Let \[A + 2B = \left[ {\begin{array}{*{20}{c}}
1&2&0\\
6&{ - 3}&3\\
{ - 5}&3&1
\end{array}} \right]\] and \[2A - B = \left[ {\begin{array}{*{20}{c}}
2&{ - 1}&5\\
2&{ - 1}&6\\
0&1&2
\end{array}} \right],\] then $$tr\left( A \right) - tr\left( B \right)$$ is
A.
1
B.
3
C.
2
D.
0
Answer :
2
Solution :
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 $$tr\left( A \right) = 1{\text{ and }}tr\left( B \right) = - 1$$
$$ \Rightarrow tr\left( A \right) - tr\left( B \right) = 2$$
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has