Matrices and Determinants MCQ Questions & Answers in Algebra | Maths

As the determinant has to vanish for all $$a, b,$$  we have at least two rows or two columns identical.
After filling the first column in $$2 \times 2 \times 2$$   ways and filling another column likewise, the remaining column can be filled in $$2 \times 2 \times 2$$   ways. So, the number of ways $$ = {2^3} \times {2^3} = {2^6}.$$
But each of the two ways give the same determinant.
∴ the required number of determinants $$= 32.$$

$$\eqalign{ & {\left( {aI + bA} \right)^2} = {a^2}{I^2} + {b^2}{A^2} + 2abAI \cr & = {a^2}{I^2} + {b^2}{A^2} + 2abA \cr} $$
But, \[{A^2} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}} \right]\]
$$\therefore {\left( {aI + bA} \right)^2} = {a^2}I + 2abA.$$

Let \[A\left[ \begin{array}{l} {a_1}\,\,\,{b_1}\,\,\,{c_1}\\ {a_2}\,\,\,{b_2}\,\,{c_2}\\ {a_3}\,\,\,{b_3}\,\,\,{c_3} \end{array} \right]\] $$\eqalign{ & {\text{where }}{a_i},{b_i},{c_i}\,{\text{have}} \cr & {\text{values 0 or 1 for }}i = 1,2,3. \cr} $$
Then the given system is equivalent to
$$\eqalign{ & {a_1}x + {b_1}y + {c_1}z = 1, \cr & {a_2}x + {b_2}y + {c_2}z = 0, \cr & {a_3}x + {b_3}y + {c_3}z = 0. \cr} $$
Which represents three distinct planes. But three planes can not intersect at two distinct points, therefore no such system exists.

\[\begin{array}{l} {\rm{Given,\, }}{\left[ {\begin{array}{*{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0\\ \Rightarrow \left[ {1 + 2x + 15\,\,\,3 + 5x + 3\,\,\,2 + x + 2} \right]\left[ \begin{array}{l} 1\\ 2\\ x \end{array} \right] = 0\\ \Rightarrow \left[ {16 + 2x + 15\,\,\,3 + 5x + 3\,\,\,2 + x + 2} \right]\left[ \begin{array}{l} 1\\ 2\\ x \end{array} \right] = 0\\ \Rightarrow \left[ {\left( {16 + 2x} \right) \cdot 1 + \left( {6 + 5x} \right) \cdot 2 + \left( {4 + x} \right) \cdot x} \right] = 0\\ \Rightarrow \left( {16 + 2x} \right) + \left( {12 + 10x} \right) + \left( {4x + {x^2}} \right) = 0\\ \Rightarrow {x^2} + 16x + 28 = 0\\ \Rightarrow \left( {x + 14} \right)\left( {x + 2} \right) = 0\\ \Rightarrow x + 14 = 0{\rm{ \,\,or\,\, }}x + 2 = 0\\ {\rm{Hence, }}x = - 14{\rm{ }}\,{\rm{\,\,or\,\, }}x = - 2 \end{array}\]

We know, if $$A$$ is an $$n \times n$$  matrix, then $$\det \left( {\lambda A} \right) = {\lambda ^s}\det \left( A \right)$$
But given $$\det \left( {\lambda A} \right) = {\lambda ^s}\det A$$
⇒ $$s = n$$

\[\Delta = \left| {\begin{array}{*{20}{c}} {{a_1} + {b_1}\omega }&{{a_1}{\omega ^2} + {b_1}}&{{c_1} + {b_1}\bar \omega }\\ {{a_2} + {b_2}\omega }&{{a_2}{\omega ^2} + {b_2}}&{{c_2} + {b_2}\bar \omega }\\ {{a_3} + {b_3}\omega }&{{a_3}{\omega ^2} + {b_3}}&{{c_3} + {b_3}\bar \omega } \end{array}} \right|\]
Using, $${C_2} \to \omega {C_2}$$
We have,
\[\begin{array}{l} \Delta = \frac{1}{\omega }\left| {\begin{array}{*{20}{c}} {{a_1} + {b_1}\omega }&{{a_1}{\omega ^3} + {b_1}\omega }&{{c_1} + {b_1}\bar \omega }\\ {{a_2} + {b_2}\omega }&{{a_2}{\omega ^3} + {b_2}\omega }&{{c_2} + {b_2}\bar \omega }\\ {{a_3} + {b_3}\omega }&{{a_3}{\omega ^2} + {b_3}\omega }&{{c_3} + {b_3}\bar \omega } \end{array}} \right|\\ = \frac{1}{\omega }\left| {\begin{array}{*{20}{c}} {{a_1} + {b_1}\omega }&{{a_1} + {b_1}\omega }&{{c_1} + {b_1}\bar \omega }\\ {{a_2} + {b_2}\omega }&{{a_2} + {b_2}\omega }&{{c_2} + {b_2}\bar \omega }\\ {{a_3} + {b_3}\omega }&{{a_3} + {b_3}\omega }&{{c_3} + {b_3}\bar \omega } \end{array}} \right| = 0 \end{array}\]

