Question 12 Marks
Answer(1) According to given information, the system of linear equations is:
$\begin{aligned}x+y+z & =45 \\z & =x+8 \\x+z & =2 y\end{aligned}$
$\begin{aligned}\text{or}\quad x+y+z & =45 \\-x+z & =8 \\x-2 y+z & =0\end{aligned}$
In matrix form,
$\left[\begin{array}{ccc}1 & 1 & 1 \\-1 & 0 & 1 \\1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{c}45 \\8 \\0\end{array}\right]$
(2)
$\begin{aligned}\text{Let}\quad A & =\left[\begin{array}{ccc}1 & 1 & 1 \\1 & 0 & -2 \\1 & -1 & 1\end{array}\right] \\\text{Then}\quad A^{-1} & =\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & 2 \\3 & 0 & -3 \\1 & -2 & 1\end{array}\right]\end{aligned}$
We know that,
$\left(A^{\prime}\right)^{-1}=\left(A^{-1}\right)^{\prime}$
$\therefore\left(A^{-1}\right)^{\prime}=\left[\begin{array}{ccc}\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{1}{2} & 0 & -\frac{1}{2} \\\frac{1}{6} & \frac{-1}{3} & \frac{1}{6}\end{array}\right]$
$=\left[\begin{array}{ccc}\frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \frac{1}{3} & 0 & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{2} & \frac{1}{6}\end{array}\right]$
$\therefore\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 0 & -1 \\ 1 & -2 & 1\end{array}\right]^{-1}=\left[\begin{array}{ccc}\frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \frac{1}{3} & 0 & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{2} & \frac{1}{6}\end{array}\right]$
$\text{or,}\quad \left(A^{\prime}\right)^{-1}=\left(A^{-1}\right)^{\prime}$
View full question & answer→Question 22 Marks
Write the value of $\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$.
Answer$\text{Let}\quad\Delta=\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$
$=\left|\begin{array}{lll}2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0\end{array}\right| \quad\left[C_3 \rightarrow C_3-9 C_2-C_1\right]$
$=0$
[Since all values of $C_{3}$ is zero so determinant value is zero]
View full question & answer→Question 32 Marks
Write the value of the determinant:
$\left|\begin{array}{ccc}102 & 18 & 36 \\1 & 3 & 4 \\17 & 3 & 6\end{array}\right|$
AnswerLet, $\Delta=\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$
$=\left|\begin{array}{ccc}17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right| \quad\left[R_1 \rightarrow \frac{1}{6} R_1\right]$
$=0\quad \therefore (R_{1}$ and $R_{3}$ are identical$)$
View full question & answer→Question 42 Marks
If $A_{i j}$ is the cofactor of the element $a_{i j}$ of the determinant $\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$, then write the value of $a_{32} A_{32}$.
Answer
$\begin{aligned} a_{32} A_{32}=-5\left|\begin{array}{ll}2 & 5 \\ 6 & 4\end{array}\right| & \\ & =-5(8-30) \\ & =-5(-22)=110\end{aligned}$
View full question & answer→Question 52 Marks
Without expanding at any stage, find the value of the determinant:
$\Delta=\left|\begin{array}{lll}2 & x & y+z \\2 & y & z+x \\2 & z & x+y\end{array}\right|$
AnswerGiven, $\Delta=\left|\begin{array}{lll}2 & x & y+z \\ 2 & y & z+x \\ 2 & z & x+y\end{array}\right|$
By applying, $\text{C}_{3} \Rightarrow \text{C}_{2}+\text{C}_{3'}$ we get
$\Delta=\left|\begin{array}{lll}2 & x & x+y+z \\ 2 & y & x+y+z \\ 2 & z & x+y+z\end{array}\right|$
$=2(x+y+z)\left|\begin{array}{lll}1 & x & 1 \\ 1 & y & 1 \\ 1 & z & 1\end{array}\right|$
$=2\times 0=0$
[Since, $\text{C}_{1}=\text{C}_{2}\therefore \Delta =0$]
View full question & answer→Question 62 Marks
Without expanding at any stage, find the value of $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|.$
AnswerLet, $\Delta=\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|$
Applying $R_1 \Rightarrow R_1+2 R_3$, we get
$\Delta=\left|\begin{array}{ccc}a+2 x & b+2 y & c+2 z \\a+2 x & b+2 y & c+2 z \\x & y & z\end{array}\right|=0$
Since, $\text{R}_1$ and $\text{R}_2$ are identical, therefore value of $\Delta$ is zero.
View full question & answer→Question 72 Marks
If $A=\left[\begin{array}{ll}p & 2 \\ 2 & p\end{array}\right]$ and $\left|A^3\right|=125$, then find the value(s) of $p$.
