightarrow\text{x}^{2} + 2 \text{x} + \text{x} + 2 - \text{x}^{2} + 3 \text{x} + \text{x} - 3 = 13$
$\Rightarrow7\text{x} - 1 = 13$
$\Rightarrow 7\text{x} = 14$
$\Rightarrow\text{x} = 2.$
View full question & answer→Question 151 Mark
If $\begin{pmatrix} 2 & 3 \\ 5 & 7 \\ \end{pmatrix}\begin{pmatrix} 1 & -3 \\ -2 & 4 \\ \end{pmatrix}=\begin{pmatrix} -4 & 6 \\ -9 & x \\ \end{pmatrix}\text{ write the value of x.}$
View full question & answer→Question 161 Mark
If $\triangle = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$, write the minor of the element $a_{23}$.
View full question & answer→Question 171 Mark
Simplify: $\text{cos}\theta\begin{bmatrix} \text{cos}\theta & \text{sin}\theta \\ \text{-sin}\theta & \text{cos}\theta \\ \end{bmatrix}+ \text{sin}\theta\begin{bmatrix} \text{sin}\theta & \text{-cos}\theta \\ \text{cos}\theta & \text{sin}\theta \\ \end{bmatrix}$.
Answer$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$.
View full question & answer→Question 181 Mark
What is the value of the determinant $\begin{vmatrix} 0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6 \end{vmatrix}?$
View full question & answer→Question 191 Mark
Write the value of the determinant $\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6\text{x} & \text{9x} & 12\text{x} \end{vmatrix} $
AnswerGiven determinant $|A| = \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6\text{x} & \text{9x} & 12\text{x} \end{vmatrix} $$\Rightarrow |A| = 3x \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4 \end{vmatrix} = 0 $
View full question & answer→Question 201 Mark
Find the co-factor of $a_{12}$ in the following: $ \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} $
AnswerMinor of $+a_{12}$ is $M_{12}$ $= \begin{vmatrix} 6 & 4 \\ 1 & -7 \\ \end{vmatrix} = - 42 - 4 = -46 $Cofactor $C_{12} = ( -1)^{1 + 2} M_{12} = (-1)^{3}(-46) = 46$
View full question & answer→Question 211 Mark
$\text{Evaluate:} \begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \\ \end{vmatrix} $
Answer$\begin{vmatrix} a + ib & c + id \\ -c + id & a - ib \\ \end{vmatrix} $$= (a + ib) (a - ib) - (c + id) (- c + id)$
$= [a^{2} - i^{2}b^{2} ] - [i^{2} b^{2} - c ^{2} ]$
$= (a^{2} + b ^{2} ) - (- d^{2} - c ^{2} )$
$= a^{2} + b^{2} + c^{2}+ d^{2}$
View full question & answer→Question 221 Mark
If $\text{A}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix},$ then find A (adj A).
