Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

$$\because 20 = 1 \times 20 = 2 \times 10 = 4 \times 5 = 5 \times 4 = 10 \times 2 = 20 \times 1$$
Out of this number of rows is even so possible cases are $$2 \times 10 , 4 \times 5 , 10 \times 2$$     and $$20 \times 1.$$  So number of matrices are 4.

\[{C_2} \to {C_2} + {C_1}\,\,{\text{given }}\vartriangle = \left| {\begin{array}{*{20}{c}} {^{10}{C_4}}&{^{11}{C_5}}&{^{11}{C_m}} \\ {^{11}{C_6}}&{^{12}{C_7}}&{^{12}{C_{m + 2}}} \\ {^{12}{C_8}}&{^{13}{C_9}}&{^{13}{C_{m + 4}}} \end{array}} \right|\]
For $$m = 5,{C_2} \equiv {C_3}.$$

Let \[\Delta = \left| {\begin{array}{*{20}{c}} \lambda &1&1 \\ 1&\lambda &1 \\ 1&1&\lambda \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\lambda + 2}&1&1 \\ {\lambda + 2}&\lambda &1 \\ {\lambda + 2}&1&\lambda \end{array}} \right|\left[ {{C_1} \to {C_1} + {C_2} + {C_3}} \right]\]
\[ = \left( {\lambda + 2} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&\lambda &1 \\ 1&1&\lambda \end{array}} \right| = \left( {\lambda + 2} \right)\left| {\begin{array}{*{20}{c}} 1&0&0 \\ 1&{\lambda - 1}&0 \\ 1&0&{\lambda - 1} \end{array}} \right|\]
$$ = \left( {\lambda + 2} \right){\left( {\lambda - 1} \right)^2}$$     [ using $${{C_2} \to {C_2} - {C_1}}$$   and $${{C_3} \to {C_3} - {C_1}}$$   ]
If $$\Delta = 0,$$  then $$\lambda = - 2{\text{ or }}\lambda = 1.$$
But when $$\lambda = 1,$$  the system of equation becomes $${x_1} + {x_2} + {x_3} = 1$$    which has infinite number of solutions. When $$\lambda = - 2,$$  by adding three equations, we obtain $$0 = 3$$  and thus, the system of equations is inconsistent.

$$\eqalign{ & {\text{Given }}{A^2} - A + I = 0 \cr & {A^{ - 1}}{A^2} - {A^{ - 1}}A + {A^{ - 1}}.I = {A^{ - 1}}.0 \cr} $$
(Multiplying $${A^{ - 1}}$$ on both sides)
$$ \Rightarrow \,\,A - 1 + {A^{ - 1}} = 0\,\,{\text{or }}{A^{ - 1}} = 1 - A.$$

Let $$A$$ be a symmetric matrix.
Then $$A' = A$$
Now, $$\left( {B'AB} \right)' = B'A'\left( {B'} \right)'.\left[ {\because \left( {AB} \right)' = B'A'} \right]$$
$$\eqalign{ & = B'A'B\left[ {\because \left( {B'} \right)' = B} \right] \cr & = B'AB\left[ {\because A' = A} \right] \cr} $$
⇒ $$B'AB$$  is a symmetric matrix. Now, let $$A$$ be a skew-symmetric matrix.
Then, $$A' = - A$$
$$\eqalign{ & \therefore \left( {B'AB} \right)' = B'A'\left( {B'} \right)'\left[ {\because \left( {AB} \right)' = B'A'} \right] \cr & = B'A'B\left[ {\because \left( {B'} \right)' = B} \right] \cr & = B'\left( { - A} \right)B\left[ {\because A' = - A} \right] \cr & = - B'AB \cr} $$
∴ $$B'AB$$  is a skew-symmetric matrix.

