5 Marks Questions

Question 15 Marks

If $A=\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$ and I is the identity matrix of order 2 , show that
$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

Answer

L.H.S. $I + A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&0 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&1 \end{array}} \right]$
Now, $I - A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&0 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{\tan \frac{\alpha }{2}} \\ { - \tan \frac{\alpha }{2}}&1 \end{array}} \right]$
R.H.S. $ = (I - A)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} 1&{\tan \frac{\alpha }{2}} \\ { - \tan \frac{\alpha }{2}}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos \alpha + \sin \alpha \tan \frac{\alpha }{2}}&{ - \sin \alpha + \cos \alpha \tan \frac{\alpha }{2}} \\ { - \cos \alpha \tan \frac{\alpha }{2} + \sin \alpha }&{\sin \alpha \tan \frac{\alpha }{2} + \cos \alpha } \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos \alpha + \sin \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{ - \sin \alpha + \cos \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ { - \cos \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} + \sin \alpha }&{\sin \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} + \cos \alpha } \end{array}} \right]$
$=\left[ {\begin{array}{*{20}{c}} {\frac{{\cos \alpha \cos \frac{\alpha }{2} + \sin \alpha \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \alpha \cos \frac{\alpha }{2} + \cos \alpha \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{ - \cos \alpha \sin \frac{\alpha }{2} + \sin \alpha \cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\sin \alpha \sin \frac{\alpha }{2} + \cos \alpha \cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\frac{{\cos \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{\sin \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\cos \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} {\frac{{\cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&1 \end{array}} \right]$
$\therefore$ L.H.S. = R.H.S. Proved.

View full question & answer→

Question 25 Marks

If $A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, prove that $\mathrm{A}^3-6 \mathrm{~A}^2+7 \mathrm{~A}+2 \mathrm{I}=0$.

Answer

L.H.S. $= A^3 - 6A^2 + 7A + 2I$
$= \left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]$$ - 6\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]$$ + 2\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6} \\ {0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3} \\ {2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9} \end{array}} \right]$$\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]$$ - 6\left[ {\begin{array}{*{20}{c}} {1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6} \\ {0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3} \\ {2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9} \end{array}} \right]$$ + \left[ {\begin{array}{*{20}{c}} 7&0&{14} \\ 0&{14}&7 \\ {14}&0&{21} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&0&0 \\ 0&2&0 \\ 0&0&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {5 + 0 + 16}&{0 + 0 + 0}&{10 + 0 + 24} \\ {2 + 0 + 10}&{0 + 8 + 0}&{4 + 4 + 15} \\ {8 + 0 + 26}&{0 + 0 + 0}&{16 + 0 + 39} \end{array}} \right]$$ - \left[ {\begin{array}{*{20}{c}} {30}&0&{48} \\ {12}&{24}&{30} \\ {48}&0&{78} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {7 + 2}&{0 + 0}&{14 + 0} \\ {0 + 0}&{14 + 2}&{7 + 0} \\ {14 + 0}&{0 + 0}&{21 + 2} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {21}&0&{34} \\ {12}&8&{23} \\ {34}&0&{55} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {30}&0&{48} \\ {12}&{24}&{30} \\ {48}&0&{78} \end{array}} \right]$$ + \left[ {\begin{array}{*{20}{c}} 9&0&{14} \\ 0&{16}&7 \\ {14}&0&{23} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} {21 - 30}&{0 - 0}&{34 - 48} \\ {12 - 12}&{8 - 24}&{23 - 30} \\ {34 - 48}&{0 - 0}&{55 - 78} \end{array}} \right]$$ + \left[ {\begin{array}{*{20}{c}} 9&0&{14} \\ 0&{16}&7 \\ {14}&0&{23} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} { - 9}&0&{ - 14} \\ 0&{ - 16}&{ - 7} \\ { - 14}&0&{ - 23} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 9&0&{14} \\ 0&{16}&7 \\ {14}&0&{23} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} { - 9 + 9}&{0 + 0}&{ - 14 + 14} \\ {0 + 0}&{ - 16 + 16}&{ - 7 + 7} \\ { - 14 + 14}&{0 + 0}&{ - 23 + 23} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0$ (Zero matrix)
= R.H.S. Proved.

View full question & answer→

Question 35 Marks

If A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right]$ verify that

(A′)′ = A
(A + B)′ = A′ + B′
(kB)′ = kB′, where k is any constant.

