Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\begin{array}{l} f\left( x \right) = \left| \begin{array}{l} \,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x + 1\\ \,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( {x - 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 1} \right)x\\ 3x\left( {x - 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( {x - 1} \right)\left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {x + 1} \right)x\left( {x - 1} \right) \end{array} \right|\\ \,\,\,\,\,\,\,\,\,\,\,\, = \left| \begin{array}{l} \,\,\,\,\,\,\,\,\,\,x + 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x + 1\\ \,\,\,\,\,\,\left( {x + 1} \right)x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\left( {x - 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 1} \right)x\\ \left( {x + 1} \right)x\left( {x - 1} \right)\,\,\,\,\,\,\,\,\,\,\,x\left( {x - 1} \right)\left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 1} \right)x\left( {x - 1} \right) \end{array} \right| \end{array}\]
$$\left[ {{C_1} \to {C_1} + {C_2}} \right]$$
$$ = 0\left[ {\because {C_1}\,{\text{and }}{C_2}\,{\text{are identical}}} \right]$$
which is free of $$x,$$ so the function is true for all values of $$x.$$ Therefore, at
$$x = 100,f\left( x \right) = 0,{\text{i}}{\text{.e}}{\text{., }}f\left( {100} \right) = 0$$

$$\eqalign{ & a = \frac{x}{{\left( {y - z} \right)}} \cr & \Rightarrow \,x - ay + az = 0\,\,\,.....\left( 1 \right) \cr & b = \frac{y}{{\left( {z - x} \right)}} \cr & \Rightarrow \,bx + y - bz = 0\,\,\,.....\left( 2 \right) \cr & c = \frac{z}{{\left( {x - y} \right)}} \cr & \Rightarrow \, - cx + cy + z = 0\,\,\,.....\left( 3 \right) \cr} $$
As $$x, y, z$$  are not all zero, the above system has a non-trivial
solution so, \[\Delta = \left| {\begin{array}{*{20}{c}} 1&{ - a}&a\\ b&{ - 1}&{ - b}\\ { - c}&c&1 \end{array}} \right|\]
∴ $$1 + ab + bc + ca = 0$$

$${\text{Let}}\,\alpha \, = \omega \,\,{\text{and}}\,\beta = {\omega ^2} = 2$$      are roots of $${x^2} + x + 1 = 0$$
\[\& \,{\text{Let}}\,\Delta = \left| {\begin{array}{*{20}{c}} {y + 1}&\omega &{{\omega ^2}}\\ \omega &{y + {\omega ^2}}&1\\ {{\omega ^2}}&1&{y + \omega } \end{array}} \right| = \Delta \]
$${\text{Applying}}\,{C_1} \to {C_1} + {C_2} + {C_3},{\text{we}}\,{\text{get}}$$
\[\Delta {\rm{ = }}\left| {\begin{array}{*{20}{c}} {y + 1 + \omega + {\omega ^2}}&\omega &{{\omega ^2}}\\ {y + 1 + \omega + {\omega ^2}}&{y + {\omega ^2}}&1\\ {1 + \omega + {\omega ^2} + y}&1&{y + \omega } \end{array}} \right|\]
\[\Delta = \left| {\begin{array}{*{20}{c}} y&\omega &{{\omega ^2}} \\ y&{y + {\omega ^2}}&1 \\ y&1&{y + \omega } \end{array}} \right|\left( {\because \,\,1 + \omega + {\omega ^2} = 0} \right)\]
\[\Delta = y\left| {\begin{array}{*{20}{c}} 1&\omega &{{\omega ^2}}\\ 1&{y + {\omega ^2}}&1\\ 1&1&{y + \omega } \end{array}} \right|\]
Applying $${R_2} \to {R_2} - {R_1}\,\,\& \,\,{R_3} \to {R_3} - {R_1},\,{\text{we get}}$$
\[\Delta = y\left| {\begin{array}{*{20}{c}} {y + {\omega ^2} - \omega }&{1 - {\omega ^2}}\\ {1 - \omega }&{y + \omega - {\omega ^2}} \end{array}} \right|\]
$$\eqalign{ & \Rightarrow \,\,\Delta = y \cr & \left[ {y - \left( {\omega - {\omega ^2}} \right)\left( {y + \left( {\omega - {\omega ^2}} \right)} \right) - \left( {1 - \omega } \right)\left( {1 - {\omega ^2}} \right)} \right] \cr & \Rightarrow \,\,\Delta = y\left[ {{y^2} - {{\left( {\omega - {\omega ^2}} \right)}^2} - 1 + {\omega ^2} + \omega - {\omega ^3}} \right] \cr & \Rightarrow \,\,\Delta = y\left[ {{y^2} - {\omega ^2} - {\omega ^4} + 2{\omega ^3} - 1 + {\omega ^2} + {\omega ^4} - {\omega ^3}} \right]\,\,\left( {\because \,\,{\omega ^4} = \omega } \right) \cr & \Rightarrow \,\,\Delta = y\left( {{y^2}} \right) = {y^3} \cr} $$

There are in total 9 entries and each entry can be selected in exactly 2 ways. Hence, the total number of all possible matrices of the given type is $$2^9 .$$

