Mathematics M.C.Q (1 Marks)

$ = \left[ {\begin{array}{*{20}{c}}{\cos (\alpha + \beta )}&{ - \sin (\alpha + \beta )}\\{\sin (\alpha + \beta )}&{\cos (\alpha + \beta )}\end{array}} \right] = BA$ (verify).

$ = \left[ {\begin{array}{*{20}{c}}{{a^2}{b^2} - {a^2}{b^2}}&{a{b^3} - a{b^3}}\\{ - {a^3}b + {a^3}b}&{ - {a^2}{b^2} + {a^2}{b^2}}\end{array}} \right] = O$

$ \Rightarrow \,\,{A^3} = A.{A^2} = 0$ and ${A^n} = 0$, for all $n \ge 2$.

$ = \left[ {\begin{array}{*{20}{c}}4&0&0\\0&1&0\\0&0&9\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&0&0\\0&3&0\\0&0&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4}&0&0\\0&3&0\\0&0&{18}\end{array}} \right]$.

Therefore ${A^2} + {B^2} = AA + BB = A(BA) + B(AB)$

$ = (AB)A + (BA)B = BA + AB = A + B$,

$(\because \,\,AB = B$ and $BA = A)$.

==> ${({A^2})^{20}} = {A^{40}} = {(I)^{20}} = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$.

==> $A + B = \left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]$

${A^2} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\2&{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]$

${B^2} = \left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right]\,\,\,\left[ {\begin{array}{*{20}{c}}a&1\\b&{ - 1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2} + b}&{a - 1}\\{ab - b}&{b + 1}\end{array}} \right]$

Also, ${A^2} + {B^2} = \left[ {\begin{array}{*{20}{c}}{{a^2} + b - 1}&{a - 1}\\{ab - b}&b\end{array}} \right]$

==> ${(A + B)^2} = \left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{1 + a}&0\\{2 + b}&{ - 2}\end{array}} \right]$

Also, ${(A + B)^2} = \left[ {\begin{array}{*{20}{c}}{{{(1 + a)}^2}}&{\,\,\,\,0}\\{(2 + b)(1 + a) - 2(2 + b)}&{\,\,\,\,4}\end{array}} \right]$

Also, ${(A + B)^2} = {A^2} + {B^2}$

==> $\left[ {\begin{array}{*{20}{c}}{{{(1 + a)}^2}}&0\\{\,(2 + b)(a - 1)}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2} + b - 1}&{a - 1}\\{ab - b}&b\end{array}} \right]$

By equating, $a - 1 = 0 \Rightarrow a = 1$ and $b = 4$.

$\therefore  A = I$   

$x + y + z = 0$ ......$(i)$

$x - 2y - 2z = 3$ ......$(ii)$

$x + 3y + z = 4$.....$(iii)$

On solving $x = 1,\,y = 2,\,z = - 3$

i.e., $\left[ \begin{array}{l}{\rm{ }}1\\{\rm{ }}2\\ - 3\end{array} \right]$.

$\therefore$  $A^2 -(a + d)$ $A =$ $\left( {\begin{array}{*{20}{c}}{bc + ad}&0\\ 0&{bc + da}\end{array}} \right)$ $= (bc - ad) I$

As $A^2 - (a + d)A + kI = 0$, we get $(bc -ad)I + kI = 0 $

$==> k = ad - bc$

Now                 ${a_{ij}} = \frac{1}{2}|i - 3j|,$  $i=1,\,2,\,3$ and $j=1,\,2$

therefore          ${a_{11}} = \frac{1}{2}|1 - 3 \times 1|\, = \,1$

                         ${a_{12}} = \frac{1}{2}|1 - 3 \times 2|\, = \,\frac{5}{2}$

                         ${a_{21}} = \frac{1}{2}|2 - 3 \times 1|\, =\,\frac{1}{2}$

                         ${a_{22}} = \frac{1}{2}|2 - 3 \times 2|\, = \,2$

                         ${a_{31}} = \frac{1}{2}|3 - 3 \times 1|\, = \,0$

                         ${a_{32}} = \frac{1}{2}|3 - 3 \times 2|\, = \,\frac{3}{2}$

Hence the required matrix is given by   $A\, = \,\left[ {\begin{array}{*{20}{c}}
  1&{\frac{5}{2}} \\ 
  {\frac{1}{2}}&2 \\ 
  0&{\frac{3}{2}} 
\end{array}} \right]$

