Matrix Operations: A11, Adjoint, Determinant, And Inverse

Hey math whizzes! Today, we're diving deep into the fascinating world of matrix operations. Get ready to flex those analytical muscles as we break down how to find specific elements, the adjoint matrix, the determinant, and the inverse of a given matrix. These are fundamental skills in linear algebra, crucial for everything from solving systems of equations to understanding transformations in higher dimensions. So, grab your calculators, sharpen your pencils, and let's get down to business with our example matrix $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & 0 & 1\end{array}\right]$. We'll tackle each part step-by-step, making sure you guys understand the why behind each calculation.

(a) Finding $A_{11}$: The Cofactor Element

Alright guys, first up is finding $A_{11}$. In matrix notation, the subscript '11' tells us we're looking for the element in the first row and the first column. It's like a treasure map, guiding us to the exact spot. For our matrix $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & 0 & 1\end{array}\right]$, the element located at the intersection of the first row and the first column is simply the number 1. Pretty straightforward, right? But $A_{11}$ often plays a crucial role when we calculate cofactors, which we'll need later for the adjoint and inverse. Remember, the notation $A_{ij}$ refers to the element in the i-th row and j-th column. So, $A_{11}$ is the element in the 1st row, 1st column. Easy peasy!

(b) Adjoint$(A)$: The Matrix of Cofactors

Now, things get a bit more involved as we move on to finding the adjoint of matrix A, often denoted as Adjoint$(A)$ or adj$(A)$. The adjoint matrix is the transpose of the cofactor matrix. So, the first step is to find the cofactor matrix. Each element in the cofactor matrix, $C_{ij}$, is calculated using the formula $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$. The minor $M_{ij}$ is the determinant of the submatrix formed by deleting the i-th row and j-th column of the original matrix. Let's break this down for our matrix $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & 0 & 1\end{array}\right]$.

Calculating the Minors ($M_{ij}$):

• $M_{11}$: Delete row 1, col 1. $\det\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] = (1 \times 1) - (1 \times 0) = 1 - 0 = 1$
• $M_{12}$: Delete row 1, col 2. $\det\left[\begin{array}{ll}0 & 1 \\ 2 & 1\end{array}\right] = (0 \times 1) - (1 \times 2) = 0 - 2 = -2$
• $M_{13}$: Delete row 1, col 3. $\det\left[\begin{array}{ll}0 & 1 \\ 2 & 0\end{array}\right] = (0 \times 0) - (1 \times 2) = 0 - 2 = -2$
• $M_{21}$: Delete row 2, col 1. $\det\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] = (1 \times 1) - (2 \times 0) = 1 - 0 = 1$
• $M_{22}$: Delete row 2, col 2. $\det\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] = (1 \times 1) - (2 \times 2) = 1 - 4 = -3$
• $M_{23}$: Delete row 2, col 3. $\det\left[\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right] = (1 \times 0) - (1 \times 2) = 0 - 2 = -2$
• $M_{31}$: Delete row 3, col 1. $\det\left[\begin{array}{ll}1 & 2 \\ 1 & 1\end{array}\right] = (1 \times 1) - (2 \times 1) = 1 - 2 = -1$
• $M_{32}$: Delete row 3, col 2. $\det\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] = (1 \times 1) - (2 \times 0) = 1 - 0 = 1$
• $M_{33}$: Delete row 3, col 3. $\det\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] = (1 \times 1) - (1 \times 0) = 1 - 0 = 1$

Calculating the Cofactors ($C_{ij}$):

Now we apply the formula $C_{ij} = (-1)^{i+j} M_{ij}$:

• $C_{11} = (-1)^{1+1} M_{11} = (+1)(1) = 1$
• $C_{12} = (-1)^{1+2} M_{12} = (-1)(-2) = 2$
• $C_{13} = (-1)^{1+3} M_{13} = (+1)(-2) = -2$
• $C_{21} = (-1)^{2+1} M_{21} = (-1)(1) = -1$
• $C_{22} = (-1)^{2+2} M_{22} = (+1)(-3) = -3$
• $C_{23} = (-1)^{2+3} M_{23} = (-1)(-2) = 2$
• $C_{31} = (-1)^{3+1} M_{31} = (+1)(-1) = -1$
• $C_{32} = (-1)^{3+2} M_{32} = (-1)(1) = -1$
• $C_{33} = (-1)^{3+3} M_{33} = (+1)(1) = 1$

So, the cofactor matrix $C$ is:

$C = \left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & -3 & 2 \\ -1 & -1 & 1\end{array}\right]$

Transposing the Cofactor Matrix:

Finally, to get the adjoint matrix, we transpose the cofactor matrix $C$. Transposing means swapping rows with columns.

Adjoint$(A) = C^T = \left[\begin{array}{ccc}1 & -1 & -1 \\ 2 & -3 & -1 \\ -2 & 2 & 1\end{array}\right]$

Phew! That was a lot of steps, but understanding how to compute the adjoint is super important, especially when we're talking about finding the inverse of a matrix. It’s all about systematically calculating minors and cofactors and then taking the transpose. Keep practicing these steps, guys; they get easier with repetition!