\[{\rm{Let }}\,\,A = \left[ \begin{array}{l} a\,\,\,\,\,b\\ c\,\,\,\,\,d \end{array} \right]{\rm{then }}\,\,{A^2} = I\]
$$\eqalign{ & \Rightarrow \,\,{a^2} + bc = 1\,\,\,\,\,\,\,\,\,ab + bd = 0 \cr & \,\,\,\,\,\,ac + cd = 0\,\,\,\,\,\,\,\,\,bc + {d^2} = 1 \cr} $$
From these four relations,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,{a^2} + bc = bc + {d^2} \Rightarrow \,{a^2} = {d^2} \cr & {\text{and }}b\left( {a + d} \right) = 0 = c\left( {a + d} \right) \Rightarrow \,\,a = - d \cr} $$
We can take $$a = 1,b = 0,c = 0,d = - 1\,\,{\text{as one}}$$
possible set of values, then \[A = \left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,0\\ 0\,\,\,\, - 1 \end{array} \right]\]
Clearly $$A \ne I$$  and $$A \ne - I$$  and det $$A = - 1$$
∴ Statement 1 is true.
Also if $$A \ne I$$  then tr$$(A) = 0$$
∴ Statement 2 is false.

\[{\rm{Given\,\, that }}\,\,A = \left[ \begin{array}{l} \alpha \,\,\,\,\,\,0\\ 1\,\,\,\,\,\,\,\,1 \end{array} \right]{\rm{ and}}\,{\rm{ }}B = \left[ \begin{array}{l} 1\,\,\,\,\,\,\,0\\ 5\,\,\,\,\,\,\,1 \end{array} \right]\]
$${\text{and }}{A^2} = B$$
\[\begin{array}{l} \Rightarrow \,\,\left[ \begin{array}{l} \alpha \,\,\,\,\,\,0\\ 1\,\,\,\,\,\,\,\,1 \end{array} \right]\left[ \begin{array}{l} \alpha \,\,\,\,\,\,0\\ 1\,\,\,\,\,\,\,\,1 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,0\\ 5\,\,\,\,\,\,\,\,1 \end{array} \right]\\ \Rightarrow \,\,\left[ \begin{array}{l} {\alpha ^2}\,\,\,\,\,\,\,\,\,\,\,\,0\\ \alpha + 1\,\,\,\,\,\,\,\,1 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,\,\,0\\ 5\,\,\,\,\,\,\,\,1 \end{array} \right] \end{array}\]
$$\eqalign{ & \Rightarrow \,\,{\alpha ^2} = 1\,\,{\text{and }}\alpha + 1 = 5 \cr & \Rightarrow \,\,\alpha = \pm 1\,\,{\text{and }}\alpha = 4 \cr} $$
$$\because $$ There is no common value
∴ There is no real value of $$\alpha $$ for which $${A^2} = B$$

\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ A.A = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] = - 2\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right] \end{array}\]
$$A^2 = - 2A$$
\[\begin{array}{l} {A^2}.A = - 2\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ = - 2\left[ {\begin{array}{*{20}{c}} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] = 4\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right] \end{array}\]
$$A^3 = 4A$$
Hence, $${A^2} \ne - A,{A^3} = 4A.$$

For the given homogeneous system to have non-zero solution determinant of co-efficient matrix should be zero; i.e.,
\[ = \left| \begin{array}{l} 1\,\,\,\,\,\, - k\,\,\,\,\, - 1\\ k\,\,\,\,\, - 1\,\,\,\,\,\, - 1\\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\, - 1 \end{array} \right|\]
$$\eqalign{ & = 1\left( {1 + 1} \right) + k\left( { - k + 1} \right) - 1\left( {k + 1} \right) = 0 \cr & \Rightarrow \,\,2 - {k^2} + k - k - 1 = 0 \cr & \Rightarrow \,\,{k^2} = 1 \cr & \Rightarrow \,\,k = \pm 1 \cr} $$