Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

For infinitely many solutions the two equations become identical
$$\eqalign{ & \Rightarrow \,\,\frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}} \cr & \Rightarrow \,\,k = 1 \cr} $$

\[\left| {\begin{array}{*{20}{c}} 1&{\sin A}&{{{\sin }^2}A}\\ 1&{\sin B}&{{{\sin }^2}B}\\ 1&{\sin C}&{{{\sin }^2}C} \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow \left( {\sin A - \sin B} \right)\left( {\sin B - \sin C} \right)\left( {\sin C - \sin A} \right) = 0 \cr & \Rightarrow \sin A = \sin B{\text{ or }}\sin B = \sin C{\text{ or }}\sin C = \sin A \cr} $$
∴ atleast two of $$A, B, C$$  are equal.
Hence the triangle is isosceles or equilateral.

Since, $$a, b, c$$  are in G.P.
⇒ $$b^2 = ac$$
Explanding the determinant we get,
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} a&b&{a + b}\\ b&c&{b + c}\\ {a + b}&{b + c}&0 \end{array}} \right|\\ = a\left| {\begin{array}{*{20}{c}} c&{b + c}\\ {b + c}&0 \end{array}} \right| - b\left| {\begin{array}{*{20}{c}} b&{b + c}\\ {a + b}&0 \end{array}} \right| + \left( {a + b} \right)\left| {\begin{array}{*{20}{c}} b&c\\ {a + b}&{b + c} \end{array}} \right| \end{array}\]
$$\eqalign{ & = - a{\left( {b + c} \right)^2} + b\left( {a + b} \right)\left( {b + c} \right) + \left( {a + b} \right)\left( {{b^2} + bc - ac - bc} \right) \cr & = - a\left( {{b^2} + {c^2} + 2bc} \right) + b\left( {ab + ac + {b^2} + bc} \right) \cr & = - a{b^2} - a{c^2} - 2abc + a{b^2} + 2abc + {b^2}c\left( {\because {b^2} = ac} \right) \cr & = - a{c^2} + {b^2}c = - a{c^2} + ac \cdot c = - a{c^2} + a{c^2} = 0 \cr} $$

Here to find the value of $$tr\left( A \right) - tr\left( B \right),$$    we need not to find the matrices $$A$$ and $$B.$$
We can find $$tr\left( A \right) - tr\left( B \right)$$    using the properties of trace of matrix, i.e.,
\[A + 2B = \left[ {\begin{array}{*{20}{c}} 1&2&0\\ 6&{ - 3}&3\\ { - 5}&3&1 \end{array}} \right]\]
$$\eqalign{ & \Rightarrow tr\left( {A + 2B} \right) = - 1\left( 1 \right) \cr & {\text{or, }}tr\left( A \right) + 2tr\left( B \right) = - 1 \cr} $$
\[ \Rightarrow 2A - B = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&5\\ 2&{ - 1}&6\\ 0&1&2 \end{array}} \right]\]
$$ \Rightarrow tr\left( {2A - B} \right) = 3\,\,\,\,{\text{or, }}2tr\left( A \right) - tr\left( B \right) = 3\left( 2 \right)$$
Solving (1) and (2), we get $$tr\left( A \right) = 1{\text{ and }}tr\left( B \right) = - 1$$
$$ \Rightarrow tr\left( A \right) - tr\left( B \right) = 2$$

For homogeneous system of equations to have non-zero solution, $$\Delta = 0$$
\[\begin{array}{l} \left| \begin{array}{l} 1\,\,\,\,\,\,2a\,\,\,\,\,\,a\\ 1\,\,\,\,\,\,3b\,\,\,\,\,\,\,b\\ 1\,\,\,\,\,\,4c\,\,\,\,\,\,c \end{array} \right| = 0\,\,\,{C_2} \to {C_2} - 2{C_3}\\ \left| \begin{array}{l} 1\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,a\\ 1\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,b\\ 1\,\,\,\,\,\,2c\,\,\,\,\,\,\,c \end{array} \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}\\ \left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\ 0\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b - a\\ 0\,\,\,\,\,\,2c\, - b\,\,\,\,\,\,\,\,c - b \end{array} \right| = 0 \end{array}\]
$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$
On simplification, $$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
∴ $$a, b, c$$  are in Harmonic Progression.