(c) $\operatorname{det}(A)$: The Determinant of Matrix A

Next up, we need to find the determinant of matrix A, denoted as $\det(A)$ or $|A|$. The determinant is a scalar value that can be calculated from the elements of a square matrix. It tells us some really important properties about the matrix, like whether it's invertible. For a 3x3 matrix, we can calculate the determinant using a few methods. A common one is the cofactor expansion along any row or column. Let's use the first row for our matrix $A=\left[\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & 0 & 1\end{array}\right]$. The formula for cofactor expansion along the first row is:

$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

We already calculated the cofactors in the previous section! Let's plug in the values:

• $a_{11} = 1$, $C_{11} = 1$
• $a_{12} = 1$, $C_{12} = 2$
• $a_{13} = 2$, $C_{13} = -2$

So, $\det(A) = (1)(1) + (1)(2) + (2)(-2)$

$\det(A) = 1 + 2 - 4$

$\det(A) = -1$

Alternatively, we can use the Sarrus' rule for a 3x3 matrix. This involves repeating the first two columns to the right of the matrix and then summing the products of the diagonals:

$A = \left[\begin{array}{lll|ll}1 & 1 & 2 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 2 & 0 & 1 & 2 & 0\end{array}\right]$

• Sum of products of forward diagonals: $(1 \times 1 imes 1) + (1 imes 1 imes 2) + (2 imes 0 imes 0) = 1 + 2 + 0 = 3$

• Sum of products of backward diagonals: $(2 imes 1 imes 2) + (1 imes 1 imes 0) + (1 imes 0 imes 1) = 4 + 0 + 0 = 4$

• Determinant = (Sum of forward diagonals) - (Sum of backward diagonals) $\det(A) = 3 - 4 = -1$

Both methods give us the same result: $\mathbf{-1}$. Since the determinant is non-zero ($\det(A) \neq 0$), this matrix is invertible, which means we can proceed to find its inverse. If the determinant were zero, the matrix would be singular, and no inverse would exist. Keep this in mind, guys; the determinant is a really powerful indicator!

(d) $A^{-1}$: The Inverse of Matrix A

Finally, the moment we've all been waiting for: finding the inverse of matrix A, denoted as $A^{-1}$. The inverse of a matrix $A$ is a matrix such that when multiplied by $A$, it results in the identity matrix ($I$). The formula for finding the inverse of a square matrix $A$ is:

$A^{-1} = \frac{1}{\det(A)} \text{Adjoint}(A)$

We've already done all the heavy lifting! We found that $\det(A) = -1$ and Adjoint$(A) = \left[\begin{array}{ccc}1 & -1 & -1 \\ 2 & -3 & -1 \\ -2 & 2 & 1\end{array}\right]$. Now, we just need to plug these values into the formula.

$A^{-1} = \frac{1}{-1} \left[\begin{array}{ccc}1 & -1 & -1 \\ 2 & -3 & -1 \\ -2 & 2 & 1\end{array}\right]$

Multiplying each element of the adjoint matrix by $\frac{1}{-1}$ (which is just -1) gives us:

$A^{-1} = -1 \times \left[\begin{array}{ccc}1 & -1 & -1 \\ 2 & -3 & -1 \\ -2 & 2 & 1\end{array}\right]$

$A^{-1} = \left[\begin{array}{ccc}-1 & 1 & 1 \\ -2 & 3 & 1 \\ 2 & -2 & -1\end{array}\right]$

And there you have it – the inverse of matrix A! To double-check our work, we can multiply $A$ by $A^{-1}$ and see if we get the identity matrix $I = \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$.

$A \times A^{-1} = \left[\begin{array}{lll}1 & 1 & 2 \\ 0 & 1 & 1 \\ 2 & 0 & 1\end{array}\right] \times \left[\begin{array}{ccc}-1 & 1 & 1 \\ -2 & 3 & 1 \\ 2 & -2 & -1\end{array}\right]$

Let's compute the element in the first row, first column: $(1 \times -1) + (1 imes -2) + (2 imes 2) = -1 - 2 + 4 = 1$

First row, second column: $(1 imes 1) + (1 imes 3) + (2 imes -2) = 1 + 3 - 4 = 0$

First row, third column: $(1 imes 1) + (1 imes 1) + (2 imes -1) = 1 + 1 - 2 = 0$

...and so on. If you complete the entire multiplication, you will indeed get the identity matrix. This confirms that our calculated inverse is correct!

So, to recap, we've successfully found $A_{11}$, the adjoint of A, the determinant of A, and finally, the inverse of A. These calculations, while sometimes tedious, are the bedrock of many advanced mathematical and scientific concepts. Keep practicing these matrix operations, guys, and you'll master them in no time! Feel free to try these steps with different matrices to solidify your understanding. Happy calculating!