Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Prove that\[\left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = ab + bc + ca + abc.\]

seo-qna
Last updated date: 20th Jun 2024
Total views: 414.3k
Views today: 12.14k
Answer
VerifiedVerified
414.3k+ views
Hint: We prove the question using the formula for determinant of order $3 \times 3$ \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|.\]
Then use the following formula to calculate the determinant of order $2 \times 2.$
\[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc.\]
Foil method: The multiplication of $(x + y)(u + v)$ is given by $xu + xv + yu + yv.$

Complete step-by-step answer:
We are given\[\left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right|.\]
Using the formula\[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right| = a\left| {\begin{array}{*{20}{c}}
  e&f \\
  h&i
\end{array}} \right| - b\left| {\begin{array}{*{20}{c}}
  d&f \\
  g&i
\end{array}} \right| + c\left| {\begin{array}{*{20}{c}}
  d&e \\
  g&h
\end{array}} \right|\] expand the given determinant.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right|\] $...(1)$
Now we will calculate values of each determinant of order $2 \times 2$ separately and then substitute it back in equation (1).
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| = (1 + b)(1 + c) - (1)(1)\]
 Find the multiplication of \[(1 + b)(1 + c)\] by using foil method.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| = (1)(1) + (1)(c) + b(1) + bc - 1\]
Simplify by doing multiplication.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| = 1 + c + b + bc - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| = c + b + bc\] $...(2)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right| = (1)(1 + c) - (1)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right| = 1 + c - 1\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right| = c\] $...(3)$
Calculate the value of \[\left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right|\] using \[\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc\] formula.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right| = (1)(1) - (1 + b)(1)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right| = 1 - (1 + b)\]
Simplify by distributing negative signs over parenthesis.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right| = 1 - 1 - b\]
Simplify by adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right| = - b\] $...(4)$
Step 5: Substitute values of \[\left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right|,\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right|\] and \[\left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right|\] from equation$(2), (3)$ and $(4)$ in equation \[\left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}}
  {1 + b}&1 \\
  1&{1 + c}
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{1 + c}
\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}
  1&{1 + b} \\
  1&1
\end{array}} \right|\].
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = (1 + a)(c + b + bc) - 1(c) + 1( - b)\]
 \[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = (1)(c + b + bc) + a(c + b + bc) - c - b\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = c + b + bc + ac + ab + abc - c - b\]
Simplify adding like terms.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + a}&1&1 \\
  1&{1 + b}&1 \\
  1&1&{1 + c}
\end{array}} \right| = ab + bc + ca + abc\]
Hence, proved.

Additional Information: * Any matrix with a row having all elements as 0 will have its determinant is 0.
* The determinant of a diagonal matrix is the product of its diagonal entries.
* The determinant of a matrix is 0 if and only if its rows are linearly dependent, which means elements of a row can be written as a linear combination of elements of another row. If the rows are linearly independent, then the determinant is non-zero.

Note: Students can easily get confused while calculating the determinant if they don’t break it into three parts of $2 \times 2$ matrices. Also, many students make the mistake of not writing negative signs along with the second term of the determinant. Keep in mind we move in an alternate sign way, we take first value positive, then second negative and then third value again positive.