Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\left| {\begin{array}{*{20}{c}} { - 5}&3&2 \\ 3&2&{ - 5} \\ 4&{ - 1}&{ - 3} \end{array}} \right| = 0\,\,{\text{and }}\left| {\begin{array}{*{20}{c}} { - 5}&3 \\ 3&2 \end{array}} \right| \ne 0.\]        So, the only possible third-order minor is 0, while one second-order minor is non-zero. So, the rank is 2.

Expanding by Sarrus rule,
\[\left| {\begin{array}{*{20}{c}} 1&{{e^{\frac{{i\pi }}{3}}}}&{{e^{\frac{{i\pi }}{4}}}}\\ {{e^{ - \frac{{i\pi }}{3}}}}&1&{{e^{\frac{{i2\pi }}{3}}}}\\ {{e^{ - \frac{{i\pi }}{4}}}}&{{e^{ - \frac{{i2\pi }}{3}}}}&1 \end{array}} \right| = 1 + {e^{\frac{{i\pi }}{3}}} \times {e^{\frac{{i2\pi }}{3}}} \times {e^{ - \frac{{i\pi }}{4}}} + {e^{ - \frac{{i\pi }}{3}}} \times {e^{ - \frac{{i2\pi }}{3}}} \times {e^{ - \frac{{i\pi }}{4}}} - \left( {{e^{\frac{{i2\pi }}{4}}} \times {e^{ - \frac{{i\pi }}{4}}} + {e^{ - \frac{{i\pi }}{3}}} \times {e^{ - \frac{{i\pi }}{3}}} + {e^{ - \frac{{i\pi }}{3}}} \times {e^{ - \frac{{i2\pi }}{3}}}} \right)\]
$$\eqalign{ & = 1 + {e^{\frac{{i3\pi }}{4}}} + {e^{ - \frac{{i3\pi }}{4}}} - \left( {1 + 1 + 1} \right) \cr & = - 2 + 2\cos \left( {\frac{{3\pi }}{4}} \right) = - 2 - \sqrt 2 \cr} $$

Let
\[\begin{array}{l} M = \left[ \begin{array}{l} {a_1}\,\,\,\,\,\,{a_2}\,\,\,\,\,\,{a_3}\\ {a_4}\,\,\,\,\,\,{a_5}\,\,\,\,\,\,{a_6}\\ {a_7}\,\,\,\,\,\,{a_8}\,\,\,\,\,\,{a_9} \end{array} \right]{\rm{where }}\,\,{a_i} \in \left\{ {0,1,2} \right\}\\ {\rm{Then }}\,\,{M^T}M = \left[ \begin{array}{l} {a_1}\,\,\,\,\,\,\,\,{a_4}\,\,\,\,\,\,\,\,{a_7}\\ {a_2}\,\,\,\,\,\,\,\,{a_5}\,\,\,\,\,\,\,\,{a_8}\\ {a_3}\,\,\,\,\,\,\,\,{a_6}\,\,\,\,\,\,\,\,{a_9} \end{array} \right]\left[ \begin{array}{l} {a_1}\,\,\,\,\,\,{a_2}\,\,\,\,\,\,{a_3}\\ {a_4}\,\,\,\,\,\,{a_5}\,\,\,\,\,\,{a_6}\\ {a_7}\,\,\,\,\,\,{a_8}\,\,\,\,\,\,{a_9} \end{array} \right] \end{array}\]
Sum of the diagonal entries in $${M^T}M = 5$$
$$ \Rightarrow \,\,\left( {{a_1}^2 + {a_4}^2 + {a_7}^2} \right) + \left( {{a_2}^2 + {a_5}^2 + {a_8}^2} \right) + \left( {{a_3}^2 + {a_6}^2 + {a_9}^2} \right) = 5$$
It is possible when
Case I :
5 $${a_i}s$$ are 1 and 4 $${a_i}\,'s$$ are zero
Which can be done in
$$^9{C_4}\,{\text{ways = }}\frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}} = 126$$
Case II :
1 $${a_i}$$ is 1 and 1$${a_i}$$ is 2 and rest. 7 $${a_i}\,'s$$ are Zero
It can be done in $$^9{C_1} \times {\,^8}{C_1} = 9 \times 8 = 72\,{\text{ways}}$$
$$\therefore \,\,{\text{Total no}}{\text{. of ways}} = 126 + 72 = 198.$$

$$\eqalign{ & \left( {A + B} \right)\left( {A - B} \right) = \left( {A - B} \right)\left( {A + B} \right) \cr & \Rightarrow AB = BA \cr} $$
as $$A$$ is symmetric and $$B$$ is skew-symmetric
$${\left( {AB} \right)^t} = - AB$$
⇒ $$k$$ is an odd integer.

