If `[(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))]=[(1,2,3),(2,3,4),(3,4,5)][(-1,-2),(-2,0),(0,-4)][(-4,-5,-6),(0,0,1)]`, then the value of `a_(22)` is -

Options

A. 40
B. `-40`
C. `-20`
D. 20

Answer

To solve the problem, we need to find the value of \( a_{22} \) from the equation:

\[ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix} \begin{pmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{pmatrix} \begin{pmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 1: Multiply the first two matrices

Let \( A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix} \) and \( B = \begin{pmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{pmatrix} \).

To find the product \( AB \):

\[ AB = \begin{pmatrix} 1 \cdot (-1) + 2 \cdot (-2) + 3 \cdot 0 & 1 \cdot (-2) + 2 \cdot 0 + 3 \cdot (-4) \\ 2 \cdot (-1) + 3 \cdot (-2) + 4 \cdot 0 & 2 \cdot (-2) + 3 \cdot 0 + 4 \cdot (-4) \\ 3 \cdot (-1) + 4 \cdot (-2) + 5 \cdot 0 & 3 \cdot (-2) + 4 \cdot 0 + 5 \cdot (-4) \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 1 \cdot (-1) + 2 \cdot (-2) + 3 \cdot 0 = -1 - 4 + 0 = -5 \)
- First row, second column: \( 1 \cdot (-2) + 2 \cdot 0 + 3 \cdot (-4) = -2 + 0 - 12 = -14 \)
- Second row, first column: \( 2 \cdot (-1) + 3 \cdot (-2) + 4 \cdot 0 = -2 - 6 + 0 = -8 \)
- Second row, second column: \( 2 \cdot (-2) + 3 \cdot 0 + 4 \cdot (-4) = -4 + 0 - 16 = -20 \)
- Third row, first column: \( 3 \cdot (-1) + 4 \cdot (-2) + 5 \cdot 0 = -3 - 8 + 0 = -11 \)
- Third row, second column: \( 3 \cdot (-2) + 4 \cdot 0 + 5 \cdot (-4) = -6 + 0 - 20 = -26 \)

Thus, we have:

\[ AB = \begin{pmatrix} -5 & -14 \\ -8 & -20 \\ -11 & -26 \end{pmatrix} \]

### Step 2: Multiply the result with the third matrix

Let \( C = \begin{pmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{pmatrix} \).

Now, we need to calculate \( (AB)C \):

\[ (AB)C = \begin{pmatrix} -5 & -14 \\ -8 & -20 \\ -11 & -26 \end{pmatrix} \begin{pmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( -5 \cdot (-4) + (-14) \cdot 0 = 20 + 0 = 20 \)
- First row, second column: \( -5 \cdot (-5) + (-14) \cdot 0 = 25 + 0 = 25 \)
- First row, third column: \( -5 \cdot (-6) + (-14) \cdot 1 = 30 - 14 = 16 \)
- Second row, first column: \( -8 \cdot (-4) + (-20) \cdot 0 = 32 + 0 = 32 \)
- Second row, second column: \( -8 \cdot (-5) + (-20) \cdot 0 = 40 + 0 = 40 \)
- Second row, third column: \( -8 \cdot (-6) + (-20) \cdot 1 = 48 - 20 = 28 \)
- Third row, first column: \( -11 \cdot (-4) + (-26) \cdot 0 = 44 + 0 = 44 \)
- Third row, second column: \( -11 \cdot (-5) + (-26) \cdot 0 = 55 + 0 = 55 \)
- Third row, third column: \( -11 \cdot (-6) + (-26) \cdot 1 = 66 - 26 = 40 \)

Thus, we have:

\[ (AB)C = \begin{pmatrix} 20 & 25 & 16 \\ 32 & 40 & 28 \\ 44 & 55 & 40 \end{pmatrix} \]

### Step 3: Identify \( a_{22} \)

From the resulting matrix, we can see that:

\[ a_{22} = 40 \]

### Final Answer:

The value of \( a_{22} \) is \( 40 \).

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