\[\begin{array}{l} {A^2} = \left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \beta &\alpha \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&b\\ b&a \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a&b\\ b&a \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {{a^2} + {b^2}}&{2ab}\\ {2ab}&{{a^2} + {b^2}} \end{array}} \right];\alpha = {a^2} + {b^2};\beta = 2ab \end{array}\]

\[\vartriangle = \left| {\begin{array}{*{20}{c}} {{a_1}x + {b_1}y}&{{a_2}x + {b_2}y}&{{a_3}x + {b_3}y} \\ {{a_1}\left( {y - 1} \right)}&{{a_2}\left( {y - 1} \right)}&{{a_3}\left( {y - 1} \right)} \\ {{b_1}x + {a_1}}&{{b_2}x + {a_2}}&{{b_3}x + {a_3}} \end{array}} \right|\]
$$\left( {{R_2} \to {R_2} - {R_3}} \right)$$
\[\vartriangle = \left( {y - 1} \right)\left| {\begin{array}{*{20}{c}} {{b_1}y}&{{b_2}y}&{{b_3}y} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}x + {a_1}}&{{b_2}x + {a_2}}&{{b_3}x + {a_3}} \end{array}} \right|\]
$$\left( {{R_1} \to {R_1} - x \times {R_2}} \right)$$
\[\vartriangle = \left( {y - 1} \right)y\left| {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}&{{b_3}} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ 0&0&0 \end{array}} \right|\left( {{\text{Using }}{R_3} \to {R_3} - x \times {R_1} - {R_2}} \right).\]

We have,
\[\bar \Delta = \left| {\begin{array}{*{20}{c}} 0&{ - \bar y}&{ - \bar z}\\ y&0&{ - \bar x}\\ z&x&0 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 0&y&z\\ { - \bar y}&0&x\\ { - \bar z}&{ - \bar x}&0 \end{array}} \right|\]
[Interchanging rows and columns]
\[ = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}} 0&{ - y}&{ - z}\\ {\bar y}&0&{ - x}\\ {\bar z}&{\bar x}&0 \end{array}} \right|\]
[Taking $$–1$$ common from each row]
$$\eqalign{ & = - \Delta \cr & \therefore \bar \Delta + \Delta = 0 \cr & \Rightarrow 2\operatorname{Re} \left( \Delta \right) = 0 \cr} $$
∴ $$\Delta$$ is purely imaginary.

Let, \[A = \left[ {\begin{array}{*{20}{c}} 3&2\\ 1&4 \end{array}} \right]\]
We have,
If $$A$$ is a square matrix of order $$n$$ then
$$\eqalign{ & A\left( {{\text{adj}}\,A} \right) = \left| A \right|.{I_n} \cr & {\text{Here, }}n = 2 \cr & \therefore \,A\left( {adj\,A} \right) = {I_2}\left| A \right| \cr} $$
\[\begin{array}{l} = \,\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\,\left| {\begin{array}{*{20}{c}} 3&2\\ 1&4 \end{array}} \right| = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\left( {12 - 2} \right) = 10\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\ = \,\left[ {\begin{array}{*{20}{c}} {10}&0\\ 0&{10} \end{array}} \right] \end{array}\]

\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right].\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&1 \end{array}} \right].\left[ {\begin{array}{*{20}{c}} 1&3\\ 0&1 \end{array}} \right].\left[ {\begin{array}{*{20}{c}} 1&4\\ 0&1 \end{array}} \right]......\left[ {\begin{array}{*{20}{c}} 1&{n - 1}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{78}\\ 0&1 \end{array}} \right]\\ \Rightarrow \,\,\left[ {\begin{array}{*{20}{c}} 1&{1 + 2 + 3 + ...... + \left( {n - 1} \right)}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{78}\\ 0&1 \end{array}} \right]\\ \Rightarrow \,\,\frac{{\left( {n - 1} \right)n}}{2} = 78\\ \Rightarrow \,\,{n^2} - n - 15 = 0\\ \Rightarrow \,\,n = 13\\ {\rm{Now, \,the \,matrix }}\left[ {\begin{array}{*{20}{c}} 1&n\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{13}\\ 0&1 \end{array}} \right] \end{array}\]
Then, the required inverse of
\[\left[ {\begin{array}{*{20}{c}} 1&{13}\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 13}\\ 0&1 \end{array}} \right]\]