Question

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $B = \begin{bmatrix} x \\ y \end{bmatrix}$, $x, y \in R - \{0\}$ such that $AB = B$ and trace $(A) = 17$. Then, value of $ad - bc$ is

Answer 1

16

Answer 2

The problem involves matrix properties, specifically matrix multiplication and the trace of a matrix. It can be solved using either direct algebraic manipulation or by recognizing the concept of eigenvalues.

Method 1: Algebraic Manipulation

Given:

1. Matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$
2. Matrix $B = \begin{bmatrix} x \\ y \end{bmatrix}$, where $x, y \in R - \{0\}$
3. Condition $AB = B$
4. Condition trace$(A) = 17$

We need to find the value of $ad - bc$, which is the determinant of matrix $A$, i.e., $\det(A)$.

From the condition $AB = B$:

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}$$

Performing the matrix multiplication, we get a system of linear equations:

1. $ax + by = x$
2. $cx + dy = y$

Rearranging these equations:

1. $(a-1)x + by = 0 \quad \dots (i)$
2. $cx + (d-1)y = 0 \quad \dots (ii)$

This is a system of homogeneous linear equations in $x$ and $y$. Since $x, y \in R - \{0\}$, it implies that $B$ is a non-zero vector. For a system of homogeneous linear equations to have non-trivial solutions (i.e., solutions other than $x=0, y=0$), the determinant of the coefficient matrix must be zero.

The coefficient matrix is $\begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix}$. Setting its determinant to zero:

$$\det \begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix} = 0$$ $$(a-1)(d-1) - bc = 0$$

Expanding the expression:

$$ad - a - d + 1 - bc = 0$$

Rearranging the terms to isolate $ad - bc$:

$$ad - bc = a + d - 1 \quad \dots (iii)$$

We are given that trace$(A) = 17$. For matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the trace is the sum of its diagonal elements:

$$\text{trace}(A) = a+d$$

So, we have:

$$a+d = 17 \quad \dots (iv)$$

Substitute the value of $(a+d)$ from equation $(iv)$ into equation $(iii)$:

$$ad - bc = 17 - 1$$ $$ad - bc = 16$$

Method 2: Using Eigenvalues (More concise)

The condition $AB = B$ can be rewritten as $AB = 1 \cdot B$. This equation implies that $B$ is an eigenvector of matrix $A$ corresponding to the eigenvalue $\lambda = 1$. Since $B = \begin{bmatrix} x \\ y \end{bmatrix}$ with $x, y \in R - \{0\}$, $B$ is a non-zero vector. Therefore, $\lambda = 1$ must be an eigenvalue of matrix $A$.

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the characteristic equation is given by:

$$\lambda^2 - (\text{trace}(A))\lambda + \det(A) = 0$$

We know that $\lambda = 1$ is an eigenvalue, and trace$(A) = 17$. Also, $\det(A) = ad - bc$. Substitute these values into the characteristic equation:

$$1^2 - (17)(1) + (ad - bc) = 0$$ $$1 - 17 + (ad - bc) = 0$$ $$-16 + (ad - bc) = 0$$ $$ad - bc = 16$$

Both methods yield the same result.

The final answer is $\boxed{16}$.

Explanation of the solution: The condition $AB=B$ implies that $B$ is an eigenvector of $A$ with eigenvalue $1$. Since $B$ is a non-zero vector, $1$ must be an eigenvalue of $A$. The characteristic equation for a $2 \times 2$ matrix $A$ is $\lambda^2 - (\text{trace}(A))\lambda + \det(A) = 0$. Substituting $\lambda=1$ and the given trace$(A)=17$ into this equation yields $1^2 - 17(1) + \det(A) = 0$, which simplifies to $\det(A) = 16$. Since $\det(A) = ad-bc$, the value of $ad-bc$ is $16$.