$x+3=0$,            $z+4=6$,            $2 y-7=3 y-2$

$a-1=-3$,            $0=2 c+2$            $b-3=2 b+4$

Simplifying, we get

$a=-\,2$,  $b=-\,7$,  $c=-\,1$,  $x=-\,3$,  $y=-\,5$,  $z=2$

Thus, we have

$a_{11}=2 \times 1-1=2-1=1$

$a_{21}=2 \times 2-1=4-1=3$

$a_{31}=2 \times 3-1=6-1=5$

$a_{12}=2 \times 1-2=2-2=0$

$a_{22}=2 \times 2-2=4-2=2$

$a_{12}=2 \times 3-2=6-2=4$

$a_{13}=2 \times 1-3=2-3=-1$

$a_{2 n}=2 \times 2-3=4-3=1$

$a_{13}=2 \times 3-3=6-3=3$

$a_{14}=2 \times 1-4=2-4=-2$

$a_{24}=2 \times 2-4=4-4=0$

$a_{14}=2 \times 3-4=6-4=2$

Therefore, the required matrix is $A=\left[\begin{array}{cccc}1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2\end{array}\right]$

or         $2A+3X-2A=5B-2A$

or         $2 A-2A+3X=5B-2A$          $($ Matrix addition is commutative $)$

or         $O+3 X=5B-2 A$        $(-2A$ is the additive inverse of $2A)$

or         $3 \mathrm{X}=5 \mathrm{B}-2 \mathrm{A}$                  ( $O$ is the additive identity)

or         $X=\frac{1}{3}(5 B-2 A)$

or         ${\text{X}} = $ $\frac{1}{3}\left( {5\left[ {\begin{array}{*{20}{c}}
  2&{ - 2} \\ 
  4&2 \\ 
  { - 5}&1 
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{l}}
  8&0 \\ 
  4&{ - 2} \\ 
  3&6 
\end{array}} \right]} \right)$ $ = \frac{1}{3}\left( {\left[ {\begin{array}{*{20}{c}}
  {10}&{ - 10} \\ 
  {20}&{10} \\ 
  { - 25}&5 
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  { - 16}&0 \\ 
  { - 8}&4 \\ 
  { - 6}&{ - 12} 
\end{array}} \right]} \right)$

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  {10 - 16}&{ - 10 + 0} \\ 
  {20 - 8}&{10 + 4} \\ 
  { - 25 - 6}&{5 - 12} 
\end{array}} \right]$ 

$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
  { - 6}&{ - 10} \\ 
  {12}&{14} \\ 
  { - 31}&{ - 7} 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{\frac{{ - 10}}{3}} \\ 
  4&{\frac{{14}}{3}} \\ 
  {\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}} 
\end{array}} \right]$

The selling prices of a chemistry book, a physics book, and an economics book are respectively  given as  Rs. $80$, Rs. $60$ and Rs. $40 .$

The total amount of money that will be received from the sale of all these books can be erepresented in the form of a matrix as :

$12\left[\begin{array}{lll}10 & 8 & 10\end{array}\right]\left[\begin{array}{l}80 \\ 60 \\ 40\end{array}\right]$

$=12[10 \times 80+8 \times 60+10 \times 40]$

$=12(800+480+400)$

$=12(1680)$

$=20160$

Thus, the bookshop will receive Rs. $20160$ from the sale of all these books.

Therefore, matrix $P Y$ will be defined if $k=3$

Consequently, $P Y$ will be of the order $p \times k$. Matrices $W$ and $Y$ are of the orders $n \times 3$ and $3 \times k$ respectively.

since the number of columns in $W$ is equal to the number of rows in $Y$, matrix $W Y$ is welldefined and is of the order $n\times k$.

Matrices $P Y$ and $W Y$ can be added only when their orders are the same.

However, $P Y$ is of the order $p \times k$ and $W Y$ is of the order $n \times k .$ Therefore. we must have

$p=n$

Thus, $k=3$ and $p=n$. are the restrictions on $n, \,k,$ and $p$ so that $P Y+W Y$ will be defined.

Therefore, matrix $7 X$ is also of the same order.

Matrix $Z$ is of the order $2 \times p,$   i.e, $2 \times n \quad[\text { since } n=p]$

Therefore, matrix $5 Z$ is also of the same order.

Now, both the matrices $7 X$ and $5 Z$ are of the order $2 \times n$.

Thus, matrix $7 X-5 Z$ is well-defined and is of the order $2 \times n$.

