Solveeit Logo
Login

Question

Question: If A = $\begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix}$ and det A = 190 then Ad...

If A = [k522−k552−k]\begin{bmatrix} k & 5 & 2 \\ 2 & -k & 5 \\ 5 & 2 & -k \end{bmatrix} and det A = 190 then Adj A =

Answer

[−1193131−19−111919−19]\begin{bmatrix} -1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19 \end{bmatrix}

Explanation

Solution

Here's how to find the adjugate (adjoint) of matrix A:

  1. Find the value of k:

    • Compute the determinant of A in terms of k: det⁡(A)=k∣−k52−k∣−5∣255−k∣+2∣2−k52∣.\det (A)= k\begin{vmatrix} -k & 5 \\ 2 & -k \end{vmatrix} - 5\begin{vmatrix} 2 & 5 \\ 5 & -k \end{vmatrix} + 2\begin{vmatrix} 2 & -k \\ 5 & 2 \end{vmatrix}.

    • Evaluate the 2x2 determinants:

      • M11=∣−k52−k∣=(−k)(−k)−5⋅2=k2−10M_{11}=\begin{vmatrix} -k & 5 \\ 2 & -k \end{vmatrix}= (-k)(-k)-5\cdot2=k^2-10 .
      • M12=∣255−k∣=2(−k)−5⋅5=−2k−25M_{12}=\begin{vmatrix} 2 & 5 \\ 5 & -k \end{vmatrix}=2(-k)-5\cdot5=-2k-25 .
      • M13=∣2−k52∣=2⋅2−(−k)⋅5=4+5kM_{13}=\begin{vmatrix} 2 & -k \\ 5 & 2 \end{vmatrix}=2\cdot2-(-k)\cdot5=4+5k .

    • Substitute back into the determinant: det⁡(A)=k(k2−10)−5(−2k−25)+2(4+5k).\det (A)= k(k^2-10)-5(-2k-25)+2(4+5k).

      Simplify: det⁡(A)=k3−10k+10k+125+8+10k=k3+10k+133.\det (A) = k^3-10k+10k+125+8+10k = k^3+10k+133.

    • Since det⁡(A)=190\det (A)=190, we have: k3+10k+133=190  ⟹  k3+10k−57=0.k^3+10k+133=190 \implies k^3+10k-57=0.

      By trying k=3k=3, we find that 33+10⋅3−57=27+30−57=03^3+10\cdot3-57=27+30-57=0. Thus, k=3k=3.

  2. Write the matrix A with k=3:

    A=[3522−3552−3].A=\begin{bmatrix} 3 & 5 & 2 \\ 2 & -3 & 5 \\ 5 & 2 & -3 \end{bmatrix}.

  3. Find Adj A:

    The adjugate of A is the transpose of the cofactor matrix. Compute the cofactors Cij=(−1)i+jMijC_{ij} = (-1)^{i+j}M_{ij}:

    • First Row Cofactors:

      • C11=det⁡[−352−3]=(−3)(−3)−5⋅2=9−10=−1C_{11}=\det\begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix} = (-3)(-3)-5\cdot2 = 9-10=-1 .
      • C12=−det⁡[255−3]=−(2(−3)−5⋅5)=−(−6−25)=31C_{12}=-\det\begin{bmatrix} 2 & 5 \\ 5 & -3 \end{bmatrix} = -\Big(2(-3)-5\cdot5\Big) = -(-6-25)=31 .
      • C13=det⁡[2−352]=2⋅2−(−3)⋅5=4+15=19C_{13}=\det\begin{bmatrix} 2 & -3 \\ 5 & 2 \end{bmatrix} =2\cdot2-(-3)\cdot5=4+15=19 .

    • Second Row Cofactors:

      • C21=−det⁡[522−3]=−(5(−3)−2⋅2)=−(−15−4)=19C_{21}=-\det\begin{bmatrix} 5 & 2 \\ 2 & -3 \end{bmatrix} = -\Big(5(-3)-2\cdot2\Big) = -(-15-4)=19 .
      • C22=det⁡[325−3]=3(−3)−2⋅5=−9−10=−19C_{22}=\det\begin{bmatrix} 3 & 2 \\ 5 & -3 \end{bmatrix} =3(-3)-2\cdot5=-9-10=-19 .
      • C23=−det⁡[3552]=−(3⋅2−5⋅5)=−(6−25)=19C_{23}=-\det\begin{bmatrix} 3 & 5 \\ 5 & 2 \end{bmatrix} = -\Big(3\cdot2-5\cdot5\Big) = -(6-25)=19 .

    • Third Row Cofactors:

      • C31=det⁡[52−35]=5⋅5−2(−3)=25+6=31C_{31}=\det\begin{bmatrix} 5 & 2 \\ -3 & 5 \end{bmatrix} =5\cdot5-2(-3)=25+6=31 .
      • C32=−det⁡[3225]=−(3⋅5−2⋅2)=−(15−4)=−11C_{32}=-\det\begin{bmatrix} 3 & 2 \\ 2 & 5 \end{bmatrix} = -\Big(3\cdot5-2\cdot2\Big) = -(15-4)=-11 .
      • C33=det⁡[352−3]=3(−3)−5⋅2=−9−10=−19C_{33}=\det\begin{bmatrix} 3 & 5 \\ 2 & -3 \end{bmatrix} = 3(-3)-5\cdot2=-9-10=-19 .

    So the cofactor matrix is:

    [−1311919−191931−11−19].\begin{bmatrix} -1 & 31 & 19 \\ 19 & -19 & 19 \\ 31 & -11 & -19 \end{bmatrix}.

    Taking the transpose, we get:

    Adj⁡A=[−1193131−19−111919−19].\operatorname{Adj} A=\begin{bmatrix} -1 & 19 & 31 \\ 31 & -19 & -11 \\ 19 & 19 & -19 \end{bmatrix}.