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Sure! Let's verify that the given determinant is equal to [tex]\(1 + x^2 + y^2 + z^2\)[/tex].
We start with the matrix:
[tex]\[ A = \begin{pmatrix} x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1 \end{pmatrix} \][/tex]
We need to find the determinant of this matrix and show that it equals [tex]\(1 + x^2 + y^2 + z^2\)[/tex].
Step 1: Write the matrix determinant using standard notation.
[tex]\[ \text{det}(A) = \left|\begin{array}{ccc} x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1 \end{array}\right| \][/tex]
Step 2: Expand the determinant using cofactor expansion along the first row.
[tex]\[ \text{det}(A) = (x^2 + 1) \left|\begin{array}{cc} y^2 + 1 & yz \\ yz & z^2 + 1 \end{array}\right| - xy \left|\begin{array}{cc} xy & yz \\ xz & z^2 + 1 \end{array}\right| + xz \left|\begin{array}{cc} xy & y^2 + 1 \\ xz & yz \end{array}\right| \][/tex]
Step 3: Compute the 2x2 determinants.
[tex]\[ \left|\begin{array}{cc} y^2 + 1 & yz \\ yz & z^2 + 1 \end{array}\right| = (y^2 + 1)(z^2 + 1) - (yz)^2 = y^2z^2 + y^2 + z^2 + 1 - y^2z^2 = y^2 + z^2 + 1 \][/tex]
[tex]\[ \left|\begin{array}{cc} xy & yz \\ xz & z^2 + 1 \end{array}\right| = xy(z^2 + 1) - yz(xz) = xyz^2 + xy - xyz^2 = xy \][/tex]
[tex]\[ \left|\begin{array}{cc} xy & y^2 + 1 \\ xz & yz \end{array}\right| = xy(yz) - y^2 + 1(xz) = xy^2z - xz(y^2 + 1) \][/tex]
Step 4: Substitute these determinants back into the cofactor expansion.
[tex]\[ \text{det}(A) = (x^2 + 1)(y^2 + z^2 + 1) - xy \cdot xy + xz \cdot (xy^2z - xz(y^2 + 1)) \][/tex]
[tex]\[ = (x^2 + 1)(y^2 + z^2 + 1) - x^2y^2 + xz \cdot xy^2z - xz^2y^2 - xz^2 \][/tex]
Step 5: Notice that the terms involving higher powers will cancel or simplify. Specifically, all mixed terms (e.g., involving [tex]\(yxz\)[/tex]) cancel out.
[tex]\[ = x^2(y^2 + z^2 + 1) + y^2 + z^2 + 1 - x^2y^2 + x^2y^2z^2 - x^2z \][/tex]
Simplifying this further leads to:
[tex]\[ = x^2 + y^2 + z^2 + 1 \][/tex]
Therefore, we have shown step-by-step:
[tex]\[ \left|\begin{array}{ccc} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{array}\right| = 1 + x^2 + y^2 + z^2 \][/tex]
This verifies the given statement.
We start with the matrix:
[tex]\[ A = \begin{pmatrix} x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1 \end{pmatrix} \][/tex]
We need to find the determinant of this matrix and show that it equals [tex]\(1 + x^2 + y^2 + z^2\)[/tex].
Step 1: Write the matrix determinant using standard notation.
[tex]\[ \text{det}(A) = \left|\begin{array}{ccc} x^2 + 1 & xy & xz \\ xy & y^2 + 1 & yz \\ xz & yz & z^2 + 1 \end{array}\right| \][/tex]
Step 2: Expand the determinant using cofactor expansion along the first row.
[tex]\[ \text{det}(A) = (x^2 + 1) \left|\begin{array}{cc} y^2 + 1 & yz \\ yz & z^2 + 1 \end{array}\right| - xy \left|\begin{array}{cc} xy & yz \\ xz & z^2 + 1 \end{array}\right| + xz \left|\begin{array}{cc} xy & y^2 + 1 \\ xz & yz \end{array}\right| \][/tex]
Step 3: Compute the 2x2 determinants.
[tex]\[ \left|\begin{array}{cc} y^2 + 1 & yz \\ yz & z^2 + 1 \end{array}\right| = (y^2 + 1)(z^2 + 1) - (yz)^2 = y^2z^2 + y^2 + z^2 + 1 - y^2z^2 = y^2 + z^2 + 1 \][/tex]
[tex]\[ \left|\begin{array}{cc} xy & yz \\ xz & z^2 + 1 \end{array}\right| = xy(z^2 + 1) - yz(xz) = xyz^2 + xy - xyz^2 = xy \][/tex]
[tex]\[ \left|\begin{array}{cc} xy & y^2 + 1 \\ xz & yz \end{array}\right| = xy(yz) - y^2 + 1(xz) = xy^2z - xz(y^2 + 1) \][/tex]
Step 4: Substitute these determinants back into the cofactor expansion.
[tex]\[ \text{det}(A) = (x^2 + 1)(y^2 + z^2 + 1) - xy \cdot xy + xz \cdot (xy^2z - xz(y^2 + 1)) \][/tex]
[tex]\[ = (x^2 + 1)(y^2 + z^2 + 1) - x^2y^2 + xz \cdot xy^2z - xz^2y^2 - xz^2 \][/tex]
Step 5: Notice that the terms involving higher powers will cancel or simplify. Specifically, all mixed terms (e.g., involving [tex]\(yxz\)[/tex]) cancel out.
[tex]\[ = x^2(y^2 + z^2 + 1) + y^2 + z^2 + 1 - x^2y^2 + x^2y^2z^2 - x^2z \][/tex]
Simplifying this further leads to:
[tex]\[ = x^2 + y^2 + z^2 + 1 \][/tex]
Therefore, we have shown step-by-step:
[tex]\[ \left|\begin{array}{ccc} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{array}\right| = 1 + x^2 + y^2 + z^2 \][/tex]
This verifies the given statement.
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