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Given:

[tex]\[ A=\left(\begin{array}{ccc}1 & -2 & 3 \\ -1 & 2 & 1\end{array}\right) \][/tex]
[tex]\[ B=\left(\begin{array}{ll}2 & 3 \\ 3 & 1 \\ 1 & 2\end{array}\right) \][/tex]

Find the determinant of the matrix [tex]\(AB - 3I\)[/tex], where [tex]\(I\)[/tex] is a [tex]\(2 \times 2\)[/tex] identity matrix.


Sagot :

Let's work through the problem step-by-step.

1. Matrix Multiplication [tex]\( AB \)[/tex]:
- First, we multiply matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].

[tex]\[ A = \begin{pmatrix} 1 & -2 & 3 \\ -1 & 2 & 1 \end{pmatrix} \][/tex]

[tex]\[ B = \begin{pmatrix} 2 & 3 \\ 3 & 1 \\ 1 & 2 \end{pmatrix} \][/tex]

To find [tex]\( AB \)[/tex], we compute each element of the resulting matrix by taking the dot product of the corresponding row of [tex]\( A \)[/tex] with the corresponding column of [tex]\( B \)[/tex].

[tex]\[ AB = \begin{pmatrix} (1 \cdot 2) + (-2 \cdot 3) + (3 \cdot 1) & (1 \cdot 3) + (-2 \cdot 1) + (3 \cdot 2) \\ (-1 \cdot 2) + (2 \cdot 3) + (1 \cdot 1) & (-1 \cdot 3) + (2 \cdot 1) + (1 \cdot 2) \end{pmatrix} \][/tex]

[tex]\[ AB = \begin{pmatrix} 2 - 6 + 3 & 3 - 2 + 6 \\ -2 + 6 + 1 & -3 + 2 + 2 \end{pmatrix} \][/tex]

[tex]\[ AB = \begin{pmatrix} -1 & 7 \\ 5 & 1 \end{pmatrix} \][/tex]

2. Matrix Subtraction [tex]\( AB - 3I \)[/tex]:
- Next, we calculate [tex]\( AB - 3I \)[/tex], where [tex]\( I \)[/tex] is the [tex]\( 2 \times 2 \)[/tex] identity matrix:

[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]

So, [tex]\( 3I \)[/tex] is:

[tex]\[ 3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \][/tex]

Then, subtract [tex]\( 3I \)[/tex] from [tex]\( AB \)[/tex]:

[tex]\[ AB - 3I = \begin{pmatrix} -1 & 7 \\ 5 & 1 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \][/tex]

[tex]\[ AB - 3I = \begin{pmatrix} -1 - 3 & 7 - 0 \\ 5 - 0 & 1 - 3 \end{pmatrix} \][/tex]

[tex]\[ AB - 3I = \begin{pmatrix} -4 & 7 \\ 5 & -2 \end{pmatrix} \][/tex]

3. Finding the Determinant:
- Finally, we find the determinant of the resultant matrix [tex]\( AB - 3I \)[/tex]:

[tex]\[ \text{Det}(AB - 3I) = \text{Det}\begin{pmatrix} -4 & 7 \\ 5 & -2 \end{pmatrix} \][/tex]

The determinant of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is given by [tex]\( ad - bc \)[/tex]. Applying this formula:

[tex]\[ \text{Det}\begin{pmatrix} -4 & 7 \\ 5 & -2 \end{pmatrix} = (-4)(-2) - (7)(5) \][/tex]

[tex]\[ = 8 - 35 \][/tex]

[tex]\[ = -27 \][/tex]

Therefore, the determinant of [tex]\( AB - 3I \)[/tex] is [tex]\(-27\)[/tex].