Answer
$\begin{array}{l}\text{Given,}\quad A=\left[\begin{array}{ll}p & 2 \\2 & p\end{array}\right] \text { and }\left|A^3\right|=125 \\\text { Since, } \quad\left|A^n\right|=|A|^n \\\therefore \quad\left|A^3\right|=|A|^3\end{array}$
$\text{Now,}\quad|A|=\left|\begin{array}{ll}p & 2 \\2 & p\end{array}\right|=p^2-4$
According to given condition,
$|A^{3}|=125$
$\Rightarrow |A|^{3}=125$
$\Rightarrow (p^{2}-4)^{3}=125$
$\Rightarrow (p^{2}-4)^{3}=5^{3}$
$\Rightarrow p^{2}-4=5$
$\Rightarrow p^{2}=9$
$\Rightarrow p=\pm3$
Hence, values of $p=\pm3.$
View full question & answer→Question 82 Marks
If $A$ is a skew-symmetric matrix of order 3, then prove that $\operatorname{det} A=0$.
AnswerSince $A$ is a skew-symmetric matrix
$\begin{array}{l}\therefore \quad A^T=-A \\\therefore\left|A^T\right|=|-A|=(-1)^3 \cdot|A| \\\text { or }\left|A^T\right|=-|A| \\\text { or } 2|A|=0 \text { or }|A|=0 \text {. }\end{array}$
View full question & answer→Question 92 Marks
Find the inverse of the matrix $\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]$ Hence, find the matrix $P$ satisfying the matrix equation $P\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$.
Answer
$\begin{array}{l}{\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]^{-1}=\frac{1}{9-10}\left[\begin{array}{ll}-3 & -2 \\ -5 & -3\end{array}\right]=\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right]} \\ \therefore P=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right]=\left[\begin{array}{cc}13 & 8 \\ 1 &1\end{array}\right]\end{array}$
View full question & answer→Question 102 Marks
If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$, be such that $A^{-1}=k A$, then find the value of $k$.
Answer$\text{Finding}{~} A^{-1}=\frac{-1}{19}\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]$
$\Rightarrow \frac{-1}{19}\left[\begin{array}{cc}-2 & -3 \\ -5 & 2\end{array}\right]=\left[\begin{array}{cc}2 k & 3 k \\ 5 k & -2 k\end{array}\right]$
$\Rightarrow k=\frac {1}{19}$
View full question & answer→Question 112 Marks
Given $A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]$, compute $A^{-1}$ and show that $2 A^{-1}=9 I-A$.
Answer$|A|=2,$
$\begin{aligned} A^{-1} & =\frac{1}{2}\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right] \\ \text{LHS} & =2 A^{-1}=\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right],\end{aligned}$
$\begin{aligned} \text { RHS } & =9\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right] \\ & =\left[\begin{array}{ll}7 & 3 \\ 4 & 2\end{array}\right] \\ \text { LHS } & =\text { RHS }\end{aligned}$
View full question & answer→Question 122 Marks
If $x=-9$ is a root of $\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$, then find the other two roots.
AnswerSince $\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$
Expanding along $R_1$,
$\Rightarrow x(x^{2}-12)-3(2x-14)+7(12-7x)=0$
$\Rightarrow x^{3}-12x-6x+42+84-49x=0$
$\Rightarrow x^{3}-67x+126=0\quad\ldots\text{(i)}$
Here, $126\times 1=9\times 2\times 7$
For $x=2,$
$\Rightarrow 2^{3}-67\times 2+126=134-134=0$
Hence, $x=2$ is a root.
For $x=7$
$\Rightarrow 7^{3}-67\times 7+126=469-469=0$
Hence, $x=7$ is also a root.
View full question & answer→Question 132 Marks
If $A=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$, write the minor of the element $a_{23}$.
AnswerGiven, $A=\left|\begin{array}{lll}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{array}\right|$
$\begin{aligned} \therefore \text { Minor of } a_{23}=\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right| & \\ & =10-3=7\end{aligned}$
View full question & answer→Question 142 Marks
Check whether $(l+m+n)$ is a factor of the determinant $\left|\begin{array}{ccc}l+m & m+n & n+l \\ n & l & m \\ 2 & 2 & 2\end{array}\right|$ or not. Give reason.
AnswerApply $R_{1} \rightarrow R_{1}+R_{2}$
$\left|\begin{array}{ccc}l+m+n & m+n+l & n+l+m \\ n & l & m \\ 2 & 2 & 2\end{array}\right|$
$=2(l+m+n)\left|\begin{array}{ccc}1 & 1 & 1 \\ n & l & m \\ 1 & 1 & 1\end{array}\right|;$ yes $(l+m+n)$ is a factor.
