The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
x, y, z are in G.P.
Understanding the Determinant Problem
The question asks for the condition under which a specific \(3 \times 3\) determinant evaluates to zero. Determinants are important in linear algebra and represent a scalar value that can tell us about the properties of a matrix, such as invertibility.
The given determinant is:
\[ \Delta = \left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| \]
We need to find the relationship between x, y, and z that makes \(\Delta = 0\).
Expanding the Determinant
To find the value of the determinant, we can expand it along any row or column. The third row has a zero, which simplifies the calculation. Let's expand along the third row:
\[ \Delta = 0 \cdot C_{31} - (xp + y) \cdot C_{32} + (yp + z) \cdot C_{33} \]
where \(C_{ij}\) is the cofactor of the element in the i-th row and j-th column.
The cofactors are calculated as \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the minor (the determinant of the submatrix obtained by removing the i-th row and j-th column).
\[ C_{31} = (-1)^{3+1} \left| {\begin{array}{*{20}{c}} x&y\\ y&z \end{array}} \right| = (xz - y^2) \] (This term is multiplied by 0, so it won't affect the result)
\[ C_{32} = (-1)^{3+2} \left| {\begin{array}{*{20}{c}} {xp + y}&y\\ {yp + z}&z \end{array}} \right| = -((xp+y)z - y(yp+z)) = -(xpz + yz - y^2p - yz) = -(xpz - y^2p) = -p(xz - y^2) \] (Mistake in calculation above in thought process, fixing here. \(yz\) terms cancel)
\[ C_{33} = (-1)^{3+3} \left| {\begin{array}{*{20}{c}} {xp + y}&x\\ {yp + z}&y \end{array}} \right| = ((xp+y)y - x(yp+z)) = (xpy + y^2 - xyp - xz) = (y^2 - xz) \] (Mistake in calculation above in thought process, fixing here. \(xpy\) terms cancel)
Now substitute these cofactors back into the expansion:
\[ \Delta = 0 \cdot (xz - y^2) - (xp + y) \cdot (-p(xz - y^2)) + (yp + z) \cdot (y^2 - xz) \]
\[ \Delta = 0 + p(xp + y)(xz - y^2) + (yp + z)(y^2 - xz) \]
Notice that \((y^2 - xz) = -(xz - y^2)\). Let's factor out \((y^2 - xz)\) (or \((xz - y^2)\)). Factoring out \((y^2 - xz)\):
\[ \Delta = -p(xp + y)(y^2 - xz) + (yp + z)(y^2 - xz) \]
\[ \Delta = (y^2 - xz) [-(xp + y)p + (yp + z)] \]
\[ \Delta = (y^2 - xz) [-xp^2 - yp + yp + z] \]
\[ \Delta = (y^2 - xz) [-xp^2 + z] \]
\[ \Delta = (y^2 - xz) (z - xp^2) \]
(Correcting the expansion again: \(C_{32}\) was \( -(-(xpz - y^2p))\) which is \(+(xpz - y^2p)\). Let's re-calculate carefully.)
Expanding along row 3:
\[ \Delta = 0 \cdot M_{31} - (xp+y) M_{32} + (yp+z) M_{33} \]
\[ M_{32} = \left| {\begin{array}{*{20}{c}} {xp + y}&y\\ {yp + z}&z \end{array}} \right| = (xp+y)z - y(yp+z) = xpz + yz - y^2p - yz = xpz - y^2p = p(xz - y^2) \]
\[ M_{33} = \left| {\begin{array}{*{20}{c}} {xp + y}&x\\ {yp + z}&y \end{array}} \right| = (xp+y)y - x(yp+z) = xpy + y^2 - xyp - xz = y^2 - xz \]
So the determinant is:
\[ \Delta = 0 - (xp+y) [p(xz - y^2)] + (yp+z) [y^2 - xz] \]
\[ \Delta = -p(xp+y)(xz - y^2) + (yp+z)(y^2 - xz) \]
\[ \Delta = p(xp+y)(y^2 - xz) + (yp+z)(y^2 - xz) \]
Factor out \((y^2 - xz)\):
\[ \Delta = (y^2 - xz) [p(xp+y) + (yp+z)] \]
\[ \Delta = (y^2 - xz) [xp^2 + yp + yp + z] \]
\[ \Delta = (y^2 - xz) [xp^2 + 2yp + z] \]
Condition for Determinant to be Zero
The determinant is zero, i.e., \(\Delta = 0\), if the product of the two factors is zero:
\[ (y^2 - xz) (xp^2 + 2yp + z) = 0 \]
This equation holds if either of the factors is zero:
- \(y^2 - xz = 0\)
- \(xp^2 + 2yp + z = 0\)
The problem asks for a condition on x, y, z. Looking at the options, they provide relationships between x, y, and z that are independent of p. The first condition, \(y^2 - xz = 0\), is a condition solely on x, y, and z. The second condition involves p, and it being zero would depend on the value of p or a more complex relationship between x, y, z, and p.
Therefore, the condition that makes the determinant zero, independent of the value of p, is \(y^2 - xz = 0\).
Connecting the Condition to Progression Types
We found the condition \(y^2 = xz\).
- If x, y, z are in Arithmetic Progression (A.P.), the condition is \(2y = x + z\).
- If x, y, z are in Geometric Progression (G.P.), the condition is \(y^2 = xz\).
- If x, y, z are in Harmonic Progression (H.P.), the condition is \(1/y = (1/x + 1/z)/2\), which simplifies to \(2xz = y(x+z)\).
Comparing our condition \(y^2 = xz\) with the standard conditions for A.P., G.P., and H.P., we see that it matches the condition for x, y, and z to be in Geometric Progression (G.P.).
Thus, the determinant is zero if x, y, z are in G.P.
Important Questions from Determinants
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\det(A^2 B (A^T)^3) = 16$ and $\det(A^3 B^{-2}) = 32$, then $\det(B^2 A^{-1} (B^T)^3)$ is equal to:
The value of determinant \(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) is:
If \(\left| {\begin{array}{*{20}{c}} 5&a\\ a&2 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2&1\\ 3&2 \end{array}} \right|\), then the values of a are:
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {\sqrt {13} + \sqrt 3 }&{2\sqrt 5 }&{\sqrt 5 }\\ {\sqrt {15} + \sqrt {26} }&5&{\sqrt {10} }\\ {3 + \sqrt {65} }&{\sqrt {15} }&5 \end{array}} \right|\) is
If Δ(a, b, c, α) = 0 for every α > 0, then which one of the following is correct ?
