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Question: If \(A = diag(2\;\, - 5\,{\text{ }}9),B = diag(1\,\,1\,\, - 4){\text{ and }}C = diag( - 6\,\,3\,\,4)...

If A=diag(2  5 9),B=diag(114) and C=diag(634)A = diag(2\;\, - 5\,{\text{ }}9),B = diag(1\,\,1\,\, - 4){\text{ and }}C = diag( - 6\,\,3\,\,4) then find B+C2AB + C - 2A.

Explanation

Solution

Hint : Identify the type of matrix then put the value in the given equation to get the value.

Here diag(a  b c)diag(a\;\,b\,{\text{ }}c) represents diagonal matrix whose diagonal elements are a,b,ca,b,c
Therefore,
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+C2AB + C - 2A
On putting the values of A,B,CA,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(91418)diag( - 9\,\,14\,\, - 18) which is a diagonal matrix of diagonal elements 9,14,18 - 9,\,\,14,\,\, - 18.

Note :- To solve these types of problems we have to remember that diag(a  b c)diag(a\;\,b\,{\text{ }}c) represents diagonal elements whose diagonal elements are a,b,ca,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.