Mejia Estrada
05/30/2023 · Senior High School
i.) Prove that:- \( \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y \)
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Step-by-step Solution
To prove that
\[
\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|=x y,
\]
we will compute the determinant of the given matrix.
Let
\[
A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{pmatrix}.
\]
We can compute the determinant using the formula for the determinant of a \(3 \times 3\) matrix:
\[
\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg),
\]
where the matrix is
\[
\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}.
\]
For our matrix \(A\), we have:
- \(a = 1\), \(b = 1\), \(c = 1\)
- \(d = 1\), \(e = 1+x\), \(f = 1\)
- \(g = 1\), \(h = 1\), \(i = 1+y\)
Now, substituting these values into the determinant formula:
\[
\det(A) = 1 \cdot ((1+x)(1+y) - 1 \cdot 1) - 1 \cdot (1 \cdot (1+y) - 1 \cdot 1) + 1 \cdot (1 \cdot 1 - 1 \cdot (1+x)).
\]
Calculating each term:
1. The first term:
\[
(1+x)(1+y) - 1 = 1 + x + y + xy - 1 = x + y + xy.
\]
2. The second term:
\[
1 \cdot (1+y) - 1 = 1 + y - 1 = y.
\]
3. The third term:
\[
1 \cdot 1 - 1 \cdot (1+x) = 1 - (1+x) = -x.
\]
Putting it all together, we have:
\[
\det(A) = 1 \cdot (x + y + xy) - 1 \cdot y + 1 \cdot (-x).
\]
This simplifies to:
\[
\det(A) = x + y + xy - y - x = xy.
\]
Thus, we have shown that:
\[
\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right| = xy.
\]
Therefore, the proof is complete:
\[
\boxed{xy}.
\]
Quick Answer
The determinant of the matrix is \( xy \).
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