Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

For any square matrix $$X,$$ we have $$X\left( {adj\,X} \right) = \left| X \right|{I_n}$$
Taking $$X = adj\,A,$$   we get
$$\eqalign{ & \left( {adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = \left| {adj\,A} \right|{I_n} \cr & \Rightarrow \left( {adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = {\left| A \right|^{n - 1}}{I_n} \cr & \left[ {\because \left| {adj\,A} \right| = {{\left| A \right|}^{n - 1}}} \right] \cr & \Rightarrow \left( {A\,adj\,A} \right)\left[ {adj\left( {adj\,A} \right)} \right] = {\left| A \right|^{n - 1}}A \cr & \left[ {\because A\,{I_n} = A} \right] \cr & \left( {\left| A \right|{I_n}} \right)\left( {adj\left( {adj\,A} \right)} \right) = {\left| A \right|^{n - 1}}A \cr & \Rightarrow adj\left( {adj\,A} \right) = {\left| A \right|^{n - 2}}A \cr} $$

We have
\[\begin{array}{l} \left| Q \right| = \left| \begin{array}{l} {2^2}{a_{11}}\,\,\,\,\,\,\,\,{2^3}{a_{12}}\,\,\,\,\,\,\,\,\,{2^4}{a_{13}}\\ {2^3}{a_{21}}\,\,\,\,\,\,\,\,{2^4}{a_{22}}\,\,\,\,\,\,\,\,{2^5}{a_{23}}\\ {2^4}{a_{31}}\,\,\,\,\,\,\,\,{2^5}{a_{32}}\,\,\,\,\,\,\,\,{2^6}{a_{33}} \end{array} \right|\\ = {2^2}{.2^3}{.2^4}\,\left| \begin{array}{l} \,\,\,{a_{11}}\,\,\,\,\,\,\,\,\,\,{a_{12}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a_{13}}\\ \,2{a_{21}}\,\,\,\,\,\,\,2{a_{22}}\,\,\,\,\,\,\,\,\,\,\,\,2{a_{23}}\\ {2^2}{a_{31}}\,\,\,\,\,{2^2}{a_{32}}\,\,\,\,\,\,\,\,\,{2^2}{a_{33}} \end{array} \right|\\ = {2^9}{.2.2^2}\left| \begin{array}{l} {a_{11}}\,\,\,\,\,{a_{12}}\,\,\,\,\,\,{a_{13}}\\ {a_{21}}\,\,\,\,\,{a_{22}}\,\,\,\,\,{a_{23}}\\ {a_{31}}\,\,\,\,\,{a_{32}}\,\,\,\,\,{a_{33}} \end{array} \right| \end{array}\]
$$ = {2^{12}} \times \left| P \right| = {2^{12}} \times 2 = {2^{13}}$$

\[\begin{array}{l} X = BC = \left[ {\begin{array}{*{20}{c}} 3&4\\ 2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&{ - 4}\\ { - 2}&3 \end{array}} \right]\\ \Rightarrow X = BC = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = I \end{array}\]
$${\text{So, }}{X^n} = {I^n} = I$$

We know that $$\left| {{\text{adj}}\left( {{\text{adj}}\,A} \right)} \right| = {\left| {{\text{adj}}\,A} \right|^{2 - 1}}$$
$$ = {\left| A \right|^{2 - 1}} = \left| A \right|$$
∴ Both the statements are true and statement - 2 is a correct explantion for statement - 1 .

\[\left| {\begin{array}{*{20}{c}} {{x^2} + x}&{3x - 1}&{ - x + 3}\\ {2x + 1}&{2 + {x^2}}&{{x^3} - 3}\\ {x - 3}&{{x^2} + 4}&{3x} \end{array}} \right| = {a_0} + {a_1}x + {a_2}{x^2} + ..... + {a_7}{x^7}\]
Put, $$x = 0$$
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} 0&{ - 1}&3\\ 1&2&{ - 3}\\ { - 3}&4&0 \end{array}} \right| = {a_0}\]
⇒ $${a_0} = 21$$

Let $$A = {\left[ {{a_{ij}}} \right]_{m\, \times m}}$$   be a matrix and $$C = {\left[ {{c_{ij}}} \right]_{m\, \times m}}$$   be another matrix where $${c_{ij}}$$  is the cofactor of $${a_{ij}}.$$
∴ The value of $$\left| {AC} \right| = {\left| A \right|^{m + 1}}$$

