Add Question 7
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@ -212,21 +212,23 @@ $, find $ adj. A$
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Adjoint of a matrix is the transpose of the cofactor matrix of the original matrix
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\[
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A_{11} =
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Cofactor\;of\;A_{11} =
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\begin{vmatrix}
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4 & 5 \\
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-6 & -7
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\end{vmatrix}
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= 2
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\;\;
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A_{12} =
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\]
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\[
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Cofactor\;of\;A_{12} =
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\begin{vmatrix}
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3 & 5 \\
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0 & -7
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\end{vmatrix}
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= -21
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\;\;
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A_{13} =
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\]
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\[
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Cofactor\;of\;A_{13} =
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\begin{vmatrix}
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3 & 4 \\
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0 & -6
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@ -234,21 +236,23 @@ Adjoint of a matrix is the transpose of the cofactor matrix of the original matr
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= -18
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\]
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\[
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A_{21} =
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Cofactor\;of\;A_{21} =
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\begin{vmatrix}
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0 & -1 \\
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-6 & -7
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\end{vmatrix}
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= -6
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\;\;
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A_{22} =
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\]
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\[
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Cofactor\;of\;A_{22} =
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\begin{vmatrix}
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1 & -1 \\
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0 & -7
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\end{vmatrix}
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= -7
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\;\;
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A_{23} =
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\]
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\[
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Cofactor\;of\;A_{23} =
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\begin{vmatrix}
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1 & -1 \\
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0 & -6
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@ -256,21 +260,23 @@ Adjoint of a matrix is the transpose of the cofactor matrix of the original matr
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= -6
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\]
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\[
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A_{31} =
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Cofactor\;of\;A_{31} =
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\begin{vmatrix}
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0 & -1 \\
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4 & 5
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\end{vmatrix}
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= 4
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\;\;
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A_{32} =
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\]
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\[
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Cofactor\;of\;A_{32} =
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\begin{vmatrix}
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1 & -1 \\
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3 & 5
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\end{vmatrix}
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= 8
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\;\;
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A_{33} =
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\]
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\[
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Cofactor\;of\;A_{33} =
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\begin{vmatrix}
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1 & 0 \\
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3 & 4
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@ -297,4 +303,77 @@ Adjoint matrix is the transpose of Cofactor Matrix.
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\end{bmatrix}
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\]
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\section{Question 7}
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\( A =
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\begin{bmatrix}
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0 & 0 & 1 \\
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0 & 1 & 0 \\
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1 & 0 & 0
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\end{bmatrix}
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\), show that $A^{-1} = A$.
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We know that
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\[
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A^{-1} = \frac{adj.(A)}{|A|}
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\]
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\begin{align*}
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Cofactor\;of\;A_{11} &= 0 \\
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Cofactor\;of\;A_{12} &= 0 \\
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Cofactor\;of\;A_{13} &= -1 \\
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Cofactor\;of\;A_{21} &= 0 \\
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Cofactor\;of\;A_{22} &= -1 \\
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Cofactor\;of\;A_{23} &= 0 \\
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Cofactor\;of\;A_{31} &= -1 \\
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Cofactor\;of\;A_{32} &= 0 \\
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Cofactor\;of\;A_{33} &= 0
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\end{align*}
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\[
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Cofactor\;Matrix =
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\begin{bmatrix}
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0 & 0 & -1 \\
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0 & -1 & 0 \\
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-1 & 0 & 0
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\end{bmatrix}
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\]
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\[
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adj.A =
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\begin{bmatrix}
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0 & 0 & -1 \\
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0 & -1 & 0 \\
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-1 & 0 & 0
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\end{bmatrix}
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\]
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\[
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|A| =
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\begin{vmatrix}
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0 & 0 & 1 \\
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0 & 1 & 0 \\
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1 & 0 & 0
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\end{vmatrix}
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= 1 \times
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\begin{vmatrix}
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0 & 1 \\
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1 & 0
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\end{vmatrix}
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= 1 \times -1 = -1
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\]
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\[
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A^{-1} = \frac{adj.A}{|A|}
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\]
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\[
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A^{-1} =
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\begin{bmatrix}
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0 & 0 & 1 \\
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0 & 1 & 0 \\
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1 & 0 & 0
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\end{bmatrix}
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\]
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\[
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\therefore A^{-1} = A
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\]
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\end{document}
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