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:
(a + b + c)3
Evaluating the Given Determinant
We are asked to find the value of the given 3x3 determinant. Determinants can often be simplified using row or column operations, which do not change the value of the determinant or change it by a known factor.
The given determinant is:
\(\Delta = \left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\)
Applying Row Operations to Simplify the Determinant
Let's perform the operation \(R_1 \to R_1 + R_2 + R_3\). This means we add the elements of the second and third rows to the corresponding elements of the first row. Let's see what happens to the elements in the first row:
- First element: \((a - b - c) + 2b + 2c = a + b + c\)
- Second element: \(2a + (b - c - a) + 2c = a + b + c\)
- Third element: \(2a + 2b + (c - a - b) = a + b + c\)
After performing this row operation, the determinant becomes:
| \(a + b + c\) | \(a + b + c\) | \(a + b + c\) |
| \(2b\) | \(b - c - a\) | \(2b\) |
| \(2c\) | \(2c\) | \(c - a - b\) |
Factoring Out the Common Term
Now we see that the first row has a common factor of \((a + b + c)\). We can factor this term out of the determinant:
\(\Delta = (a + b + c) \left| {\begin{array}{*{20}{c}} {1}&{1}&{1}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\)
Applying Column Operations
To further simplify the determinant, we can introduce zeros in the first row by using column operations. Let's perform the operations \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\).
- For \(C_2 \to C_2 - C_1\):
- Element in R1: \(1 - 1 = 0\)
- Element in R2: \((b - c - a) - 2b = -b - c - a = -(a + b + c)\)
- Element in R3: \(2c - 2c = 0\)
- For \(C_3 \to C_3 - C_1\):
- Element in R1: \(1 - 1 = 0\)
- Element in R2: \(2b - 2b = 0\)
- Element in R3: \((c - a - b) - 2c = -c - a - b = -(a + b + c)\)
The determinant now becomes:
\(\Delta = (a + b + c) \left| {\begin{array}{*{20}{c}} {1}&{0}&{0}\\ {2b}&{-(a + b + c)}&{0}\\ {2c}&{0}&{-(a + b + c)} \end{array}} \right|\)
Evaluating the Simplified Determinant
The resulting determinant is a triangular matrix. The determinant of a triangular matrix is simply the product of its diagonal elements. Alternatively, we can expand the determinant along the first row (since it has two zeros). Expanding along \(R_1\):
\(\Delta = (a + b + c) \times \left[ 1 \times \left| {\begin{array}{*{20}{c}} {-(a + b + c)}&{0}\\ {0}&{-(a + b + c)} \end{array}} \right| - 0 + 0 \right]\)
\(\Delta = (a + b + c) \times \left[ (-(a + b + c)) \times (-(a + b + c)) - 0 \times 0 \right]\)
\(\Delta = (a + b + c) \times \left[ (a + b + c)^2 - 0 \right]\)
\(\Delta = (a + b + c) \times (a + b + c)^2\)
\(\Delta = (a + b + c)^3\)
Thus, the value of the given determinant is \((a + b + c)^3\). This step-by-step process involving determinant properties allows us to evaluate the determinant efficiently.
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:
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
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
If Δ(a, b, c, α) = 0 for every α > 0, then which one of the following is correct ?
