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If A=[(1,2,2),(2,1,2),(2,2,1)], then the...

If `A=[(1,2,2),(2,1,2),(2,2,1)],` then the value of `|A^4-18A^2-32A|` is

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1

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To solve the problem, we need to calculate the determinant of the expression \( |A^4 - 18A^2 - 32A| \) where \( A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \). ### Step 1: Calculate \( A^2 \) First, we will calculate \( A^2 \) by multiplying matrix \( A \) by itself. \[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \] Calculating each element of \( A^2 \): - First row, first column: \( 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 = 1 + 4 + 4 = 9 \) - First row, second column: \( 1 \cdot 2 + 2 \cdot 1 + 2 \cdot 2 = 2 + 2 + 4 = 8 \) - First row, third column: \( 1 \cdot 2 + 2 \cdot 2 + 2 \cdot 1 = 2 + 4 + 2 = 8 \) - Second row, first column: \( 2 \cdot 1 + 1 \cdot 2 + 2 \cdot 2 = 2 + 2 + 4 = 8 \) - Second row, second column: \( 2 \cdot 2 + 1 \cdot 1 + 2 \cdot 2 = 4 + 1 + 4 = 9 \) - Second row, third column: \( 2 \cdot 2 + 1 \cdot 2 + 2 \cdot 1 = 4 + 2 + 2 = 8 \) - Third row, first column: \( 2 \cdot 1 + 2 \cdot 2 + 1 \cdot 2 = 2 + 4 + 2 = 8 \) - Third row, second column: \( 2 \cdot 2 + 2 \cdot 1 + 1 \cdot 2 = 4 + 2 + 2 = 8 \) - Third row, third column: \( 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 1 = 4 + 4 + 1 = 9 \) Thus, \[ A^2 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} \] ### Step 2: Calculate \( A^4 \) Next, we calculate \( A^4 \) as \( A^2 \cdot A^2 \): \[ A^4 = A^2 \cdot A^2 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} \cdot \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} \] Calculating each element of \( A^4 \): - First row, first column: \( 9 \cdot 9 + 8 \cdot 8 + 8 \cdot 8 = 81 + 64 + 64 = 209 \) - First row, second column: \( 9 \cdot 8 + 8 \cdot 9 + 8 \cdot 8 = 72 + 72 + 64 = 208 \) - First row, third column: \( 9 \cdot 8 + 8 \cdot 8 + 8 \cdot 9 = 72 + 64 + 72 = 208 \) - Second row, first column: \( 8 \cdot 9 + 9 \cdot 8 + 8 \cdot 8 = 72 + 72 + 64 = 208 \) - Second row, second column: \( 8 \cdot 8 + 9 \cdot 9 + 8 \cdot 8 = 64 + 81 + 64 = 209 \) - Second row, third column: \( 8 \cdot 8 + 9 \cdot 8 + 8 \cdot 9 = 64 + 72 + 72 = 208 \) - Third row, first column: \( 8 \cdot 9 + 8 \cdot 8 + 9 \cdot 8 = 72 + 64 + 72 = 208 \) - Third row, second column: \( 8 \cdot 8 + 8 \cdot 9 + 9 \cdot 8 = 64 + 72 + 72 = 208 \) - Third row, third column: \( 8 \cdot 8 + 8 \cdot 8 + 9 \cdot 9 = 64 + 64 + 81 = 209 \) Thus, \[ A^4 = \begin{pmatrix} 209 & 208 & 208 \\ 208 & 209 & 208 \\ 208 & 208 & 209 \end{pmatrix} \] ### Step 3: Calculate \( 18A^2 \) Now we calculate \( 18A^2 \): \[ 18A^2 = 18 \cdot \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} = \begin{pmatrix} 162 & 144 & 144 \\ 144 & 162 & 144 \\ 144 & 144 & 162 \end{pmatrix} \] ### Step 4: Calculate \( 32A \) Next, we calculate \( 32A \): \[ 32A = 32 \cdot \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 32 & 64 & 64 \\ 64 & 32 & 64 \\ 64 & 64 & 32 \end{pmatrix} \] ### Step 5: Calculate \( A^4 - 18A^2 - 32A \) Now we can compute the expression \( A^4 - 18A^2 - 32A \): \[ A^4 - 18A^2 - 32A = \begin{pmatrix} 209 & 208 & 208 \\ 208 & 209 & 208 \\ 208 & 208 & 209 \end{pmatrix} - \begin{pmatrix} 162 & 144 & 144 \\ 144 & 162 & 144 \\ 144 & 144 & 162 \end{pmatrix} - \begin{pmatrix} 32 & 64 & 64 \\ 64 & 32 & 64 \\ 64 & 64 & 32 \end{pmatrix} \] Calculating each element: - First row, first column: \( 209 - 162 - 32 = 15 \) - First row, second column: \( 208 - 144 - 64 = 0 \) - First row, third column: \( 208 - 144 - 64 = 0 \) - Second row, first column: \( 208 - 144 - 64 = 0 \) - Second row, second column: \( 209 - 162 - 32 = 15 \) - Second row, third column: \( 208 - 144 - 64 = 0 \) - Third row, first column: \( 208 - 144 - 64 = 0 \) - Third row, second column: \( 208 - 144 - 64 = 0 \) - Third row, third column: \( 209 - 162 - 32 = 15 \) Thus, \[ A^4 - 18A^2 - 32A = \begin{pmatrix} 15 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 15 \end{pmatrix} \] ### Step 6: Calculate the Determinant Now we can calculate the determinant of the resulting matrix: \[ |A^4 - 18A^2 - 32A| = |15I| = 15^3 = 3375 \] ### Final Answer The value of \( |A^4 - 18A^2 - 32A| \) is \( 3375 \).

