Question

The matrix $$ \left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| $$ is singular if \(\ldots \ldots\) (a) \(a-b=0\) (b) \(a+b=0\) (c) \(a+b+c=0\) (d) \(a=0\)

Short answer

The matrix is singular if \(a+b+c=0\).

Step-by-step solution

Write down the given matrix

We are given the matrix: $$ A = \left|\begin{array}{ccc} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{array}\right| $$

Calculate the determinant of the matrix

To calculate the determinant, we can use the following formula: $$ \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) $$ Plug in the values from the given matrix: $$ \det(A) = (a-b-c)((b-c-a)(c-a-b) - 2b\cdot 2c) - 2a(2b(c-a-b) - 2b\cdot 2c) + 2a(2b\cdot 2c - (b-c-a)2c) $$

Simplify and find the condition for determinant to be zero

To simplify and find when the determinant is zero, expand and factorize the terms: \begin{align*} \det(A) &= (a-b-c)((b^2-c^2-ab-ac+bc) - 4bc) \\ &\quad -2a(-4c+2b+2c-(a-c)(-4c)) \\ &\quad +2a(-4c+2b-(a-b)(-2c)) \\ \end{align*} Now, factorize and combine similar terms: \begin{align*} \det(A) &= (a-b-c)(b^2-c^2-4bc-ab+ac+bc) + 8a^2c-8abc \\ &= (a-b-c)(b^2-5bc+c^2-ab+ac+bc) + 8a^2c-8abc \\ \end{align*} Observe that (b^2-5bc+c^2) can be written as (b-c)^2, and the expression inside the parentheses can be written as a sum of two squares: \begin{align*} \det(A) &= (a-b-c)((b-c)^2 - (a-c)(a-b)) + 8a^2c-8abc \\ \end{align*} The determinant will be zero if the following condition holds: $$ (a-b-c)((b-c)^2 - (a-c)(a-b)) + 8a^2c-8abc = 0 $$

Compare with the given options

Now, we will compare the condition obtained in Step 3 with the given options: (a) \(a-b=0\): This condition does not match our result. (b) \(a+b=0\): This condition also does not match our result. (c) \(a+b+c=0\): This condition matches the result obtained in Step 3, because if \(a+b+c=0\), then \(a-b-c = 0\) and therefore \(\det(A) = 0\). (d) \(a=0\): This condition does not match our result. Thus, the correct option is (c) \(a+b+c=0\).

Key concepts

Determinant Calculation

One of the first steps in determining if a matrix is singular is to calculate its determinant. The determinant is a special number that can be computed from a square matrix. It provides vital information about the matrix, including whether it has an inverse. If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
In calculating the determinant for a 3x3 matrix, we use the formula that involves the elements of the matrix. It is a bit lengthy but follows a specific pattern. For matrix \( A \), the determinant is given by:
\( \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \)
This formula might look intimidating, but with practice, it becomes more intuitive. By substituting the values of \( a_{ij} \) for the given matrix, you can compute the specific determinant.
The main point is to simplify the determinant until you can see if it equals zero. Only then can we establish if the matrix is singular.

Factorization Techniques

Once the determinant is calculated, the next step is to simplify it. This often involves factorization, which can simplify expressions using algebraic identities and techniques.
In our problem, simplifying the determinant involved rewriting parts of the expression as squares or products of binomials. For instance, recognizing that \((b^2 - 5bc + c^2)\) is equivalent to \((b-c)^2\) can significantly reduce complexity.
These techniques rely on spotting patterns like difference of squares and using algebraic identities. It's about rearranging and grouping terms to make the expression manageable.

• Look for common factors each term shares and factor them out.
• Recognize perfect square trinomials and rewrite them accordingly.
• Use algebraic properties to rewrite terms, such as \((a-b-c)(b-c)^2\).

The power of factorization is in transforming a bulky expression into something easier to handle, ultimately helping us pinpoint conditions for when the determinant is zero.

Conditions for Singularity

Finally, we need to understand under what conditions a matrix becomes singular, meaning its determinant is zero. This is critical because these conditions tell us under which circumstances the matrix lacks an inverse.
In this exercise, by simplifying and factoring the determinant to zeros, we discovered that the matrix is singular if \( a + b + c = 0 \). This condition essentially reflects a balance among the elements of the matrix, where their sum leads to specific cancellations in calculations.
Testing provided options against this derived condition helps confirm singularity. For instance:

• Option (a), \(a-b=0\), doesn't satisfy our determinant equation because it doesn't result in zero when plugged into our formula.
• Option (b), \(a+b=0\), similarly fails to align with the zero determinant condition.
• Option (c), \(a+b+c=0\), matches and results in a zero determinant, indicating singularity.
• Option (d), \(a=0\), also does not lead to zero without additional conditions among \(b\) and \(c\).

Understanding these conditions and testing them against the problem gives clarity on when the matrix becomes singular. This knowledge is essential for dealing with linear algebra problems involving matrix inverses.