Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

The given system of equations is
$$\eqalign{ & x + y + z = 2{\text{ }}\,\,\,.....\left( {\text{i}} \right) \cr & 2x + y - z = 3{\text{ }}\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{and }}\,3x + 2y + kz = 4\,\,\,.....\left( {{\text{iii}}} \right) \cr} $$
This system has a unique solution if
\[\left| {\begin{array}{*{20}{c}} 1&1&1\\ 2&1&{ - 1}\\ 3&2&k \end{array}} \right| \ne 0\]
Applying, $${C_2} \to {C_2} - {C_1}\,\,{\text{and }}{C_3} \to {C_3} - {C_1}$$
We get, \[\left| {\begin{array}{*{20}{c}} 1&0&0\\ 3&{ - 1}&{ - 3}\\ 3&{ - 1}&{k - 3} \end{array}} \right| \ne 0\]
$$\eqalign{ & \Rightarrow - 1\left( {k - 3} \right) - 3 \ne 0{\text{ or }} - k + 3 - 3 \ne 0 \cr & \Rightarrow k \ne 0 \cr} $$

The given system is, $$x + ay = 0$$
$$az + y = 0$$
$$ax + z = 0$$
It is system of homogeneous equations therefore, it will have infinite many solutions if determinant of co-efficient matrix is zero. i.e.,
\[\left| \begin{array}{l} 1\,\,\,\,\,\,\,a\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,1\,\,\,\,\,\,\,a\\ a\,\,\,\,\,\,\,0\,\,\,\,\,\,\,1 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \,\,1\left( {1 - 0} \right) - a\left( {0 - {a^2}} \right) = 0 \cr & \Rightarrow \,\,1 + {a^3} = 0 \cr & \Rightarrow \,\,{a^3} = - 1 \cr & \Rightarrow \,\,a = - 1 \cr} $$

Applying, $${C_1} \to {C_1} + {C_2} + {C_3}\,\,{\text{we get}}$$
\[f\left( x \right) = \left| \begin{array}{l} 1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\ 1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\,1 + {b^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\ 1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x\,\,\,\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,\,1 + {c^2}x \end{array} \right|\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\ 1\,\,\,\,\,\,\,\,\,1 + {b^2}x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {c^2}} \right)x\\ 1\,\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,\,\,\,1 + {c^2}x \end{array} \right|\]
$$\eqalign{ & \left[ {{\text{As given that }}{a^2} + {b^2} + {c^2} = - 2} \right] \cr & \therefore \,\,{a^2} + {b^2} + {c^2} + 2 = 0 \cr & {\text{Applying }}{R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3} \cr} $$
\[\therefore \,\,f\left( x \right) = \left| \begin{array}{l} 0\,\,\,\,\,\,\,\,\,x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,\,1 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x - 1\\ 1\,\,\,\,\,\,\left( {1 + {b^2}} \right)x\,\,\,\,\,\,\,\,\,1 + {c^2}x \end{array} \right|\]
$$f\left( x \right) = {\left( {x - 1} \right)^2}\,\,\,\,\,\,\,{\text{Hence degree}} = 2.$$

Let $$A$$ be a matrix such that $$3A^3 + 2A^2 + 5A + I = 0$$
Post multiply by $$A^{-1}$$ on both the sides, we get,
$$\eqalign{ & 3{A^3}{A^{ - 1}} + 2{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + I{A^{ - 1}} = 0 \cr & \Rightarrow 3{A^2} + 2A + 5I + {A^{ - 1}} = 0 \cr & \Rightarrow {A^{ - 1}} = - \left( {3{A^2} + 2A + 5I} \right) \cr} $$