Operating $${C_1} \to {C_1} + {C_2} + {C_3}$$    and taking out the common factor from $$C_1 ,$$ we have
$$\Delta = \sin \left( {x + h} \right)\left( {1 + 2\cos h} \right)$$
\[\left| {\begin{array}{*{20}{c}} 1&{\sin \left( {x + h} \right)}&{\sin \left( {x + 2h} \right)}\\ 1&{\sin x}&{\sin \left( {x + h} \right)}\\ 1&{\sin \left( {x + 2h} \right)}&{\sin x} \end{array}} \right|\]
$$\left\{ {{\text{Since, }}\sin \left( {x + h} \right) + \sin \left( x \right) + \sin \left( {x + 2h} \right) = \sin \left( {x + h} \right)\left( {1 + 2\cos h} \right)} \right\}$$
By operating $${R_2} \to {R_2} - {R_1}{\text{ and }}{R_3} \to {R_3} - {R_1}$$
we have, $$\Delta = \sin \left( {x + h} \right)\left( {1 + 2\cos h} \right)$$
\[\left| {\begin{array}{*{20}{c}} 1&{\sin \left( {x + h} \right)}&{\sin \left( {x + 2h} \right)}\\ 0&{\sin x - \sin \left( {x + h} \right)}&{\sin \left( {x + h} \right) - \sin \left( {x + 2h} \right)}\\ 0&{\sin \left( {x + 2h} \right) - \sin \left( {x + h} \right)}&{\sin x - \sin \left( {x + 2h} \right)} \end{array}} \right|\]
$$ = \left\{ {\left( {2\cos \left( {x + \frac{h}{2}} \right)\sin + \frac{h}{2}} \right)\left( {2\cos \left( {x + h} \right)\sin h} \right) + {{\left( {2\cos \left( {x + \frac{{3h}}{2}} \right)\sin \frac{h}{2}} \right)}^2}} \right\}$$
Therefore, $$\mathop {\lim }\limits_{h \to 0} \frac{\Delta }{{{h^2}}} = 3{\cos ^2}x.$$

Given that, \[A = \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{2 + 6}\\ 0&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&8\\ 0&9 \end{array}} \right]\]
Since, $$f\left( x \right)\, = {x^2} - x + 2$$
Putting $$A$$ in place of $$x$$
$$f\left( A \right)\, = {A^2} - A + 2I$$
\[\begin{array}{l} = \left[ {\begin{array}{*{20}{c}} 1&8\\ 0&9 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&2\\ 0&3 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&0\\ 0&2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {1 - 1 + 2}&{8 - 2 + 0}\\ {0 - 0 + 0}&{9 - 3 + 2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&6\\ 0&8 \end{array}} \right] \end{array}\]

The desired number of sub-matrices
$$ = {\,^4}{C_2} \times {\,^3}{C_2} = 18$$

\[{\rm{We}}\,{\rm{have}}\,\,\left| \begin{array}{l} \,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,ax + b\\ \,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,bx + c\\ ax + b\,\,\,\,\,\,bx + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \end{array} \right|\]
$$By\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$
\[ = \left| \begin{array}{l} a\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,ax + b\\ b\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,bx + c\\ 0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - \left( {a{x^2} + 2bx + C} \right) \end{array} \right|\]
$$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right) = \left( + \right)\left( - \right) = - ve{\text{.}}$$

\[\vartriangle = \left| {\begin{array}{*{20}{c}} 0&{q - p}&{r - p} \\ {p - q}&0&{r - q} \\ {p - r}&{q - r}&0 \end{array}} \right| = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}} 0&{p - q}&{p - r} \\ {q - p}&0&{q - r} \\ {r - p}&{r - q}&0 \end{array}} \right|,\]            taking $$(- 1)$$  common from all the rows
\[\left( {R \rightleftarrows C} \right)\]
$$\eqalign{ & = {\left( { - 1} \right)^3}\vartriangle \cr & \therefore \,\,\vartriangle = - \vartriangle \cr & \Rightarrow \,\,\vartriangle = 0. \cr} $$

$$\because \,\,{a_1},{a_2},{a_3},.....,$$     are in G.P.
∴ Using $${a_n} = a{r^{n - 1}},$$   we get the given determinant, as
\[\left| \begin{array}{l} \log a{r^{n - 1}}\,\,\,\,\,\,\,\,\,\log a{r^n}\,\,\,\,\,\,\,\,\,\,\,\,\log a{r^{n + 1}}\\ \log a{r^{n + 2}}\,\,\,\,\,\,\,\,\,\log a{r^{n + 3}}\,\,\,\,\,\,\,\,\log a{r^{n + 4}}\\ \log a{r^{n + 5}}\,\,\,\,\,\,\,\,\,\log a{r^{n + 6}}\,\,\,\,\,\,\,\,\log a{r^{n + 7}} \end{array} \right|\]
$${\text{Operating }}{C_3} - {C_2}\,{\text{and }}{C_2} - {C_1}$$      and using
$$\log m - \log n = \log \frac{m}{n}\,\,{\text{we get}}$$
\[ = \left| \begin{array}{l} \log a{r^{n - 1}}\,\,\,\,\,\,\,\,\,\log r\,\,\,\,\,\,\,\,\log r\\ \log a{r^{n + 2}}\,\,\,\,\,\,\,\,\,\log r\,\,\,\,\,\,\,\,\log r\\ \log a{r^{n + 5}}\,\,\,\,\,\,\,\,\,\log r\,\,\,\,\,\,\,\,\log r \end{array} \right|\]
= 0 (two columns being identical)