$\therefore A^{2}=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}9+2 & 6+2 \\ 3+1 & 2+1\end{array}\right]=\left[\begin{array}{cc}11 & 8 \\ 4 & 3\end{array}\right]$

Now,

$A^{2}+a A+b I=0$

$\Rightarrow(A A) A^{-1}+a A A^{-1}+b L A^{-1}=0 \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]$

$\Rightarrow A\left(A A^{-1}\right)+a I+b\left(L A^{-1}\right)=0$

$\Rightarrow A I+a I+b A^{-1}=0$

$\Rightarrow A+a I=-b A^{-1}$

$\Rightarrow A^{-1}=\frac{1}{b}(A+a I)$

Now,

$A^{-1}=\frac{1}{|A|}$ adj. $A=\frac{1}{1}\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]$

We have:

$\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]=-\frac{1}{b}\left(\left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}a & 0 \\ 0 & a\end{array}\right]\right)=-\frac{1}{b}\left[\begin{array}{cc}3+a & 2 \\ 1 & 1+a\end{array}\right]=\left[\begin{array}{cc}\frac{-3-a}{b} & -\frac{2}{b} \\ -\frac{1}{b} & \frac{-1-a}{b}\end{array}\right]$

Comparing the corresponding elements of the two matrices, we have:

$-\frac{1}{b}=-1 \Rightarrow b=1$

$\frac{-3-a}{b}=1 \Rightarrow-3-a \Rightarrow a=-4$

Hence, $-4$ and $1$ are the required values of $a$ and $b$ respectively.

$=I+A^{3}+3 A+3 A^{2}-7 A$

$=I+A^{2} \cdot A+3 A+3 A-7 A$                       $\left[A^{2}=A\right]$

$=I+A \cdot A-A$

$=I+A^{2}-A$

$=I+A-A$

$=I$

$\therefore(I+A)^{3}-7 A=I$

$\therefore$ Option $(a)$ is not true.

$\Rightarrow A^{\prime}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

Now $A+A^{\prime}=1$

$\therefore $    $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow $   $\left[\begin{array}{cc}2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Comparing the corresponding elements of the two matrices, we have :

$\cos \alpha=\frac{1}{2}$

$\alpha=\cos ^{-1}\left(\frac{1}{2}\right)$

$\therefore   $ $\alpha=\frac{\pi}{3}$

[${C_1} \to {C_1} - {C_2}$    ;  ${C_2} \to {C_2} - {C_3}$]

=  $\left| {\,\begin{array}{*{20}{c}}
1&2&1\\
1&5&3\\
1&9&6
\end{array}\,} \right|\,$   by ${C_2} \to {C_2} + {C_3}$

= $\left| {\,\begin{array}{*{20}{c}}3&1&1\\6&2&3\\{10}&3&6\end{array}\,} \right|$, by ${C_1} \to {C_1} + {C_2} + {C_3}$.

But $ \ne \left| {\,\begin{array}{*{20}{c}}2&1&1\\2&2&3\\2&3&6\end{array}\,} \right|$.

=$(2 + i)$$\left| {\,\begin{array}{*{20}{c}}0&{ - 2i}&{ - 1}\\0&{ - 1 + 2i}&{2i}\\1&2&{1 - i}\end{array}\,} \right|$          by $\begin{array}{l}{R_1} \to {R_1} - {R_2}\\{R_2} \to {R_2} - {R_3}\end{array}$

= $(2 + i)\,\,\{ - 4{i^2} + ( - 1 + 2i)\} = (2 + i)\,(4 - 1 + 2i)$

= $(2 + i)\,(3 + 2i) = 4 + 7i$.

we have $(9 + x)$ $\left| {\,\begin{array}{*{20}{c}}1&3&5\\1&{x + 2}&5\\1&3&{x + 4}\end{array}\,} \right|$ = 0

$ \Rightarrow $ $(x + 9)$ $\left| {\,\begin{array}{*{20}{c}}0&{1 - x}&0\\0&{ - (1 - x)}&{1 - x}\\1&3&{x + 4}\end{array}\,} \right| = 0$

$ \Rightarrow $ $(x + 9)$ ${(1 - x)^2}\left| {\,\begin{array}{*{20}{c}}0&1&0\\0&{ - 1}&1\\1&3&{x + 4}\end{array}\,} \right| = 0$

$ \Rightarrow $ $x = 1,\,1,\, - 9$, 

Hence $(a - b),\,(b - c),\,(c - a)$ are factors of $\Delta $. Since $\Delta $ is symmetric in $a,b,c $ and of $4th$  degree, $(a + b + c)$ is also a factor, so that we can write

$\Delta$=$k(a-b)(b-c)(c-a)(a+b+c)  $     ......................$(i)$

Where by comparing the coefficients of the leading term $b{c^3}$ on both the sides of identity   $(i).$ We get $1 = k( - 1)\,( - 1) \Rightarrow k = 1$

$\Delta$=$(a-b)(b-c)(c-a)(a+b+c)    $

Trick : Put $a = 1,\,b = 2,\,c = 3$, so that determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&8&{27}\end{array}\,} \right| = 1(30) - 1(24) + 1(8 - 2) = 12$

which is given by $(c)$ . i.e. $(1 + 2 + 3)\,(1 - 2)\,(2 - 3)(3 - 1) = 12$.