AnswerGiven,
$\text{A}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix}$
$\text{M}_{11}=(-1)^2\begin{bmatrix}2&3\\3&5 \end{bmatrix}=10-9=1$
$\text{M}_{12}=-\begin{bmatrix}1&3\\3&5 \end{bmatrix}=-[-5-9]=14$
$\text{M}_{13}=(-1)^4.\begin{bmatrix}-1&2\\3&3 \end{bmatrix}=[-3-6]=-9$
$\text{M}_{21}=-\begin{bmatrix}0&0\\3&5\end{bmatrix}=0$
$\text{M}_{22}=\begin{bmatrix}2&0\\3&5\end{bmatrix}=10$
$\text{M}_{23}=-\begin{bmatrix}2&0\\3&3\end{bmatrix}=-6$
$\text{M}_{31}=\begin{bmatrix}0&0\\2&3\end{bmatrix}=0$
$\text{M}_{32}=-\begin{bmatrix}2&0\\-1&3\end{bmatrix}=-6$
$\text{M}_{33}=\begin{bmatrix}2&0\\-1&2\end{bmatrix}=4$
$\text{Adj.(A)}=\begin{bmatrix}1&0&0\\14&10&-6\\-9&-6&4\end{bmatrix}$
Then,
$\text{A.(adj A)}=\begin{bmatrix}2&0&0\\-1&2&3\\3&3&5\end{bmatrix}.\begin{bmatrix}1&0&0\\14&10&-6\\-9&-6&4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\times1+(0)\times14+0\times9&2\times0+0+0&2\times0+0\times0\\-1\times1+2\times14+3\times-9&-1\times0+2\times10+3\times-6&-1\times0+2\times-6+14\\3\times1+3\times14+5\times-9&3\times0+3\times10+5\times-6&3\times0+3\times-6+5\times4\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}=2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A.(adj A)}=2\text{I}$
View full question & answer→Question 231 Mark
Find values of x, if:
$\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$
Answer$\begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}\text{x}&3\\2\text{x}&5\end{vmatrix}$
$\Rightarrow2\times5-3\times4=\text{x}\times5-3\times2\text{x}$
$\Rightarrow10-12=5\text{x}-6\text{x}$
$ \Rightarrow-2=-\text{x} $
$\Rightarrow \text{x}=2$
View full question & answer→Question 241 Mark
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}2&7&65\\3&8&75\\5&9&86\end{vmatrix}=0$
Answer$\text{On}\begin{vmatrix}2&7&65\\3&8&75\\5&9&86\end{vmatrix},\ \text{operating}\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
$=\begin{vmatrix}2&7&63\\3&8&72\\5&9&81\end{vmatrix}=9\begin{vmatrix}2&7&7\\3&8&8\\5&9&9\end{vmatrix}=9\times0=0$ $\left[\because\text{two columns are identical}\right]$ Proved.
View full question & answer→Question 251 Mark
Find value of x, if: $\begin{vmatrix}2&4\\2&1\end{vmatrix}=\begin{vmatrix}2x&4\\6&x\end{vmatrix}$
Answer$\begin{vmatrix}2&4\\2&1\end{vmatrix}=\begin{vmatrix}2x&4\\6&x\end{vmatrix}$
$\Rightarrow2\times1-2\times4=2x^2-24$
$ \Rightarrow2-8=2x^2-24 $
$\Rightarrow-6=2x^2-24$
$\Rightarrow-6+24=2x^2$
$\Rightarrow 18=2x^2$
$ \Rightarrow x^2=9$
$ \Rightarrow x=\pm3$
View full question & answer→MCQ 261 Mark
If $ \begin{vmatrix}\text{x} &\text{amp; y}\\ 4 &\text{amp; } 2 \end{vmatrix}=7$ and $ \begin{vmatrix}2 &\text{amp; }3\\ \text{y} &\text{amp; } \text{x} \end{vmatrix}=4$ then,
- A
$\text{x}=-3, \text{ y}=-\dfrac {5}{2}$
- ✓
$\text{x}=-\dfrac {5}{2}, \text{y}=-3$
- C
$\text{x}=-3, \text{ y}=\dfrac {5}{2}$
- D
$\text{x}=-\dfrac {5}{2}, \text{ y}=3$
AnswerCorrect option: B. $\text{x}=-\dfrac {5}{2}, \text{y}=-3$
$ \begin{vmatrix}\text{x} &\text{amp; y}\\ 4 &\text{amp; } 2 \end{vmatrix}=7$
$\Rightarrow2\text{x}-4\text{y}=7$
$ \begin{vmatrix}2 &\text{amp; }3\\ \text{y} &\text{amp; } \text{x} \end{vmatrix}=4$
$\Rightarrow2\text{x}-3\text{y}=4$
$\therefore\text{x}=-\dfrac {5}{2}, \text{y}=-3$
View full question & answer→MCQ 271 Mark
System of simultaneous equations $kx + 2y - z = 1, (k - 1)y - 2z = 2$ and $(k + 2)z = 3$ have a unique solution, if $k$ is equal to :
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