If A=[[2,-3],[-4,1]], then adj(3A^(2)+12 A) is equal to

Find A^2: Given matrix A: A = [[2, -3], [-4, 1]] First, we need to find A^2, which is the matrix A multiplied by itself. A^2 = A * A A^2 = [[2, -3], [-4, 1]] * [[2, -3], [-4, 1]] To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of the second matrix. A^2 = [[(2*2 + (-3)*(-4)), (2*(-3) + (-3)*1)], [(-4)*2 + 1*(-4), (-4)*(-3) + 1*1]] A^2 = [[4 + 12, -6 - 3], [-8 - 4, 12 + 1]] A^2 = [[16, -9], [-12, 13]]Calculate 3A^2: Now, we need to calculate 3A^2. 3A^2 = 3 * [[16, -9], [-12, 13]] 3A^2 = [[3*16, 3*(-9)], [3*(-12), 3*13]] 3A^2 = [[48, -27], [-36, 39]]Calculate 12A: Next, we calculate 12A. 12A = 12 * [[2, -3], [-4, 1]] 12A = [[12*2, 12*(-3)], [12*(-4), 12*1]] 12A = [[24, -36], [-48, 12]]Add 3A^2 and 12A: Now, we add 3A^2 and 12A to get 3A^2 + 12A. 3A^2 + 12A = [[48, -27], [-36, 39]] + [[24, -36], [-48, 12]] 3A^2 + 12A = [[48 + 24, -27 - 36], [-36 - 48, 39 + 12]] 3A^2 + 12A = [[72, -63], [-84, 51]]Find adjugate of 3A^2 + 12A: The next step is to find the adjugate of the matrix 3A^2 + 12A. The adjugate of a 2x2 matrix is found by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements. adj(3A^2 + 12A) = [[51, 63], [84, 72]]

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

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Algebra 2

Evaluate expression when a complex numbers and a variable term is given

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Q. If

A=\left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right]

, then

\text{adj}(3A^{2}+12 A)

is equal to

Find $A^2$ : Given matrix $A$ :
$A = \left[\begin{array}{cc}2 & -3 \ -4 & 1\end{array}\right]$

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

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