Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

\[\left| {\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {a + b + c}&b&c\\ {a + b + c}&c&a\\ {a + b + c}&a&b \end{array}} \right|\]
$$\left( {\because {C_1} \to {C_1} + {C_2} + {C_3}} \right)$$
\[ = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}} 1&b&c\\ 1&c&a\\ 1&a&b \end{array}} \right|\]
[ on taking $$(a + b + c)$$   common from $$C_1$$ ]
$$\eqalign{ & = \left( {a + b + c} \right)\left[ {1\left( {bc - {a^2}} \right) - b\left( {b - a} \right) + c\left( {a - c} \right)} \right] \cr & = \left( {a + b + c} \right)\left[ {bc - {a^2} - {b^2} + ab + ac - {c^2}} \right] \cr & = \left( {a + b + c} \right)\left[ { - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right] \cr & = - \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr} $$
Hence, the determinant is negative value.

Using $${C_2} \to {C_2} - {C_1}$$   and $${C_3} \to {C_3} - {C_1}$$   in the given determinant, we have
\[\Delta = \left| {\begin{array}{*{20}{c}} 1&0&0\\ {1 + \sin A}&{\sin B - \sin A}&{\sin \,C - \sin \,A}\\ {\sin \,A + {{\sin }^2}A}&{{{\sin }^2}B - {{\sin }^2}A}&{{{\sin }^2}C - {{\sin }^2}A} \end{array}} \right|\]
Now taking $$\sin \,B - \sin \,A$$   common from $$C_2$$ and $${\sin \,C - \sin \,A}$$   common from $$C_3,$$ we have
$$\Delta = \left( {\sin \,B - \sin \,A} \right)\,\left( {\sin \,C - \sin \,A} \right)$$
\[\left| {\begin{array}{*{20}{c}} 1&0&0\\ {1 + \sin A}&1&1\\ {\sin \,A + {{\sin }^2}A}&{\sin \,B + \sin \,A}&{\sin \,C + \sin \,A} \end{array}} \right|\]
$$ = \,\left( {\sin \,B - \sin \,A} \right)\,\left( {\sin \,C - \sin \,A} \right)\,\left( {\sin \,C - \sin \,B} \right).$$
As the determinant is zero, we must have $$\sin B = \sin A{\text{ or }}\sin A{\text{ or }}\sin C = \sin A{\text{ or }}\sin C = \sin B,{\text{ that is, }}B = A{\text{ or }}C = A{\text{ or }}C = B.$$
In all three cases we will have an isosceles triangle.

$$\eqalign{ & A \to 3 \times 3,B \to 3 \times 2,C \to 3 \times 1 \cr & AB \to 3 \times 2 \cr & \Rightarrow \,{\left( {AB} \right)^T} = 2 \times 3 \cr & \Rightarrow \,{\left( {AB} \right)^T}C\,{\text{is}}\,{\text{defined}} \cr & \Rightarrow \,{C^T} \to 1 \times 3, \cr & \Rightarrow \,{C^T}C \to 1 \times 1 \cr} $$
Hence, $${C^T}C{\left( {AB} \right)^T}$$   is not defined. Now, $$C^T AB$$  is also defined.
$$\eqalign{ & {A^T} \to 3 \times 3,{B^T} \to 2 \times 3;\,\,\,{A^T}A \to 3 \times 3 \cr & B{B^T} \to 3 \times 3 \cr & \Rightarrow \,{A^T}AB{B^T} \to 3 \times 3 \cr & \Rightarrow \,{A^T}AB{B^T}C\,{\text{is}}\,{\text{defined}} \cr} $$

$$\eqalign{ & {\text{Given that }}\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2},{\omega ^2} = - \frac{1}{2} - i\frac{{\sqrt 3 }}{2} \cr & {\text{Also 1}} + \omega + {\omega ^2} = 0\,\,{\text{and }}{\omega ^3} = 1 \cr & {\text{Now given det}}{\text{. is}} \cr} $$
\[\Delta = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\, - 1 - {\omega ^2}\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^4} \end{array} \right| = \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ 1\,\,\,\,\,\,\,\,\omega \,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^2}\\ 1\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\,\,\,\omega \end{array} \right|\]
$$\eqalign{ & \left[ {{\text{Using }}\omega = - 1 - {\omega ^2}\,\,{\text{and }}{\omega ^3} = 1} \right] \cr & {\text{Operating }}\,{C_1} \to {C_1} + {C_2} + {C_3} \cr} $$
\[\Delta = \left| \begin{gathered} \,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1 \hfill \\ \,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\omega \,\,\,\,\,\,\,\,\,\,{\omega ^2} \hfill \\ 0\,\,\,\,\,\,\,\,\,{\omega ^2}\,\,\,\,\,\,\,\,\omega \hfill \\ \end{gathered} \right|\,\,\left( {{\text{as 1 + }}\omega {\text{ + }}{\omega ^2} = 0} \right)\]
Expanding along $${C_1},$$ we get
$$3\left( {{\omega ^2} - {\omega ^4}} \right) = 3\left( {{\omega ^2} - \omega } \right) = 3\omega \left( {\omega - 1} \right)$$

