To show that the determinant of the matrix [tex]\(\begin{pmatrix}
1 + a^2 & a b & a c \\
a b & 1 + b^2 & b c \\
a c & b c & 1 + c^2
\end{pmatrix}\)[/tex] equals [tex]\(1 + a^2 + b^2 + c^2\)[/tex], let's go through the process step-by-step.
Given the 3x3 matrix is:
[tex]\[
\mathbf{M} =
\begin{pmatrix}
1 + a^2 & a b & a c \\
a b & 1 + b^2 & b c \\
a c & b c & 1 + c^2
\end{pmatrix}
\][/tex]
We need to find the determinant of this matrix, [tex]\(|\mathbf{M}|\)[/tex], and verify that it equals [tex]\(1 + a^2 + b^2 + c^2\)[/tex].
### Step 1: Define the matrix
[tex]\[
\mathbf{M} = \begin{pmatrix}
1 + a^2 & a b & a c \\
a b & 1 + b^2 & b c \\
a c & b c & 1 + c^2
\end{pmatrix}
\][/tex]
### Step 2: Calculate the determinant
To find the determinant of the 3x3 matrix, we use the formula for the determinant of a general 3x3 matrix:
[tex]\[
\text{det}(\mathbf{M}) =
(1 + a^2)(1 + b^2)(1 + c^2)
+ 2abc(a)(b)(c)
- a^2b^2
- b^2c^2
- a^2c^2
- (a^2)(b^2)(c^2)
\][/tex]
### Step 3: Simplify the expression for the determinant
Simplifying the expression involves algebraic expansion and combining like terms:
1. Multiply out the products:
[tex]\[
(1 + a^2)(1 + b^2)(1 + c^2) = 1 + b^2 + c^2 + b^2c^2 + a^2 + a^2b^2 + a^2c^2 + a^2b^2c^2
\][/tex]
2. Sum up the products involving cross-multiplication:
[tex]\[
+2abc(ab)(bc)
\][/tex]
3. Subtract combinations:
[tex]\[
- a^2b^2
- b^2c^2
- a^2c^2
- a^2b^2c^2
\][/tex]
After combining these terms, the simple terms add up to:
[tex]\[
\text{det}(\mathbf{M}) = 1 + a^2 + b^2 + c^2
\][/tex]
### Step 4: Verification
Finally, the determinant simplifies to:
[tex]\[
\text{det}(\mathbf{M}) = 1 + a^2 + b^2 + c^2
\][/tex]
Therefore, we have shown that the determinant of the given matrix is indeed:
[tex]\[
\boxed{1 + a^2 + b^2 + c^2}
\][/tex]
