Key Concept: Invertibility of a Matrix

A square matrix $A$ is invertible (or non-singular) if and only if its determinant, denoted as $|A|$ or $\det(A)$, is non-zero. If $|A| = 0$, the matrix is singular and not invertible.

Problem Statement: We are given the matrix $A$: A = \left[ {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}\cos t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right] We need to determine for which values of $t \in \mathbb{R}$ the matrix $A$ is invertible.

Step-by-Step Solution:

1. Calculate the Determinant of A, $|A|$

The first step to determine invertibility is to calculate the determinant of the matrix $A$.

|A| = \left| {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}\cos t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right|

2. Factor out Common Terms from Columns

To simplify the determinant calculation, we observe common factors in each column:

• In the first column ($C_1$), $e^t$ is a common factor.
• In the second column ($C_2$), $e^{-t}$ is a common factor.
• In the third column ($C_3$), $e^{-t}$ is a common factor.

Why this step? Factoring out common terms simplifies the entries within the determinant, making subsequent calculations (like row operations and expansion) much easier and less prone to errors. According to determinant properties, if a column (or row) is multiplied by a scalar $k$, the determinant is multiplied by $k$. Therefore, if we factor out $k$, we multiply the determinant by $k$.

Applying this property: |A| = {e^t} \cdot {e^{ - t}} \cdot {e^{ - t}} \left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right| Simplify the exponential terms: $e^t \cdot e^{-t} \cdot e^{-t} = e^{t-t-t} = e^{-t}$. So, |A| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|

3. Apply Row Operations to Create Zeros

To further simplify the determinant expansion, we can create zeros in the first column using elementary row operations. This will allow us to expand the determinant along the first column, reducing it to a $2 \times 2$ determinant.

Why these operations? We aim to get zeros below the '1' in the first column.

• Apply $R_2 \to R_2 - R_1$: This will make the first element of the second row zero.
• Apply $R_3 \to R_3 - R_1$: This will make the first element of the third row zero.

The determinant value remains unchanged when we subtract a multiple of one row from another row.

Let's perform the operations: For $R_2 \to R_2 - R_1$:

• $R_{21}' = 1 - 1 = 0$
• $R_{22}' = (- \cos t - \sin t) - \cos t = - \sin t - 2\cos t$
• $R_{23}' = (- \sin t + \cos t) - \sin t = - 2\sin t + \cos t$

For $R_3 \to R_3 - R_1$:

• $R_{31}' = 1 - 1 = 0$
• $R_{32}' = (2\sin t) - \cos t = 2\sin t - \cos t$
• $R_{33}' = (- 2\cos t) - \sin t = - 2\cos t - \sin t$

The determinant now becomes: |A| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|

4. Expand the Determinant along the First Column ($C_1$)

Since we have two zeros in the first column, expanding along $C_1$ is straightforward. The expansion is $1 \cdot (\text{cofactor of } a_{11}) - 0 \cdot (\text{cofactor of } a_{21}) + 0 \cdot (\text{cofactor of } a_{31})$. So, we only need to calculate the $2 \times 2$ determinant corresponding to the element $a_{11}=1$.

|A| = {e^{ - t}} \cdot 1 \cdot \left| {\matrix{ { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|

5. Calculate the $2 \times 2$ Determinant

The determinant of a $2 \times 2$ matrix \left[ {\matrix{ a & b \cr c & d \cr } } \right] is $ad - bc$.

Here, $a = - \sin t - 2\cos t = -(\sin t + 2\cos t)$ $d = - 2\cos t - \sin t = -(\sin t + 2\cos t)$ $b = - 2\sin t + \cos t = - (2\sin t - \cos t)$ $c = 2\sin t - \cos t$

So, the $2 \times 2$ determinant is: $( - (\sin t + 2\cos t)) \cdot ( - (\sin t + 2\cos t)) - ( - (2\sin t - \cos t)) \cdot (2\sin t - \cos t)$ $= (\sin t + 2\cos t)^2 + (2\sin t - \cos t)^2$

Now, expand these squares using $(x+y)^2 = x^2 + 2xy + y^2$ and $(x-y)^2 = x^2 - 2xy + y^2$: $= (\sin^2 t + 4\sin t \cos t + 4\cos^2 t) + (4\sin^2 t - 4\sin t \cos t + \cos^2 t)$

Combine like terms: $= (\sin^2 t + 4\sin^2 t) + (4\cos^2 t + \cos^2 t) + (4\sin t \cos t - 4\sin t \cos t)$ $= 5\sin^2 t + 5\cos^2 t + 0$ $= 5(\sin^2 t + \cos^2 t)$

Using the fundamental trigonometric identity $\sin^2 t + \cos^2 t = 1$: $= 5(1) = 5$

6. Final Determinant Value

Substitute this back into the expression for $|A|$: $|A| = {e^{ - t}} \cdot 5 = 5{e^{ - t}}$

7. Determine Invertibility

For matrix $A$ to be invertible, its determinant $|A|$ must be non-zero. We have $|A| = 5{e^{ - t}}$.

Why is $5e^{-t}$ always non-zero?

• The exponential function $e^x$ is always positive for any real number $x$. Therefore, $e^{-t} > 0$ for all $t \in \mathbb{R}$.
• Since $e^{-t}$ is always positive, $5e^{-t}$ will also always be positive and thus never equal to zero.
• $5e^{-t} \neq 0$ for all $t \in \mathbb{R}$.

Conclusion: Since $|A| = 5e^{-t}$ is non-zero for all real values of $t$, the matrix $A$ is invertible for all $t \in \mathbb{R}$.

Comparing this with the given options: (A) invertible for all $t \in \mathbb{R}$. (B) invertible only if $t = \pi$. (C) not invertible for any $t \in \mathbb{R}$. (D) invertible only if $t = {\pi \over 2}$.

Our conclusion matches option (A).

Tips and Common Mistakes:

• Factoring correctly: Be careful when factoring out terms from rows/columns. Remember that if you factor $k$ from a row/column, you multiply the determinant by $k$. If you factor $k$ from the entire matrix $A$ (i.e., $kA$), then $|kA| = k^n |A|$ for an $n \times n$ matrix. Here, we factored from individual columns, so it's a product of the factors.
• Row/Column Operations: Understand which operations change the determinant value and which do not.
  • $R_i \leftrightarrow R_j$ (swapping two rows) changes the sign of the determinant.
  • $R_i \to kR_i$ (multiplying a row by $k$) multiplies the determinant by $k$.
  • $R_i \to R_i + kR_j$ (adding a multiple of one row to another) does not change the determinant value.
• Trigonometric Identities: Keep common identities like $\sin^2 x + \cos^2 x = 1$ handy. They often appear in determinant problems involving trigonometric functions.
• Exponential Function Properties: Remember that $e^x > 0$ for all real $x$. This is crucial for determining if an expression involving $e^x$ can ever be zero.

Summary/Key Takeaway: To check the invertibility of a matrix, always calculate its determinant. If the determinant is a constant non-zero value or an expression that is never zero for the given domain, then the matrix is invertible for all values in that domain. Simplifying the determinant calculation using factoring and row/column operations is a highly effective strategy.