\[\left| {\begin{array}{*{20}{c}} {{{\sin }^2}{{13}^ \circ }}&{{{\cos }^2}{{13}^ \circ }}&{ - 1}\\ {{{\cos }^2}{{13}^ \circ }}&{ - 1}&{{{\sin }^2}{{13}^ \circ }}\\ { - 1}&{{{\sin }^2}{{13}^ \circ }}&{{{\cos }^2}{{13}^ \circ }} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 0&{{{\cos }^2}{{13}^ \circ }}&{ - 1}\\ 0&{ - 1}&{{{\sin }^2}{{13}^ \circ }}\\ 0&{{{\sin }^2}{{13}^ \circ }}&{{{\cos }^2}{{13}^ \circ }} \end{array}} \right| = 0\]
$${\text{Applying }}{C_1} \to {C_1} + {C_2} + {C_3}$$

$$Z$$ is idempotent then $${Z^2} = Z$$
$$\eqalign{ & \Rightarrow {Z^3},{Z^4},.....,{Z^n} = Z \cr & {\left( {I + Z} \right)^n} = {\,^n}{C_0}{I^n} + {\,^n}{C_1}{I^{n - 1}}Z + {\,^n}{C_2}{I^{n - 2}}{Z^2} + ..... + {\,^n}{C_n}{Z^n} \cr & = {\,^n}{C_0}I + {\,^n}{C_1}Z + {\,^n}{C_2}Z + {\,^n}{C_3}Z + ..... + {\,^n}{C_n}Z \cr & = I + \left( {^n{C_1} + {\,^n}{C_2} + {\,^n}{C_3} + ..... + {\,^n}{C_n}} \right)Z \cr & = I + \left( {{2^n} - 1} \right)Z \cr} $$

Equations are solvable when they are consistent.

We have, $${A^2} + {B^2} = AA + BB = A\left( {BA} \right) + B\left( {AB} \right)\left( {\therefore AB = B{\text{ and }}BA + A} \right)$$
$$\eqalign{ & = \left( {AB} \right)A + \left( {BA} \right)B \cr & = BA + AB = A + B\left( {\therefore AB = B{\text{ and }}BA = A} \right) \cr} $$

\[{\rm{Here, }}\left| \begin{array}{l} x - 4\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,x - 4 \end{array} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}\]
$${\text{Put }}x = 0$$
\[ \Rightarrow \,\,\left| \begin{array}{l} - 4\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,0\,\,\,\,\,\,\, - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\\ \,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - 4 \end{array} \right| = {A^3}\]
$$\eqalign{ & \Rightarrow \,\,{A^3} = {\left( { - 4} \right)^3} \cr & \Rightarrow \,\,A = - 4 \cr} $$
\[ \Rightarrow \,\,\left| \begin{array}{l} x - 4\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,x - 4\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,x - 4 \end{array} \right| = \left( {Bx - 4} \right){\left( {x + 4} \right)^2}\]
Now take $$x$$ common from both the sides
\[\therefore \,\,\left| \begin{array}{l} 1 - \frac{4}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \frac{4}{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\\ \,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \frac{4}{x} \end{array} \right| = \left( {B - \frac{4}{x}} \right){\left( {1 + \frac{4}{x}} \right)^2}\]
$${\text{Now take }}x \to \infty ,\,{\text{then }}\frac{1}{x} \to 0$$
\[ \Rightarrow \,\,\left| \begin{array}{l} 1\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2\\ 2\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,1 \end{array} \right| = B\]
$$ \Rightarrow \,\,B = 5$$
∴ Ordered pair $$(A, B)$$  is $$(- 4, 5)$$