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If A=[23 41]A=\left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right], then adj(3A2+12A)\text{adj}(3A^{2}+12 A) is equal to

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Q. If A=[23 41]A=\left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right], then adj(3A2+12A)\text{adj}(3A^{2}+12 A) is equal to
  1. Find A2A^2: Given matrix AA:
    A=[23 41]A = \left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right]

    First, we need to find A2A^2, which is the matrix AA multiplied by itself.
    A2=A×AA^2 = A \times A
    A2=[23 41]×[23 41]A^2 = \left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right] \times \left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right]

    To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of the second matrix.
    A2=[(2×2+(3)×(4)),(2×(3)+(3)×1) (4)×2+1×(4),(4)×(3)+1×1]A^2 = \left[\begin{array}{cc}(2\times2 + (-3)\times(-4)), (2\times(-3) + (-3)\times1)\ (-4)\times2 + 1\times(-4), (-4)\times(-3) + 1\times1\end{array}\right]
    A2=[4+12,63 84,12+1]A^2 = \left[\begin{array}{cc}4 + 12, -6 - 3\ -8 - 4, 12 + 1\end{array}\right]
    A2=[16,9 12,13]A^2 = \left[\begin{array}{cc}16, -9\ -12, 13\end{array}\right]
  2. Calculate 3A23A^2: Now, we need to calculate 3A23A^2.
    3A^2 = 3 \times \begin{bmatrix}16 & -9\-12 & 13\end{bmatrix}
    3A^2 = \begin{bmatrix}3\times16 & 3\times(-9)\3\times(-12) & 3\times13\end{bmatrix}
    3A^2 = \begin{bmatrix}48 & -27\-36 & 39\end{bmatrix}
  3. Calculate 12A12A: Next, we calculate 12A12A.
    12A=12×[23 41]12A = 12 \times \begin{bmatrix} 2 & -3 \ -4 & 1 \end{bmatrix}
    12A=[12×212×(3) 12×(4)12×1]12A = \begin{bmatrix} 12\times2 & 12\times(-3) \ 12\times(-4) & 12\times1 \end{bmatrix}
    12A=[2436 4812]12A = \begin{bmatrix} 24 & -36 \ -48 & 12 \end{bmatrix}
  4. Add 3A23A^2 and 12A12A: Now, we add 3A23A^2 and 12A12A to get 3A2+12A3A^2 + 12A. \newline3A2+12A=[[48,27],[36,39]]+[[24,36],[48,12]]3A^2 + 12A = [[48, -27], [-36, 39]] + [[24, -36], [-48, 12]]\newline3A2+12A=[[48+24,2736],[3648,39+12]]3A^2 + 12A = [[48 + 24, -27 - 36], [-36 - 48, 39 + 12]]\newline3A2+12A=[[72,63],[84,51]]3A^2 + 12A = [[72, -63], [-84, 51]]
  5. Find adjugate of 3A2+12A3A^2 + 12A: The next step is to find the adjugate of the matrix 3A2+12A3A^2 + 12A. The adjugate of a 2×22\times2 matrix is found by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements.\newlineadj(3A2+12A)=[5163 8472]\text{adj}(3A^2 + 12A) = \begin{bmatrix}51 & 63 \ 84 & 72\end{bmatrix}

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