To solve the problem, we need to calculate the determinant of the expression \( |A^4 - 18A^2 - 32A| \) where \( A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \). ### Step 1: Calculate \( A^2 \) First, we will calculate \( A^2 \) by multiplying matrix \( A \) by itself. \[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} ...
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OBJECTIVE RD SHARMA ENGLISH-MATRICES-Chapter Test
  1. If A=[(1,2,2),(2,1,2),(2,2,1)], then the value of |A^4-18A^2-32A| is

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  2. If A is an invertible matrix and B is a matrix, then

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  3. What is the order of the product [x" "y" "z][{:(a,h,g),(h,b,f),(g,f,c)...

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  4. If {:A=[(a,0,0),(0,b,0),(0,0,c)]:}," then "A^(-1), is

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  5. The inverse of the matrix {:[(1,3),(3,10)]:} is equal to

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  6. If {:A=[(5,2),(3,1)]:}," then "A^(-1)=

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  7. If {:X=[(3,-4),(1,-1)]:}, the value of X^n is equal to

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  8. If {:A=[(5,2),(3,1)]:}," then "A^(-1)=

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  9. For the system of equations: x+2y+3z=1 2x+y+3z=2 5x+5y+9z=4

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  10. If {:A=[(3,1),(-1,2)]:}," then "A^(2)=

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  11. if A=[(4,x+2),(2x-3,x+1)] is symmetric, then x is equal to

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  12. If A+B={:[(1,0),(1,1)]:}andA-2B={:[(-1,1),(0,-1)]:}, then A is equal t...

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  13. {:[(-6,5),(-7,6)]^(-1)=:}

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  14. From the matrix equation AB=AC, we conclude B=C provided.

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  15. If I3 is the identily matrix of order 3, then (I3)^(-1)=

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  16. Let a ,b , c be real numbers. The following system of equations in x ,...

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  17. If A and B are two matrices such that A+B and AB are both defind, then

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  18. A and B are tow square matrices of same order and A' denotes the tran...

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  19. STATEMENT-1: The lines a(1)x+b(1)y+c(1)=0a(2)x+b(2)y+c(2)=0,a(3)x+b(3)...

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  20. The system of linear equations x+y+z=2,2x+y-z=3, 3x+2y+kz=4 has a uniq...

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  21. If A and B ar square matrices of order 3 such that |A|=-1|B|=3, then |...

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