Problem 75 - College Algebra

Question

Find values of \(a\) for which the following matrix is not invertible: $$ \left[\begin{array}{rr} 1 & a+1 \\ a-2 & 4 \end{array}\right] $$

Step-by-Step Solution

Answer

There are no real values of \(a\) for which the given matrix is not invertible.

Step 1: Find the Determinant of the Matrix

The determinant of the 2x2 matrix \(\left[\begin{array}{rr} 1 & a+1 \ a-2 & 4 \end{array}\right]\) is calculated using the formula \(ad-bc\), where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix. Hence, determinant = \((1*4) - ((a+1)*(a-2)) = 4 - a^2 - a + 2a - 2\). Simplifying this gives the determinant as \(a^2 - a + 2\)

Step 2: Set the Determinant Equal to Zero

For a matrix to not be invertible, the determinant must be equal to zero. So, set the determinant expression to zero, i.e., \(a^2 - a + 2 = 0\)

Step 3: Solve the Quadratic Equation

This is a quadratic equation in the form \(ax^2 + bx + c = 0\) and it can be solved using the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\). Plugging in the given values gives \(a = [1 ± sqrt((1)^2 - 4*1*2)] / 2*1\). This simplifies to \(a = [1 ± sqrt(-7)] / 2\).

Step 4: Explanation of the Result

The term under the square root, called the discriminant, is negative (i.e., \(-7\)). When the discriminant is negative, there are no real solutions. Thus, the equation \(a^2 - a + 2 = 0\) has no real solutions.

Key Concepts

Determinant

The determinant is a special number calculated from certain square matrices. It plays a vital role in understanding whether a matrix is invertible or not. If you have a matrix that looks like this:

• Row 1: elements are 1 and \(a+1\)
• Row 2: elements are \(a-2\) and 4

For a 2x2 matrix, the determinant can be calculated using the formula: \[adsr- bcdn\]where \(a, b, c, d\) are the elements of the matrix. By plugging in the values, for our matrix:Determinant = \( (1 \cdot 4) - ((a + 1) \cdot (a - 2)) \).When simplified further, we obtain an expression in terms of \(a\). Here, solving gives:Determinant = \(a^2 - a + 2\).When a matrix is invertible, its determinant is not equal to zero. However, if the determinant is zero, the matrix is not invertible.

Quadratic Equation

A quadratic equation is a type of polynomial equation in which the highest power of the variable is squared. The general format is:\(ax^2 + bx + c = 0\).The equation we derived from the determinant \(a^2 - a + 2 = 0\) fits perfectly into this form, where:

• \(a = 1\)
• \(b = -1\)
• \(c = 2\)

Quadratic equations are typically solved using several methods:

• Factoring (if possible)
• Completing the square
• Quadratic formula

In this case, we use the quadratic formula:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]This formula allows us to find the solutions, or roots, of the quadratic equation. These solutions tell us the values of \(a\) that make the determinant zero, and hence, when the matrix is not invertible.

Discriminant

The discriminant is a component of the quadratic formula, specifically the part inside the square root: \(b^2 - 4ac\). It gives crucial insight into the nature of the solutions of a quadratic equation:

• If the discriminant is positive, the equation has two distinct real solutions.
• If the discriminant is zero, it has exactly one real solution (a repeated root).
• If the discriminant is negative, the equation has no real solutions and instead has two complex solutions.

In our problem, the discriminant calculated is:
\((1)^2 - 4 \times 1 \times 2 = -7\)Since \(-7\) is negative, the quadratic equation \(a^2 - a + 2 = 0 \) has no real solutions. Therefore, there are no real values for \(a\) that make the matrix non-invertible. This underlines the importance of the discriminant in judging the nature of the solutions without necessarily calculating them.