The rank of the matrix $\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ where a = b ≠ c is:

Question

The rank of the matrix $\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ where a = b &ne; c is:

Prepp Team · Accepted Answer

Understanding Matrix Rank

The rank of a matrix is a fundamental concept in linear algebra. It tells us the maximum number of linearly independent rows or columns in the matrix. For a square matrix, the rank is related to its determinant. If the determinant of an n x n matrix is non-zero, its rank is n. If the determinant is zero, the rank is less than n.

Analyzing the Given Matrix with the Condition a = b ≠ c

The given matrix is:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix} \]

We are given the condition that $a = b \ne c$. Let's substitute $b$ with $a$ in the matrix:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ a & a & c \\ a^2 & a^2 & c^2 \end{bmatrix} \]

Calculating the Determinant with a = b ≠ c

To find the rank, we can first calculate the determinant of this 3x3 matrix. If the determinant is non-zero, the rank is 3. We can expand the determinant along the first row:

\[ \det(A) = 1 \cdot \begin{vmatrix} a & c \\ a^2 & c^2 \end{vmatrix} - 1 \cdot \begin{vmatrix} a & c \\ a^2 & c^2 \end{vmatrix} + 1 \cdot \begin{vmatrix} a & a \\ a^2 & a^2 \end{vmatrix} \] \[ \det(A) = (a \cdot c^2 - c \cdot a^2) - (a \cdot c^2 - c \cdot a^2) + (a \cdot a^2 - a \cdot a^2) \] \[ \det(A) = (ac^2 - a^2c) - (ac^2 - a^2c) + (a^3 - a^3) \] \[ \det(A) = 0 - 0 + 0 = 0 \]

The determinant of the matrix is 0. This tells us that the rank of the matrix is less than 3. The columns (or rows) are linearly dependent.

Alternatively, observing the matrix after substituting $b=a$:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ a & a & c \\ a^2 & a^2 & c^2 \end{bmatrix} \]

Notice that the first column $\begin{bmatrix} 1 \\ a \\ a^2 \end{bmatrix}$ and the second column $\begin{bmatrix} 1 \\ a \\ a^2 \end{bmatrix}$ are identical. If a matrix has two identical columns (or rows), they are linearly dependent, and the determinant is zero. This confirms our determinant calculation.

Checking for Rank 2

Since the determinant is 0, the rank is less than 3. Now, we need to check if the rank is 2. The rank is 2 if there is at least one 2x2 submatrix whose determinant is non-zero.

Let's consider the submatrix formed by the first two rows and the first and third columns:

\[ M_{13} = \begin{vmatrix} 1 & 1 \\ a & c \end{vmatrix} \]

The determinant of this submatrix is:

\[ \det(M_{13}) = (1 \cdot c) - (1 \cdot a) = c - a \]

Given the condition $a \ne c$, the value $c - a$ is non-zero.

Therefore, we have found a 2x2 submatrix with a non-zero determinant. This means that the rank of the matrix is at least 2.

Conclusion on Matrix Rank

We found that the determinant of the 3x3 matrix is 0, so the rank is less than 3. We also found a 2x2 submatrix with a non-zero determinant, so the rank is at least 2.

Combining these results, the rank of the matrix must be exactly 2.

Result

The rank of the matrix with the condition $a = b \ne c$ is 2.

Revision Table: Key Matrix Concepts

Concept	Definition	Relevance to Rank
Matrix Rank	Maximum number of linearly independent rows or columns.	The property we are determining.
Determinant	A scalar value calculated from a square matrix.	Non-zero determinant means full rank for square matrices.
Linear Independence	A set of vectors where none can be written as a linear combination of others.	Rank equals the size of the largest set of linearly independent rows/columns.
Submatrix	A matrix obtained by deleting rows and/or columns from a larger matrix.	Used to find rank if the main determinant is zero; check determinants of submatrices.

Additional Information: Finding Matrix Rank

Besides calculating determinants of submatrices, the rank of a matrix can also be found using elementary row operations (Gaussian elimination) to transform the matrix into row-echelon form. The number of non-zero rows in the row-echelon form is equal to the rank of the matrix.

In our case, with $b=a$, the matrix is:

\[ \begin{bmatrix} 1 & 1 & 1 \\ a & a & c \\ a^2 & a^2 & c^2 \end{bmatrix} \]

Performing row operations:

$R_2 \leftarrow R_2 - a R_1$

$R_3 \leftarrow R_3 - a^2 R_1$

\[ \begin{bmatrix} 1 & 1 & 1 \\ a - a(1) & a - a(1) & c - a(1) \\ a^2 - a^2(1) & a^2 - a^2(1) & c^2 - a^2(1) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & c - a \\ 0 & 0 & c^2 - a^2 \end{bmatrix} \]

We know $c \ne a$, so $c - a \ne 0$. Let's simplify the third row element:

$c^2 - a^2 = (c-a)(c+a)$

The matrix becomes:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & c - a \\ 0 & 0 & (c-a)(c+a) \end{bmatrix} \]

Now, if $c+a \ne 0$, we can perform $R_3 \leftarrow R_3 - (c+a) R_2$:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & c - a \\ 0 & 0 & (c-a)(c+a) - (c+a)(c-a) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & c - a \\ 0 & 0 & 0 \end{bmatrix} \]

Since $c-a \ne 0$, the second row is a non-zero row. The first row is also non-zero. The third row is zero. We have 2 non-zero rows.

What if $c+a = 0$? Then $c = -a$. Since $c \ne a$, this is possible if $a \ne 0$. If $c = -a$, the matrix after the first step of row operations was:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & c - a \\ 0 & 0 & (c-a)(c+a) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & -a - a \\ 0 & 0 & (-a-a)(-a+a) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & -2a \\ 0 & 0 & 0 \end{bmatrix} \]

Since we are given $a \ne c$ and $c = -a$, it implies $a \ne -a$, which means $2a \ne 0$, so $a \ne 0$. Thus, $-2a \ne 0$. We still have 2 non-zero rows.

In both cases ($c+a \ne 0$ or $c+a = 0$), the row-echelon form has 2 non-zero rows.

Thus, the rank of the matrix is 2.

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