                                                       $({R_1} \to {R_1} + {R_2} + {R_3})$

=$(a + b + c)$ $\left| {\,\begin{array}{*{20}{c}}{1}&{1}&{1}\\b&c&a\\c&a&b\end{array}\,} \right|$  =$(a + b + c)$  $\left| {\,\begin{array}{*{20}{c}}   1&1&1 \\    b&c&a \\    c&a&b  \end{array}\,} \right|$  =$(a + b + c)$    $\left| {\,\begin{array}{*{20}{c}}   1&0&0 \\    b&{b - c}&{c - a} \\    c&{c - a}&{a - b}  \end{array}\,} \right|$

= $3abc - {a^3} - {b^3} - {c^3}$,                    (After simplification).

= $ - \frac{1}{2}\left| {\,\begin{array}{*{20}{c}}0&\omega &{{\omega ^2}}\\0&1&{ - 2}\\0&{ - 1}&0\end{array}\,} \right| = 0$,       (Apply ${C_1} \to {C_1} + {C_2} + {C_3})$.

$ = a[{a^2} + ab - 2{a^2} - ab] = - {a^3} = i$.

$ = pqr({a^3} + {b^3} + {c^3}) - abc({p^3} + {q^3} + {r^3})$

= $pqr(3abc) - abc(3pqr) = 0$,

$\left( \begin{gathered}
  \because \,p + q + r = 0\,,\,\therefore \,\,{p^3} + {q^3} + {r^3} = 3pqr \hfill \\
  \because \,\,a + b + c = 0\,,\therefore \,{a^3} + {b^3} + {c^3} = 3abc \hfill \\ 
\end{gathered}  \right)$

$ \Rightarrow $$\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\{x + 1}&{x - 1}&1\\{x + 1}&1&{x - 1}\end{array}\,} \right|\, = 0$,

{Applying ${C_1} \to {C_1} + {C_2} + {C_3}$}

$ \Rightarrow $$(x + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{x - 1}&1\\1&1&{x - 1}\end{array}\,} \right|$= 0

$ \Rightarrow $ $(x + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\0&{x - 2}&0\\0&0&{x - 2}\end{array}\,} \right| = 0$

{Applying ${R_2} \to {R_2} - {R_1},\,{R_3} \to {R_3} - {R_1}$}

$\Rightarrow $ $(x+1) (x-2)^2 = 0 => x =-1,2. $

$ = (abc)(abc)\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right| = {a^2}{b^2}{c^2}( - 1)( - 4)$

$ = 4{a^2}{b^2}{c^2} = K{a^2}{b^2}{c^2}$,

$(given) ==> K = 4.$

==> $a[ - (b\alpha - c)] - b[ - 2(b\alpha - c)] + [a\alpha - b)(b - 2c)] = 0$

==>$ - ab\alpha + ac + 2{b^2}\alpha - 2bc + ab\alpha - 2ac\alpha - {b^2} + 2bc = 0$

==> $ac + 2{b^2}\alpha - 2ac\alpha - {b^2} = 0$

==> $(ac - {b^2}) - 2\alpha (ac - {b^2}) = 0$

==> $ac - {b^2} = 0$or $1 - 2\alpha = 0$ $ \Rightarrow $ ${b^2} = ac$   or $\alpha = \frac{1}{2}$

(As given in question)

So, ${b^2} = ac$ i.e, $a,b,c$ are in $G.P.$

$ = \left| {\,\begin{array}{*{20}{c}}{17}&9\\9&5\end{array}\,} \right| = (3x - 2) - (x + 6)$

==> $85 - 81 = 2x - 8$

==> $4 + 8 = 2x$

==> $x = 6$.

Hence, the determinant is divisible by $x$,${x^3}$ and $(14 + {x^2})$,

but not divisible by ${x^5}$.

==>$x(5x - 2x) - 2(2x + x) - 1(4 + 5) = 0$

==> $3{x^2} - 6x - 9 = 0$,

==> ${x^2} - 2x - 3 = 0$,

==> $(x + 1)(x - 3) = 0$

$ \Rightarrow x = - 1,\,3$.

$x = \frac{{{D_x}}}{D} = \left| {\,\begin{array}{*{20}{c}}7&{\,\,3}&{ - 5}\\6&{\,\,1}&{\,\,1}\\1&{ - 4}&{\,\,2}\end{array}\,} \right|\, \div \left| {\,\begin{array}{*{20}{c}}2&{\,\,3}&{ - 5}\\1&{\,\,1}&{\,\,1}\\3&{ - 4}&{\,\,2}\end{array}\,} \right|$.

$\Delta = \left| {\,\begin{array}{*{20}{c}}1&{\,\,1}&{ - 1}\\3&{ - 1}&{ - 1}\\1&{ - 3}&{\,\,1}\end{array}\,} \right| = 1( - 1 - 3) - 1(3 + 1) - 1( - 9 + 1)$

$ = - 4 - 4 + 8 = 0$. There are infinite number of solutions.