View full question & answer→Question 152 Marks
If $A$ is a square matrix satisfying $A^{\prime} A= I$, write value of $| A |.$
View full question & answer→Question 162 Marks
If $A$ and $B$ are square matrices of the same order $3,$ such that $| A |=2$ and $A B=2 I$, write the value of $| B |.$
Answer$\text {Given, } A B=2 I$
$\begin{array}{l}\text { or } A B=2\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}2 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 2\end{array}\right] \\\therefore \quad|A B|=2(4-0)-0+0=8\end{array}$
Since, $|A B|=|A||B|$
$\therefore \quad|B|=\frac{|A B|}{|A|}=\frac{8}{2}=4$
$[$given, $|A|=2]$
View full question & answer→Question 172 Marks
Find $|A B|,$ if $A=\left[\begin{array}{cc}0 & -1 \\ 0 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 5 \\ 0 & 0\end{array}\right].$
AnswerGiven, $A=\left[\begin{array}{cc}0 & -1 \\ 0 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 5 \\ 0 & 0\end{array}\right]$
$\therefore \quad A B=\left[\begin{array}{cc}0 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{ll}3 & 5 \\ 0 & 0\end{array}\right]$
$\text{or}\quad A B=\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 0+0\end{array}\right]$
$\text{or}\quad A B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Hence, $|AB|=0$
View full question & answer→Question 182 Marks
For what value of $x$, the given matrix $A=\left|\begin{array}{cc}3-2 x & x+1 \\ 2 & 4\end{array}\right|$ is a singular matrix?
Answer$A=\left|\begin{array}{cc}3-2 x & x+1 \\ 2 & 4\end{array}\right|$
Since $A$ is a singular matrix. $\text{i.e.,} |A|=0$
$\left|\begin{array}{cc}3-2 x & x+1 \\ 2 & 4\end{array}\right|=0$
$\text{or,}\quad 4(3-2x)-2(x+1)=0$
$\text{or,}\quad 12-8x-2x-2=0$
$\text{or,}-10x+10=0 ~\text{or,}~x=1$
View full question & answer→Question 192 Marks
If $A$ is square matrix of order 3 such that $|\operatorname{adj} A|=64$, then find $|A|$.
Answer$|\operatorname{adj} A|=|A|^{n-1}$, where $n$ is the order of the matrix
$|A|^2=64$
$\text{or,}|A|= \pm 8$
View full question & answer→Question 202 Marks
Given $A=\left(\begin{array}{ccc}4 & 2 & 5 \\ 2 & 0 & 3 \\ -1 & 1 & 0\end{array}\right),$ write the value of det. $\left(2 A A^{-1}\right).$
Answer
$\begin{array}{rlrl}\left|2 A A^{-1}\right| & =(2)^3 & {\left[\because A A^{-1}= I \right]} \\ & =8 &\end{array}$
View full question & answer→Question 212 Marks
If $A$ and $B$ are invertible matrices of order $3,$ $|A|=2$ and $\left|(A B)^{-1}\right|=-\frac{1}{6}$, then find $|B|.$
Answer
$\begin{array}{l}\frac{1}{|A B|}=-\frac{1}{6} \\ \Rightarrow \frac{1}{|A||B|}=-\frac{1}{6} \\ \Rightarrow|B|=-3 .\end{array}$
View full question & answer→Question 222 Marks
If $A$ is a $3 \times 3$ matrix, $|A| \neq 0$ and $|3 A|=k|A|$, then find the value of $k$.
AnswerSince $|kA|=k^{n}|A|,$ where $n$ is the order of matrix.
$\therefore |3A|=3^{3}|A|$
$\quad\quad\quad=27|A|$
$\text{or}\quad k=27$
View full question & answer→Question 232 Marks
If $\left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & -2 \\ 7 & 3\end{array}\right|$, then find the value of $x$.
Answer$\left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & -2 \\ 7 & 3\end{array}\right|$
$2 x^2-40=18+14$
$\text{or } 2 x^2=32+40$
$\text{or }2 x^2=72$
$\text{or }x^2=36$
$\text{or}\quad x= \pm 6$
View full question & answer→Question 242 Marks
Write the value of the determinant $\left|\begin{array}{cc}p & p+1 \\ p-1 & p\end{array}\right|.$
Answer$\left|\begin{array}{cc}p & p+1 \\ p-1 & p\end{array}\right|=p^2-\left(p^2-1\right)$
$=p^{2}-p^{2}+1$
$=1$
View full question & answer→Question 252 Marks
If $x \in N$ and $\left|\begin{array}{cc}x+3 & -2 \\ -3 x & 2 x\end{array}\right|=8$, then find the value of $x.$
AnswerGiven, $\left|\begin{array}{cc}x+3 & -2 \\ -3 x & 2 x\end{array}\right|=8$
$\text{or}\quad 2x(x+3)-(-2)(-3x)=8$
$\text{or}\quad 2x^{2}+6x-6x=8$
$\text{or}\quad x^{2}=\frac {8}{2}$
$\text{or}\quad x^{2}=4$
$\text{or}\quad x=\pm2$
$\text{Since}\quad$ $x \in N$, so, $x=2$
View full question & answer→Question 262 Marks
If $A$ and $B$ are matrices of order $3$ and $|A|=5$ $|B|=3$, then find $|3 A B|.$
AnswerWe know that,
$\begin{aligned}|A B| & =|A| \cdot|B| \\|3 A B| & =(3)^3|A B| \\& =27|A B| \\& =27|A| \cdot|B| \\& =27 \times 5 \times 3 \\& =405\end{aligned}$
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