Cofactor matrix \[ = \,\left| {\begin{array}{*{20}{c}} {{a^2}}&0&0\\ 0&{{a^2}}&0\\ 0&0&{{a^2}} \end{array}} \right|\]
\[\begin{array}{l} \therefore \,adj\,A = \left( {{\rm{cofactor}}\,{\rm{matrix}}} \right)' = \left[ {\begin{array}{*{20}{c}} {{a^2}}&0&0\\ 0&{{a^2}}&0\\ 0&0&{{a^2}} \end{array}} \right]\\ \therefore \,\left| {adj\,A} \right| = \left[ {\begin{array}{*{20}{c}} {{a^2}}&0&0\\ 0&{{a^2}}&0\\ 0&0&{{a^2}} \end{array}} \right] = {a^6} \end{array}\]

\[\begin{gathered} \because f\left( x \right) = \left| {\begin{array}{*{20}{c}} n&{n + 1}&{n + 2} \\ {{\,^n}{P_n}}&{^{n + 1}{P_{n + 1}}}&{^{n + 2}{P_{n + 2}}} \\ {{\,^n}{C_n}}&{^{n + 1}{C_{n + 1}}}&{^{n + 2}{C_{n + 2}}} \end{array}} \right| \hfill \\ = \,\left| {\begin{array}{*{20}{c}} n&{n + 1}&{n + 2} \\ {n!}&{\left( {n + 1} \right)!}&{\left( {n + 2} \right)!} \\ 1&1&1 \end{array}} \right|\,\left( {\because {\,^n}{P_n} = n!{,^n}{C_n} = 1} \right) \hfill \\ \end{gathered} \]
Applying, $${C_2} \to {C_2} - {C_1}\,\,{\text{and}}\,\,{C_3} \to {C_3} - {C_1}$$
Then, \[f\left( x \right) = \left| {\begin{array}{*{20}{c}} n&1&2\\ {n!}&{n.n!}&{\left( {{n^2} + 3n + 1} \right)n!}\\ 1&0&0 \end{array}} \right|\]
\[ = \,\left| {\begin{array}{*{20}{c}} 1&2\\ {n.n!}&{\left( {{n^2} + 3n + 1} \right)n!} \end{array}} \right| = n!\left( {{n^2} + n + 1} \right)\]

\[{\text{Using }}{C_3} \to {C_3} - {C_1},\,\vartriangle = \left| {\begin{array}{*{20}{c}} 1&\alpha &{{\alpha ^2} - 1} \\ {\cos \left( {p - d} \right)a}&{\cos pa}&0 \\ {\sin \left( {p - d} \right)a}&{\sin pa}&0 \end{array}} \right| = \left( {{\alpha ^2} - 1} \right)\sin \left\{ {pa - \left( {p - d} \right)a} \right\} = \left( {{\alpha ^2} - 1} \right)\sin da.\]

$$\eqalign{ & {\text{Given }}{P^3} = {Q^3}\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}{P^2}Q = {Q^2}P\,\,\,.....\left( 2 \right) \cr} $$
Subtracting (1) and (2), we get
$$\eqalign{ & {P^3} - {P^2}Q = {Q^3} - {Q^2}P \cr & \Rightarrow \,\,{P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0 \cr & \Rightarrow \,\,\left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0 \cr & \Rightarrow \,\,\left| {{P^2} + {Q^2}} \right| = 0\,\,{\text{as }}P \ne Q \cr} $$

Given
\[\begin{array}{l} {\Delta _1} = \left| {\begin{array}{*{20}{c}} 7&x&2\\ { - 5}&{x + 1}&3\\ 4&x&7 \end{array}} \right|\\ = 7\left( {7x + 7 - 3x} \right) - x\left( { - 35 - 12} \right) + 2\left( { - 5x - 4x - 4} \right)\\ = 28x + 49 + 47x - 18x - 8\\ = 57x + 41\\ {\Delta _2} = \left| {\begin{array}{*{20}{c}} x&2&7\\ {x + 1}&3&{ - 5}\\ x&7&4 \end{array}} \right|\\ = x\left( {12 + 35} \right) - 2\left( {4x + 4 + 5x} \right) + 7\left( {7x + 7 - 3x} \right)\\ = 47x - 18x - 8\ + 28x + 49\\ = 57x + 41\\ {\rm{Now,\, }}{\Delta _1} - {\Delta _2} = 0\\ {\rm{Since\, }}{\Delta _1} = {\Delta _2} \end{array}\]
So, the equation is correct for all real values of $$x.$$
So, the correct option is 'B'.

$$\eqalign{ & \left| {{A^2}} \right| = 25 \cr & \Rightarrow \,\,{\left| A \right|^2} = 25 \cr & \Rightarrow \,\,{\left( {25\alpha } \right)^2} = 25 \cr & \Rightarrow \,\,\left| \alpha \right| = \frac{1}{5} \cr} $$