\[\begin{array}{l} {\rm{Given,}}\\ \left[ {\begin{array}{*{20}{c}} {x + y}&y\\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 2y}&{ - y}\\ {4x}&{ - x + y} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\\ \Rightarrow \left[ \begin{array}{l} 2x + y\\ 3x + y \end{array} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right] \end{array}\]
$$\eqalign{ & {\text{Comparing the corresponding elements}} \cr & 2x + y = 3......\left( {\text{i}} \right) \cr & 3x + y = 2......\left( {{\text{ii}}} \right) \cr & {\text{Subtracting, we get}} \cr & - x = 1 \cr & \Rightarrow x = - 1 \cr & {\text{Substituting the vale of }}x{\text{ in equation }}\left( {\text{i}} \right) \cr & 2\left( { - 1} \right) + y = 3 \cr & \Rightarrow - 2 + y = 3 \cr & \Rightarrow y = 3 + 2 = 5 \cr & {\text{Hence, }}x = - 1,\,y = 5 \cr & \therefore \,x \cdot y = - 5 \cr} $$

\[\begin{array}{l} {\rm{Let, }}\,\,A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&0\\ 1&{ - 1}&2 \end{array}} \right],\,\,{\rm{then}}\\ A' = \left[ {\begin{array}{*{20}{c}} 1&2&1\\ { - 1}&1&{ - 1}\\ 1&0&2 \end{array}} \right]\\ \therefore AA' = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&0\\ 1&{ - 1}&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&2&1\\ { - 1}&1&{ - 1}\\ 1&0&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&1&4\\ 1&5&1\\ 4&1&4 \end{array}} \right] \end{array}\]

\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} 0&c&{ - b}\\ { - c}&0&a\\ b&{ - a}&0 \end{array}} \right],\,B = \left[ {\begin{array}{*{20}{c}} {{a^2}}&{ab}&{ac}\\ {ab}&{{b^2}}&{bc}\\ {ac}&{bc}&{{c^2}} \end{array}} \right]\\ AB = \left[ {\begin{array}{*{20}{c}} 0&c&{ - b}\\ { - c}&0&a\\ b&{ - a}&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{a^2}}&{ab}&{ac}\\ {ab}&{{b^2}}&{bc}\\ {ac}&{bc}&{{c^2}} \end{array}} \right]\\ AB = \left[ \begin{array}{l} 0 + abc - abc\,\,\,\,\,\,\,\,\,\,0 + {b^2}c - {b^2}c\,\,\,\,\,\,\,\,\,\,\,\,0 + b{c^2} - b{c^2}\\ - {a^2}c + 0 + {a^2}c\,\,\,\,\,\,\,\,\, - abc + 0 + abc\,\,\,\,\,\,\,\, - a{c^2} + 0 + a{c^2}\\ {a^2}b - {a^2}b + 0\,\,\,\,\,\,\,\,\,\,a{b^2} - a{b^2} + 0\,\,\,\,\,\,\,\,\,\,\,abc - abc + 0 \end{array} \right]\\ AB = \left[ \begin{array}{l} 0\,\,\,\,\,0\,\,\,\,\,0\\ 0\,\,\,\,\,0\,\,\,\,\,0\\ 0\,\,\,\,\,0\,\,\,\,\,0 \end{array} \right]\\ AB = {0_{3 \times 3}}......\left( 1 \right)\\ BA = \left[ {\begin{array}{*{20}{c}} {{a^2}}&{ab}&{ac}\\ {ab}&{{b^2}}&{bc}\\ {ac}&{bc}&{{c^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&c&{ - b}\\ { - c}&0&a\\ b&{ - a}&0 \end{array}} \right]\\ BA = \left[ \begin{array}{l} 0 + abc - abc\,\,\,\,\,\,\,\,\,\,0 + {b^2}c - {b^2}c\,\,\,\,\,\,\,\,\,\,\,\,0 + b{c^2} - b{c^2}\\ - {a^2}c + 0 + {a^2}c\,\,\,\,\,\,\,\,\, - abc + 0 + abc\,\,\,\,\,\,\,\, - a{c^2} + 0 + a{c^2}\\ {a^2}b - {a^2}b + 0\,\,\,\,\,\,\,\,\,\,a{b^2} - a{b^2} + 0\,\,\,\,\,\,\,\,\,\,\,abc - abc + 0 \end{array} \right]\\ BA = \left[ \begin{array}{l} 0\,\,\,\,\,0\,\,\,\,\,0\\ 0\,\,\,\,\,0\,\,\,\,\,0\\ 0\,\,\,\,\,0\,\,\,\,\,0 \end{array} \right]\\ BA = {0_{3 \times 3}}......\left( 2 \right)\\ {\rm{From\, equation\, }}\left( 1 \right){\rm{ \,and\, }}\left( 2 \right)\\ AB = BA = {0_{3 \times 3}} \end{array}\]

$$\eqalign{ & ^5{C_0}{ + ^5}{C_1}{ + ^5}{C_2}{ = ^5}{C_5}{ + ^5}{C_4}{ + ^5}{C_3} = {2^{5 - 1}} = 16. \cr & {\text{Use }}{R_1} \to {R_1} + {R_2} + {R_3}. \cr} $$