The set of all values of t ∈ ℝ, for which the matrix [[e^t, e⁻ᵗ(sin t - 2cos t), e⁻ᵗ(-2sin t - cos t

Q: Wait, why can we just factor out $e^t$ and $e^{-t}$ from the determinant? Is that allowed?

Yes, it is! A fundamental property of determinants states that if you multiply a row or a column by a scalar $k$, the determinant is multiplied by $k$. Conversely, you can factor a common scalar out of any row or column. Here, we factored $e^t$ from the first column and $e^{-t}$ from the second and third columns, which is perfectly valid.

Q: Why do we assume $\cos t \neq 0$ when dividing by $\cos^2 t$?

That is a great catch! In mathematics, we must be rigorous. When we divide by $\cos^2 t$, we are implicitly assuming $\cos t \neq 0$. However, we must check the case where $\cos t = 0$ separately. If $\cos t = 0$, then $\sin^2 t = 1$. Substituting this into our expression $7\sin^2 t - 3\sin t \cos t + 6\cos^2 t$ gives $7(1) - 0 + 0 = 7$, which is not zero. So, the result holds even when $\cos t = 0$.

Q: Why does a negative discriminant mean the matrix is always invertible?

Think about the graph of a quadratic function $g(x) = ax^2 + bx + c$. If the discriminant $\Delta = b^2 - 4ac$ is negative, the parabola never crosses the x-axis. If the leading coefficient $a$ is positive, the entire parabola lies above the x-axis, meaning $g(x)$ is always positive. Since our determinant is the product of an exponential (which is always positive) and this quadratic (which is also always positive), the determinant can never be zero, thus the matrix is always invertible.

Question

The set of all values of $t \in \mathbb{R}$, for which the matrix $\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \ e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}$ is invertible, is

Options:

(A) $\{(2k+1)\frac{\pi}{2}, k \in \mathbb{Z}\}$

(B) $\{k\pi + \frac{\pi}{4}, k \in \mathbb{Z}\}$

(C) $\{k\pi, k \in \mathbb{Z}\}$

(D) $\mathbb{R}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks for the set of all values of $t \in \mathbb{R}$ for which the given matrix is invertible. A matrix is invertible if and only if its determinant is non-zero.

Our primary objective is to calculate the determinant $D$ and demonstrate that it never equals zero. We begin by expressing the determinant as:

$$D = \begin{vmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t) \ e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t) \ e^t & e^{-t}\cos t & e^{-t}\sin t \end{vmatrix}$$

### The Power of Factoring

Observe the columns of the matrix. The first column contains a common factor of $e^t$, while the second and third columns contain a common factor of $e^{-t}$.

By the properties of determinants, we can extract these factors. This simplifies the expression significantly:

$$D = (e^t)(e^{-t})(e^{-t}) \begin{vmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \ 1 & 2\sin t + \cos t & \sin t - 2\cos t \ 1 & \cos t & \sin t \end{vmatrix}$$

The exponential terms simplify to $e^{-t}$. The matrix is now in a much more manageable form.

### Creating Zeros: The Strategist's Move

The presence of a column of ones is a strategic advantage. We can simplify the determinant by performing row operations to create zeros.

We apply the operations $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:

$$D = e^{-t} \begin{vmatrix} 0 & \sin t - 3\cos t & -3\sin t - \cos t \ 0 & 2\sin t & \sin t - 2\cos t \ 1 & \cos t & \sin t \end{vmatrix}$$

Expanding along the first column, we only need to evaluate the minor associated with the element in the third row, first column:

$$D = e^{-t} [1 \cdot \{(\sin t - 3\cos t)(\sin t - 2\cos t) - (2\sin t)(-3\sin t - \cos t)\}]$$

### The Algebraic Finale

We now expand the terms within the brackets carefully. The first product is:

$$(\sin t - 3\cos t)(\sin t - 2\cos t) = \sin^2 t - 5\sin t \cos t + 6\cos^2 t$$

The second product, including the subtraction, is:

$$-(2\sin t)(-3\sin t - \cos t) = 6\sin^2 t + 2\sin t \cos t$$

Combining these results, the determinant simplifies to:

$$D = e^{-t} [7\sin^2 t - 3\sin t \cos t + 6\cos^2 t]$$

### The Final Proof

Since $e^{-t}$ is strictly positive for all $t \in \mathbb{R}$, the invertibility depends entirely on the trigonometric expression $7\sin^2 t - 3\sin t \cos t + 6\cos^2 t$.

Assuming $\cos t 
eq 0$, we divide by $\cos^2 t$ to obtain the quadratic expression $7	an^2 t - 3	an t + 6$. Let $x = 	an t$. The discriminant $\Delta$ of this quadratic is:

$$\Delta = (-3)^2 - 4(7)(6) = 9 - 168 = -159$$

Because $\Delta < 0$ and the leading coefficient is positive, the quadratic is always positive. Consequently, the determinant is never zero, and the matrix is **invertible for all $t \in \mathbb{R}$**.

Animated Solution for Mathematics - Matrices and Determinants: The set of all values of $t \in R$ , for which the matrix $e^{t} e^{t} e^{t} e^{- t} (sin t - 2 cos t) e^{- t} (2 sin t + cos t) e^{- t} cos t e^{- t} (- 2 sin t - cos t) e^{- t} (sin t - 2 cos t) e^{- t} sin t$ is invertible, is

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Visualized Solution

Invertibility Condition

Setting up the Determinant

Factoring Out $e^{t}$ and $e^{- t}$

Applying Row Operations

Simplified Determinant Matrix

Expanding Along Column 1

Algebraic Expansion

Simplifying the Expression

Analyzing the Quadratic Form

Checking the Discriminant

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Factoring

Creating Zeros

The Algebraic Finale

The Final Proof

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