Answer

We have
A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right] $
$\Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc} {3} & {4} \\ {\sqrt{3}} & {2} \\ {2} & {0} \end{array}\right] $
$\Rightarrow$ (A')' = $\left[\begin{array}{ccc} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right]$ = A
Thus (A′)′ = A
We have
A = $\left[\begin{array}{lll} {3} & {\sqrt{3}} & {2} \\ {4} & {2} & {0} \end{array}\right], B=\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right] \Rightarrow A+B=\left[\begin{array}{rrr} {5} & {\sqrt{3}-1} & {4} \\ {5} & {4} & {4} \end{array}\right]$
Therefore (A + B)' = $\left[\begin{array}{ccc} {5} & {5} \\ {\sqrt{3}} {-1} & {4} \\ {4} & {4} \end{array}\right]$
Now A' = $\left[\begin{array}{cc} {3} & {4} \\ {\sqrt{3}} & {2} \\ {2} & {0} \end{array}\right], B^{\prime}=\left[\begin{array}{cc} {2} & {1} \\ {-1} & {2} \\ {2} & {4} \end{array}\right]$
So A' + B' = $\left[\begin{array}{ccc} {5} & {5} \\ {\sqrt{3}} {-1} & {4} \\ {4} & {4} \end{array}\right]$
Thus (A + B)′ = A′ + B′
We have,
$kB$ = k$\left[\begin{array}{rrr} {2} & {-1} & {2} \\ {1} & {2} & {4} \end{array}\right]=\left[\begin{array}{ccc} {2 k} & {-k} & {2 k} \\ {k} & {2 k} & {4 k} \end{array}\right]$
$\therefore$ $(kB)'$ = $\left[\begin{array}{cc} {2 k} & {k} \\ {-k} & {2 k} \\ {2 k} & {4 k} \end{array}\right]=k\left[\begin{array}{cc} {2} & {1} \\ {-1} & {2} \\ {2} & {4} \end{array}\right]=k \mathrm{B}^{\prime}$
Thus $(kB)' = kB'$

View full question & answer→

Question 45 Marks

If A = $\left[\begin{array}{ccc} {1} & {1} & {-1} \\ {2} & {0} & {3} \\ {3} & {-1} & {2} \end{array}\right], B=\left[\begin{array}{cc} {1} & {3} \\ {0} & {2} \\ {-1} & {4} \end{array}\right]$ and C = $\left[\begin{array}{cccc} {1} & {2} & {3} & {-4} \\ {2} & {0} & {-2} & {1} \end{array}\right]$ find A(BC), (AB)C and show that (AB)C = A(BC).

Answer

We have $\mathrm{AB}=\left[\begin{array}{rrr} {1} & {1} & {-1} \\ {2} & {0} & {3} \\ {3} & {-1} & {2} \end{array}\right]\left[\begin{array}{rr} {1} & {3} \\ {0} & {2} \\ {-1} & {4} \end{array}\right]=\left[\begin{array}{cc} {1+0+1} & {3+2-4} \\ {2+0-3} & {6+0+12} \\ {3+0-2} & {9-2+8} \end{array}\right]=\left[\begin{array}{cc} {2} & {1} \\ {-1} & {18} \\ {1} & {15} \end{array}\right]$
(AB) (C) = $\left[\begin{array}{cc} {2} & {1} \\ {-1} & {18} \\ {1} & {15} \end{array}\right]\left[\begin{array}{cccc} {1} & {2} & {3} & {-4} \\ {2} & {0} & {-2} & {1} \end{array}\right]=\left[\begin{array}{cccc} {2+2} & {4+0} & {6-2} & {-8+1} \\ {-1+36} & {-2+0} & {-3-36} & {4+18} \\ {1+30} & {2+0} & {3-30} & {-4+15} \end{array}\right]$
= $\left[\begin{array}{cccc} {4} & {4} & {4} & {-7} \\ {35} & {-2} & {-39} & {22} \\ {31} & {2} & {-27} & {11} \end{array}\right]$
Now, BC = $\left[\begin{array}{cc} {1} & {3} \\ {0} & {2} \\ {-1} & {4} \end{array}\right]\left[\begin{array}{cccc} {1} & {2} & {3} & {-4} \\ {2} & {0} & {-2} & {1} \end{array}\right]=\left[\begin{array}{cccc} {1+6} & {2+0} & {3-6} & {-4+3} \\ {0+4} & {0+0} & {0-4} & {0+2} \\ {-1+8} & {-2+0} & {-3-8} & {4+4} \end{array}\right]$
= $\left[\begin{array}{cccc} {7} & {2} & {-3} & {-1} \\ {4} & {0} & {-4} & {2} \\ {7} & {-2} & {-11} & {8} \end{array}\right]$
Therefore, A(BC) = $\left[\begin{array}{rrr} {1} & {1} & {-1} \\ {2} & {0} & {3} \\ {3} & {-1} & {2} \end{array}\right]\left[\begin{array}{rrrr} {7} & {2} & {-3} & {-1} \\ {4} & {0} & {-4} & {2} \\ {7} & {-2} & {-11} & {8} \end{array}\right]$
= $\left[\begin{array}{cccc} {7+4-7} & {2+0+2} & {-3-4+11} & {-1+2-8} \\ {14+0+21} & {4+0-6} & {-6+0-33} & {-2+0+24} \\ {21-4+14} & {6+0-4} & {-9+4-22} & {-3-2+16} \end{array}\right]$
= $\left[\begin{array}{cccc} {4} & {4} & {4} & {-7} \\ {35} & {-2} & {-39} & {22} \\ {31} & {2} & {-27} & {11} \end{array}\right]$,
Clearly, (AB) C = A (BC)

View full question & answer→

Maths 5 Marks Questions

Take a timed test