We know that, $$adj \,A$$  and $$A$$ has same value of determinant, if $$\det \,{A = 0},$$   then $$\det \left( {adj\,A} \right) = 0$$
so, statement (1) is correct.
Also If $$A$$ is a matrix the determinant of $$A^{- 1}$$ equals inverse of determinant $$A,$$ so, $$\det \left( {{A^{ - 1}}} \right) = {\left( {\det \,A} \right)^{ - 1}},$$     if $$A$$ is non singular; Statement 2 is correct.
Thus both (1) and (2) are correct.

$$\eqalign{ & A\left( {adj\,A} \right) = A\,{A^T} \cr & \Rightarrow \,\,{A^{ - 1}}A\left( {adj\,A} \right) = {A^{ - 1}}A\,{A^T} \cr & adj\,A = {A^T} \cr} $$
\[ \Rightarrow \,\,\left[ \begin{array}{l} \,\,2\,\,\,\,\,\,\,\,\,b\\ - 3\,\,\,\,\,\,5a \end{array} \right] = \left[ \begin{array}{l} 5a\,\,\,\,\,\,\,3\\ - b\,\,\,\,\,\,2 \end{array} \right]\]
$$\eqalign{ & \Rightarrow \,a = \frac{2}{5}\,\,{\text{and }}b = 3 \cr & \Rightarrow \,\,5a + b = 5 \cr} $$

The given system of equations is
$$\eqalign{ & {a_1}x + {b_1}y + {c_1}z = {d_1} \cr & {a_2}x + {b_2}y + {c_2}z = {d_2} \cr & {\text{and }}{a_3}x + {b_3}y + {c_3}z = {d_3} \cr} $$
Let \[\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|\]
This system has a unique solution $$x_0 , y_0 , z_0$$
If $$\Delta \ne 0\,\,{\rm{ and }}\,\,{x_0} = \frac{{\Delta x}}{\Delta } = 0$$
$$ \Rightarrow \Delta x = 0$$
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\]

\[\begin{array}{l} f\left( x \right) = \left| {\begin{array}{*{20}{c}} {{2^{ - x}}}&{{2^x}}&{{x^2}}\\ {{2^{ - 3x}}}&{{2^{3x}}}&{{x^4}}\\ {{2^{ - 5x}}}&{{2^{5x}}}&1 \end{array}} \right|\\ \therefore f\left( { - x} \right) = \left| {\begin{array}{*{20}{c}} {{2^x}}&{{2^{ - x}}}&{{x^2}}\\ {{2^{3x}}}&{{2^{ - 3x}}}&{{x^4}}\\ {{2^{5x}}}&{{2^{ - 5x}}}&1 \end{array}} \right| = - f\left( x \right) \end{array}\]

\[\vartriangle = \left| {\begin{array}{*{20}{c}} {{a^3}}&{{b^3}}&{{c^3}} \\ a&b&c \\ {{a^2}}&{{b^2}}&{{c^2}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} { - 1}&{ - 1}&{ - 1} \\ a&b&c \\ {{a^2}}&{{b^2}}&{{c^2}} \end{array}} \right|\]
\[\vartriangle = abc\left| {\begin{array}{*{20}{c}} {{a^2}}&{{b^2}}&{{c^2}} \\ 1&1&1 \\ a&b&c \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 1&1&1 \\ a&b&c \\ {{a^2}}&{{b^2}}&{{c^2}} \end{array}} \right|\]
\[\vartriangle = \left( {abc - 1} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ a&b&c \\ {{a^2}}&{{b^2}}&{{c^2}} \end{array}} \right|\]
$$\vartriangle = \left( {abc - 1} \right)\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right).\,{\text{But }}a \ne b \ne c.$$

If $$A$$ is non-singular and $$B$$ is singular, then $$AB$$  and $$A^{–1}B$$  are non-singular. Statements (2) and (4) are correct.