\[\left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,1\,\,\,\,\,\, - 2\\ a\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,b \end{array} \right]\left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,a\\ 2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,b \end{array} \right] = \left[ \begin{array}{l} 9\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,9\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,9 \end{array} \right]\]
\[ \Rightarrow \,\,\left[ \begin{array}{l} \,\,1 + 4 + 4\,\,\,\,\,\,\,\,\,\,\,\,2 + 2 - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a + 4 + 2b\\ \,2 + 2 - 4\,\,\,\,\,\,\,\,\,\,\,\,\,4 + 1 + 4\,\,\,\,\,\,\,\,\,\,\,\,\,2a + 2 - 2b\\ a + 4 + 2b\,\,\,\,\,\,\,\,2a + 2 - 2b\,\,\,\,\,\,\,\,\,\,\,{a^2} + 4 + {b^2} \end{array} \right] = \left[ \begin{array}{l} 9\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,9\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,9 \end{array} \right]\]
$$⇒ a + 4 + 2b = 0$$
$$\eqalign{ & \Rightarrow \,\,a + 2b = - 4\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr & 2a + 2 - 2b = 0 \cr & \Rightarrow \,\,2a - 2b = - 2 \cr & \Rightarrow \,\,a - b = - 1\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
On solving (i) and (ii) we get
$$\eqalign{ & - 1 + b + 2b = - 4 \cr & b = - 1\,\,\,{\text{and }}a = - 2 \cr & \left( {a,b} \right) = \left( { - 2, - 1} \right) \cr} $$

We have,
\[\begin{array}{l} AB = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta }&{\cos \theta \sin \theta }\\ {\cos \theta \sin \theta }&{{{\sin }^2}\theta } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\phi }&{\cos \phi \sin \phi }\\ {\cos \phi \sin \phi }&{{{\sin }^2}\phi } \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\theta \,{{\cos }^2}\phi + \cos \theta \,\cos \phi \,\sin \theta \,\sin \phi \,{{\cos }^2}\theta \,\cos \phi \,\sin \phi + \cos \theta \,\sin \theta \,{{\sin }^2}\phi }\\ {\cos \theta \,\sin \theta\, {{\cos }^2}\phi + {{\sin }^2}\theta \,\cos \phi \,\sin \phi \,\cos \theta \,\cos \phi \,\sin\theta \,\sin\phi + {\sin^2}\theta \,{{\sin }^2}\phi } \end{array}} \right]\\ = \,\cos \left( {\theta - \phi } \right)\left[ {\begin{array}{*{20}{c}} {\cos \theta \cos \phi }&{\cos \theta \sin \phi }\\ {\sin \theta \cos \phi }&{\sin \theta \sin \phi } \end{array}} \right] \end{array}\]
$$\eqalign{ & {\text{Since,}}\,AB = 0 \cr & \therefore \,\cos \left( {\theta - \phi } \right) = 0 \cr} $$
$$\therefore \,\theta - \phi $$   is an odd multiple of $$\frac{\pi }{2}$$

\[\left| {\begin{array}{*{20}{c}} a&1&1\\ 1&b&1\\ 1&1&c \end{array}} \right| > 0\]
$$\eqalign{ & \Rightarrow a\left( {bc - 1} \right) - 1\left( {c - 1} \right) + 1\left( {1 - b} \right) > 0 \cr & \Rightarrow abc - a - c + 1 + 1 - b > 0 \cr & \Rightarrow abc + 2 - \left( {a + b + c} \right) > 0 \cr & \Rightarrow abc > \left( {a + b + c} \right) - 2 \cr & {\text{Let, }}a = - 1;b = 0\,\,\& \,\,c = 1 \cr} $$
Then, $$0 > - 2$$  [which is correct]
Hence, $$abc = 0$$
⇒ $$abc > - 8$$

Consider first two equations :
$$2x + 3y = - 4$$    and $$3x + 4y = - 6$$
We have, \[\Delta = \left| {\begin{array}{*{20}{c}} 2&3\\ 3&4 \end{array}} \right| = - 1 \ne 0\]
\[{\Delta _x} = \left| {\begin{array}{*{20}{c}} { - 4}&3\\ { - 6}&4 \end{array}} \right| = 2\,\,{\rm{and }}\,\,{\Delta _y} = \left| {\begin{array}{*{20}{c}} 2&{ - 4}\\ 3&{ - 6} \end{array}} \right| = 0\]
∴ $$x = - 2$$  and $$y = 0$$
Now this solution satisfies the third, so the equations are consistent with unique solution.

\[\vartriangle = {\left( {1 + x} \right)^2}\left| {\begin{array}{*{20}{c}} 1&x&{{x^2}} \\ 1&{1 + x}&{{x^2}} \\ 1&x&{1 + x} \end{array}} \right|.\]
Differentiating both sides of the given equality w.r.t. $$x,$$ we get
\[2\left( {1 + x} \right)\left| {\begin{array}{*{20}{c}} 1&x&{{x^2}} \\ 1&{1 + x}&{{x^2}} \\ 1&x&{1 + x} \end{array}} \right| + {\left( {1 + x} \right)^2}\left\{ {\left| {\begin{array}{*{20}{c}} 1&1&{{x^2}} \\ 1&1&{{x^2}} \\ 1&1&{1 + x} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&x&{2x} \\ 1&{1 + x}&{2x} \\ 1&x&1 \end{array}} \right|} \right\} = 5a{x^4} + 4b{x^3} + 3c{x^2} + 2dx + \lambda .\]
Putting $$x = 0,$$
\[2\left| {\begin{array}{*{20}{c}} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&0&0 \\ 1&1&0 \\ 1&0&1 \end{array}} \right| = \lambda \]
\[\therefore \,\,2 + 1 = \lambda .\]

$$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}$$   { $$n$$ is order of square matrix }
If $$A$$ is square matrix of order 3, then $$\left| {adj\,A} \right| = {\left| A \right|^2}$$