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Math  /  Algebra

Question3) Given A=[3−5−17]A=\left[\begin{array}{cc}3 & -5 \\ -1 & 7\end{array}\right] and B=[1483]B=\left[\begin{array}{ll}1 & 4 \\ 8 & 3\end{array}\right] find the following a) A+BA+B b) ABA B c) B⊤A⊤B^{\top} A^{\top} d) (AB)⊤(A B)^{\top} e) 2(A−B)2(A-B)

Studdy Solution

STEP 1

1. Matrix addition is performed element-wise.
2. Matrix multiplication follows the rule that the element in the i-th row and j-th column of the product is the dot product of the i-th row of the first matrix and the j-th column of the second matrix.
3. The transpose of a matrix is obtained by swapping its rows with its columns.
4. Scalar multiplication of a matrix involves multiplying each element of the matrix by the scalar.

STEP 2

1. Compute the sum A+BA + B.
2. Compute the product ABA B.
3. Compute the transposes of matrices BB and AA, then find the product B⊤A⊤B^{\top} A^{\top}.
4. Compute the transpose of the product (AB)(A B).
5. Compute the scalar multiplication 2(A−B)2(A - B).

STEP 3

Add matrices AA and BB element-wise.
A+B=[3−5−17]+[1483]=[3+1−5+4−1+87+3]A + B = \left[\begin{array}{cc} 3 & -5 \\ -1 & 7 \end{array}\right] + \left[\begin{array}{cc} 1 & 4 \\ 8 & 3 \end{array}\right] = \left[\begin{array}{cc} 3+1 & -5+4 \\ -1+8 & 7+3 \end{array}\right]
A+B=[4−1710]A + B = \left[\begin{array}{cc} 4 & -1 \\ 7 & 10 \end{array}\right]

STEP 4

Compute the product of matrices AA and BB.
AB=[3−5−17][1483]A B = \left[\begin{array}{cc} 3 & -5 \\ -1 & 7 \end{array}\right] \left[\begin{array}{cc} 1 & 4 \\ 8 & 3 \end{array}\right]
Calculate each element of the resulting matrix:
(AB)11=3⋅1+(−5)⋅8=3−40=−37(AB)12=3⋅4+(−5)⋅3=12−15=−3(AB)21=−1⋅1+7⋅8=−1+56=55(AB)22=−1⋅4+7⋅3=−4+21=17\begin{array}{lll} (AB)_{11} & = 3 \cdot 1 + (-5) \cdot 8 & = 3 - 40 = -37 \\ (AB)_{12} & = 3 \cdot 4 + (-5) \cdot 3 & = 12 - 15 = -3 \\ (AB)_{21} & = -1 \cdot 1 + 7 \cdot 8 & = -1 + 56 = 55 \\ (AB)_{22} & = -1 \cdot 4 + 7 \cdot 3 & = -4 + 21 = 17 \\ \end{array}
AB=[−37−35517]A B = \left[\begin{array}{cc} -37 & -3 \\ 55 & 17 \end{array}\right]

STEP 5

Compute the transposes of matrices BB and AA, then find the product B⊤A⊤B^{\top} A^{\top}.
B⊤=[1843],A⊤=[3−1−57]B^{\top} = \left[\begin{array}{cc} 1 & 8 \\ 4 & 3 \end{array}\right], \quad A^{\top} = \left[\begin{array}{cc} 3 & -1 \\ -5 & 7 \end{array}\right]
B⊤A⊤=[1843][3−1−57]B^{\top} A^{\top} = \left[\begin{array}{cc} 1 & 8 \\ 4 & 3 \end{array}\right] \left[\begin{array}{cc} 3 & -1 \\ -5 & 7 \end{array}\right]
Calculate each element of the resulting matrix:
(B⊤A⊤)11=1⋅3+8⋅−5=3−40=−37(B⊤A⊤)12=1⋅−1+8⋅7=−1+56=55(B⊤A⊤)21=4⋅3+3⋅−5=12−15=−3(B⊤A⊤)22=4⋅−1+3⋅7=−4+21=17\begin{array}{lll} (B^{\top}A^{\top})_{11} & = 1 \cdot 3 + 8 \cdot -5 & = 3 - 40 = -37 \\ (B^{\top}A^{\top})_{12} & = 1 \cdot -1 + 8 \cdot 7 & = -1 + 56 = 55 \\ (B^{\top}A^{\top})_{21} & = 4 \cdot 3 + 3 \cdot -5 & = 12 - 15 = -3 \\ (B^{\top}A^{\top})_{22} & = 4 \cdot -1 + 3 \cdot 7 & = -4 + 21 = 17 \\ \end{array}
B⊤A⊤=[−3755−317]B^{\top} A^{\top} = \left[\begin{array}{cc} -37 & 55 \\ -3 & 17 \end{array}\right]

STEP 6

Compute the transpose of the product (AB)(A B).
(AB)⊤=[−37−35517]⊤=[−3755−317](AB)^{\top} = \left[\begin{array}{cc} -37 & -3 \\ 55 & 17 \end{array}\right]^{\top} = \left[\begin{array}{cc} -37 & 55 \\ -3 & 17 \end{array}\right]

STEP 7

Compute the scalar multiplication 2(A−B)2(A - B).
A−B=[3−5−17]−[1483]=[3−1−5−4−1−87−3]A - B = \left[\begin{array}{cc} 3 & -5 \\ -1 & 7 \end{array}\right] - \left[\begin{array}{cc} 1 & 4 \\ 8 & 3 \end{array}\right] = \left[\begin{array}{cc} 3-1 & -5-4 \\ -1-8 & 7-3 \end{array}\right]
A−B=[2−9−94]A - B = \left[\begin{array}{cc} 2 & -9 \\ -9 & 4 \end{array}\right]
Now multiply by 2:
2(A−B)=2[2−9−94]=[4−18−188]2(A - B) = 2 \left[\begin{array}{cc} 2 & -9 \\ -9 & 4 \end{array}\right] = \left[\begin{array}{cc} 4 & -18 \\ -18 & 8 \end{array}\right]
Solutions: (a) A+B=[4−1710] A + B = \left[\begin{array}{cc} 4 & -1 \\ 7 & 10 \end{array}\right] (b) AB=[−37−35517] A B = \left[\begin{array}{cc} -37 & -3 \\ 55 & 17 \end{array}\right] (c) B⊤A⊤=[−3755−317] B^{\top} A^{\top} = \left[\begin{array}{cc} -37 & 55 \\ -3 & 17 \end{array}\right] (d) (AB)⊤=[−3755−317] (A B)^{\top} = \left[\begin{array}{cc} -37 & 55 \\ -3 & 17 \end{array}\right] (e) 2(A−B)=[4−18−188] 2(A - B) = \left[\begin{array}{cc} 4 & -18 \\ -18 & 8 \end{array}\right]

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