Question

# If $A = diag(2\;\, - 5\,{\text{ }}9),B = diag(1\,\,1\,\, - 4){\text{ and }}C = diag( - 6\,\,3\,\,4)$ then find $B + C - 2A$.

Here $diag(a\;\,b\,{\text{ }}c)$ represents diagonal matrix whose diagonal elements are $a,b,c$
$diag(2\;\, - 5\,{\text{ 9}}) = \left( {\begin{array}{*{20}{c}} 2&0&0 \\ 0&{ - 5}&0 \\ 0&0&9 \end{array}} \right) = A \\ \\ diag(1\;\,1{\text{ - 4}}) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&{ - 4} \end{array}} \right) = B \\ \\ diag( - 6\;\,3{\text{ 4}}) = \left( {\begin{array}{*{20}{c}} { - 6}&0&0 \\ 0&3&0 \\ 0&0&4 \end{array}} \right) = C \\$
We have to find $B + C - 2A$
On putting the values of $A,B,C$ in the above equation we get,
$\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&{ - 4} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 6}&0&0 \\ 0&3&0 \\ 0&0&4 \end{array}} \right) - 2\left( {\begin{array}{*{20}{c}} 2&0&0 \\ 0&{ - 5}&0 \\ 0&0&9 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 9}&0&0 \\ 0&{14}&0 \\ 0&0&{ - 18} \end{array}} \right) = diag( - 9\,\,14\,\, - 18)$
So the required value is $diag( - 9\,\,14\,\, - 18)$ which is a diagonal matrix of diagonal elements $- 9,\,\,14,\,\, - 18$.
Note :- To solve these types of problems we have to remember that $diag(a\;\,b\,{\text{ }}c)$ represents diagonal elements whose diagonal elements are $a,b,c$. Then after converting it to matrix format we have to put the value of matrix and then apply the rules of calculation in matrices.