
Using properties of determinants, show that \[\left| \begin{matrix}
a+b & a & b \\
a & a+c & c \\
b & c & b+c \\
\end{matrix} \right|=4abc\]
Answer
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Hint: Given problem is based on the concept of determinants. We can prove this by taking LHS and applying column operation and expansion of determinants to obtain the required RHS. further to solve the determinants either we can use column operation or row operation but to solve this particular problem we can use column operations.
Complete step-by-step answer:
Now let us consider LHS \[\left| \begin{matrix}
a+b & a & b \\
a & a+c & c \\
b & c & b+c \\
\end{matrix} \right|\] of the given problem.
Now changing first column as \[{{c}_{1}}\to {{c}_{1}}-({{c}_{2}}+{{c}_{3}})\] we get
\[\Rightarrow \left| \begin{matrix}
0 & a & b \\
-2c & a+c & c \\
-2c & c & b+c \\
\end{matrix} \right|\]
Now applying the property 4 (which says Multiplying all the elements of a row (or column) by a scalar (a real number) is equivalent to multiplying the determinant by that scalar) to the first column. We get
\[\Rightarrow -2\left| \begin{matrix}
0 & a & b \\
c & a+c & c \\
c & c & b+c \\
\end{matrix} \right|\]
Now changing first column as \[{{c}_{2}}\to {{c}_{2}}-{{c}_{1}}\] and \[{{c}_{3}}\to {{c}_{3}}-{{c}_{1}}\] we get
\[\Rightarrow -2\left| \begin{matrix}
0 & a & b \\
c & a & 0 \\
c & 0 & b \\
\end{matrix} \right|\]
Now on expanding the above determinant we get
\[\begin{align}
& -2\left[ 0\left( ab-0 \right)-a\left( bc-0 \right)+b\left( 0-ac \right) \right] \\
& =-2(0-abc-abc) \\
& =-2(-2abc) \\
& =4abc \\
\end{align}\]
Hence proved.
Note: To solve the given problem, we have used properties of determinant instead we can expand the given determinant directly then also we will get the same solution but simplification of terms is quite complex so we have used properties of determinant to simplify the terms and at last we expanded the given determinant. There are 10 main properties of determinants which include reflection property, all-zero property, proportionality or repetition property, switching property, scalar multiple property, sum property, invariance property, factor property, triangle property, and cofactor matrix property but we use them as per the requirement of the problem.
Complete step-by-step answer:
Now let us consider LHS \[\left| \begin{matrix}
a+b & a & b \\
a & a+c & c \\
b & c & b+c \\
\end{matrix} \right|\] of the given problem.
Now changing first column as \[{{c}_{1}}\to {{c}_{1}}-({{c}_{2}}+{{c}_{3}})\] we get
\[\Rightarrow \left| \begin{matrix}
0 & a & b \\
-2c & a+c & c \\
-2c & c & b+c \\
\end{matrix} \right|\]
Now applying the property 4 (which says Multiplying all the elements of a row (or column) by a scalar (a real number) is equivalent to multiplying the determinant by that scalar) to the first column. We get
\[\Rightarrow -2\left| \begin{matrix}
0 & a & b \\
c & a+c & c \\
c & c & b+c \\
\end{matrix} \right|\]
Now changing first column as \[{{c}_{2}}\to {{c}_{2}}-{{c}_{1}}\] and \[{{c}_{3}}\to {{c}_{3}}-{{c}_{1}}\] we get
\[\Rightarrow -2\left| \begin{matrix}
0 & a & b \\
c & a & 0 \\
c & 0 & b \\
\end{matrix} \right|\]
Now on expanding the above determinant we get
\[\begin{align}
& -2\left[ 0\left( ab-0 \right)-a\left( bc-0 \right)+b\left( 0-ac \right) \right] \\
& =-2(0-abc-abc) \\
& =-2(-2abc) \\
& =4abc \\
\end{align}\]
Hence proved.
Note: To solve the given problem, we have used properties of determinant instead we can expand the given determinant directly then also we will get the same solution but simplification of terms is quite complex so we have used properties of determinant to simplify the terms and at last we expanded the given determinant. There are 10 main properties of determinants which include reflection property, all-zero property, proportionality or repetition property, switching property, scalar multiple property, sum property, invariance property, factor property, triangle property, and cofactor matrix property but we use them as per the requirement of the problem.
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