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

Last updated date: 20th Jun 2024
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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 $xu + xv + yu + yv.$

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.
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + b}&1 \\ 1&{1 + c} \end{array}} \right| = 1 + c + b + bc - 1$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + b}&1 \\ 1&{1 + c} \end{array}} \right| = c + b + bc$ $...(2)$
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.
$\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1 \\ 1&{1 + c} \end{array}} \right| = (1)(1 + c) - (1)(1)$
$\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1 \\ 1&{1 + c} \end{array}} \right| = 1 + c - 1$
$\Rightarrow \left| {\begin{array}{*{20}{c}} 1&1 \\ 1&{1 + c} \end{array}} \right| = c$ $...(3)$
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.
$\Rightarrow \left| {\begin{array}{*{20}{c}} 1&{1 + b} \\ 1&1 \end{array}} \right| = 1 - 1 - b$
$\Rightarrow \left| {\begin{array}{*{20}{c}} 1&{1 + b} \\ 1&1 \end{array}} \right| = - b$ $...(4)$
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 $\left| {\begin{array}{*{20}{c}} {1 + a}&1&1 \\ 1&{1 + b}&1 \\ 1&1&{1 + c} \end{array}} \right| = (1 + a)\left| {\begin{array}{*{20}{c}} {1 + b}&1 \\ 1&{1 + c} \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 1&1 \\ 1&{1 + c} \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} 1&{1 + b} \\ 1&1 \end{array}} \right|$.
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + a}&1&1 \\ 1&{1 + b}&1 \\ 1&1&{1 + c} \end{array}} \right| = (1 + a)(c + b + bc) - 1(c) + 1( - b)$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + a}&1&1 \\ 1&{1 + b}&1 \\ 1&1&{1 + c} \end{array}} \right| = (1)(c + b + bc) + a(c + b + bc) - c - b$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + a}&1&1 \\ 1&{1 + b}&1 \\ 1&1&{1 + c} \end{array}} \right| = c + b + bc + ac + ab + abc - c - b$
$\Rightarrow \left| {\begin{array}{*{20}{c}} {1 + a}&1&1 \\ 1&{1 + b}&1 \\ 1&1&{1 + c} \end{array}} \right| = ab + bc + ca + abc$
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.