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

Last updated date: 27th Mar 2023

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Answer

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

Here $diag(a\;\,b\,{\text{ }}c)$ represents diagonal matrix whose diagonal elements are $a,b,c$

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 $B + C - 2A$

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.

Here $diag(a\;\,b\,{\text{ }}c)$ represents diagonal matrix whose diagonal elements are $a,b,c$

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 $B + C - 2A$

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.

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