Find A² + 6A - 5I where I is unit matrix.

Understand the Problem

The question is asking to find the expression A² + 6A - 5I, where A is a given matrix and I is the identity matrix. The problem requires matrix multiplication and addition to arrive at the solution.

Answer

$$ \begin{pmatrix} 45 & 33 & 43 \\ 32 & 8 & 59 \\ 28 & 6 & 39 \end{pmatrix} $$

Steps to Solve

Calculate (A^2)
First, we need to find the square of matrix (A). This is done by multiplying matrix (A) by itself.
Given:
$$ A = \begin{pmatrix} 4 & 3 & 2 \ 2 & 1 & 5 \ 2 & 0 & 4 \end{pmatrix} $$
Multiply (A) by (A):

[ A^2 = A \cdot A = \begin{pmatrix} 4 & 3 & 2 \ 2 & 1 & 5 \ 2 & 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 3 & 2 \ 2 & 1 & 5 \ 2 & 0 & 4 \end{pmatrix} ] Calculating the result:

Row 1:

(4 \cdot 4 + 3 \cdot 2 + 2 \cdot 2 = 16 + 6 + 4 = 26)
(4 \cdot 3 + 3 \cdot 1 + 2 \cdot 0 = 12 + 3 + 0 = 15)
(4 \cdot 2 + 3 \cdot 5 + 2 \cdot 4 = 8 + 15 + 8 = 31)

Row 2:

(2 \cdot 4 + 1 \cdot 2 + 5 \cdot 2 = 8 + 2 + 10 = 20)
(2 \cdot 3 + 1 \cdot 1 + 5 \cdot 0 = 6 + 1 + 0 = 7)
(2 \cdot 2 + 1 \cdot 5 + 5 \cdot 4 = 4 + 5 + 20 = 29)

Row 3:

(2 \cdot 4 + 0 \cdot 2 + 4 \cdot 2 = 8 + 0 + 8 = 16)
(2 \cdot 3 + 0 \cdot 1 + 4 \cdot 0 = 6 + 0 + 0 = 6)
(2 \cdot 2 + 0 \cdot 5 + 4 \cdot 4 = 4 + 0 + 16 = 20)

Thus, we have:
$$ A^2 = \begin{pmatrix} 26 & 15 & 31 \ 20 & 7 & 29 \ 16 & 6 & 20 \end{pmatrix} $$

Calculate (6A)
Now, we multiply matrix (A) by 6:
$$ 6A = 6 \cdot \begin{pmatrix} 4 & 3 & 2 \ 2 & 1 & 5 \ 2 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 24 & 18 & 12 \ 12 & 6 & 30 \ 12 & 0 & 24 \end{pmatrix} $$
Calculate (-5I)
Next, we need to find (-5I), where (I) is the identity matrix. The (3 \times 3) identity matrix is:
$$ I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} $$
Thus,
$$ -5I = -5 \cdot \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -5 & 0 & 0 \ 0 & -5 & 0 \ 0 & 0 & -5 \end{pmatrix} $$
Combine the results
Now we can find the final expression:
$$ A^2 + 6A - 5I = \begin{pmatrix} 26 & 15 & 31 \ 20 & 7 & 29 \ 16 & 6 & 20 \end{pmatrix} + \begin{pmatrix} 24 & 18 & 12 \ 12 & 6 & 30 \ 12 & 0 & 24 \end{pmatrix} + \begin{pmatrix} -5 & 0 & 0 \ 0 & -5 & 0 \ 0 & 0 & -5 \end{pmatrix} $$
Calculating it step by step:

Each element's summation:

Row 1:

(26 + 24 - 5 = 45)
(15 + 18 + 0 = 33)
(31 + 12 + 0 = 43)

Row 2:

(20 + 12 + 0 = 32)
(7 + 6 - 5 = 8)
(29 + 30 + 0 = 59)

Row 3:

(16 + 12 + 0 = 28)
(6 + 0 - 0 = 6)
(20 + 24 - 5 = 39)

Thus, the final result is:
$$ A^2 + 6A - 5I = \begin{pmatrix} 45 & 33 & 43 \ 32 & 8 & 59 \ 28 & 6 & 39 \end{pmatrix} $$

$$ A^2 + 6A - 5I = \begin{pmatrix} 45 & 33 & 43 \ 32 & 8 & 59 \ 28 & 6 & 39 \end{pmatrix} $$

More Information

The matrices involved in this operation demonstrate how matrix arithmetic follows different rules than scalar arithmetic. This example is useful for understanding matrix multiplication, addition, and the effect of the identity matrix.

Tips

Confusing matrix addition with element-wise operations. Ensure each element is added correctly.
Failing to calculate the product (A^2) correctly. Double-check the calculations for each entry in the resulting matrix.

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