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Question: Let A be any 3x3 invertible matrix. Then which one of the following is not always true? \(\begin{...

Let A be any 3x3 invertible matrix. Then which one of the following is not always true?
a) adj(adj(A)) = ∣A∣⋅(adj(A))−1 b) adj(adj(A)) = ∣A∣2⋅(adj(A))−1 c) adj(adj(A)) = ∣A∣⋅A d) adj(A) = ∣A∣⋅A−1 \begin{aligned} & \text{a) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }\left| A \right|\cdot {{\left( adj(A) \right)}^{-1}} \\\ & \text{b) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }{{\left| A \right|}^{2}}\cdot {{\left( adj(A) \right)}^{-1}} \\\ & \text{c) }adj\left( adj\left( A \right) \right)\text{ }=\text{ }\left| A \right|\cdot A \\\ & \text{d) }adj\left( A \right)\text{ }=\text{ }\left| A \right|\cdot {{A}^{-1}} \\\ \end{aligned}

Explanation

Solution

In this, to check that which option does not hold we will use the following two properties of Adjoint of the matrix. Adjoint of a matrix is a matrix formed by the cofactors of the matrix.

& \text{1) A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left( \text{adj}\left( \text{A} \right) \right)\text{A=}\left| \text{A} \right|\centerdot \text{I} \\\ & \text{2) }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ = }{{\left| \text{A} \right|}^{n-1}} \\\ \end{aligned}$$ **Complete step by step answer:** Given that the matrix A is 3x3 invertible matrix then, $$\left( \text{adj}\left( \text{A} \right) \right)\text{A=}\left| \text{A} \right|\centerdot \text{I}....\text{(1)}$$ Replacing matrix A by adj(A) in equation (1), we get $$\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I}....\text{(2)}$$ Since the matrix is 3x3 $$\Rightarrow \text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ = }{{\left| \text{A} \right|}^{2}}$$ Equation (2), becomes $$\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right)\text{=}{{\left| \text{A} \right|}^{2}}\cdot \text{I}....\text{(3)}$$ Since A is invertible matrix implies that adj(A) is a invertible matrix. Apply $${{\left( \text{adj}\left( \text{A} \right) \right)}^{\text{-1}}}$$on both sides of equation (3), we get\\\ $$\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{adj}\left( \text{A} \right){{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}\text{=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}$$ Since $$\text{adj}\left( \text{A} \right){{\left( \text{adj}\left( \text{A} \right) \right)}^{\text{-1}}}=\text{I}$$ $$\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{I=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}$$ $$\text{adj}\left( \text{adj}\left( \text{A} \right) \right)\text{=}{{\left| \text{A} \right|}^{2}}{{\left( \text{adj}\left( \text{A} \right) \right)}^{-1}}$$ This proves that option b is always true. From property (1) of adjoint of matrix we get $$\text{A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{I}....\text{(3)}$$ Replacing matrix A by adj(A) in equation (3), we get $$\text{adj}\left( \text{A} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I}....\text{(4)}$$ Multiplying equation (4) matrix A from left side, we get $$\text{A}\cdot \text{adj}\left( \text{A} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=A}\left\\{ \left| \text{adj}\left( \text{A} \right) \right|\cdot \text{I} \right\\}$$ $$\left( \text{A}\cdot \text{adj}\left( \text{A} \right) \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\left\\{ \text{A}\cdot \text{I} \right\\}$$ From property (1), we get $$\left( \left| \text{A} \right|\cdot \text{I} \right)\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\left\\{ \text{A}\cdot \text{I} \right\\}$$\\\ $$\left| \text{A} \right|\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}\left| \text{adj}\left( \text{A} \right) \right|\cdot \text{A}$$ For 3x3 matrix property (2) is $$\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ = }{{\left| \text{A} \right|}^{2}}$$ $$\left| \text{A} \right|\left( \text{adj}\left( \text{adj}\left( \text{A} \right) \right) \right)\text{=}{{\left| \text{A} \right|}^{2}}\cdot \text{A}$$ By cancelling |A| form both sides, we get $$\text{adj}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{A}$$\\\ The option c is always true. Since A is invertible implies $${{\text{A}}^{\text{-1}}}$$ exist. Applying $${{\text{A}}^{\text{-1}}}$$ on both sides of equation (1), we get $$\left( \text{adj}\left( \text{A} \right) \right)\text{A}\cdot {{\text{A}}^{-1}}\text{=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}$$ Since $$\text{A}\cdot {{\text{A}}^{-1}}\text{=I}$$ $$\Rightarrow \left( \text{adj}\left( \text{A} \right) \right)\text{I=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}$$ $$\Rightarrow \text{adj}\left( \text{A} \right)\text{=}\left| \text{A} \right|\cdot {{\text{A}}^{-1}}$$\\\ This proves that option d is always true. If the option b is always true implies that option a is always not true **So, the correct answer is “Option A”.** **Note:** proof of property (2) is given below for nxn matrix. Let A be any nxn matrix. By property (1), we have $$\text{A}\left( \text{adj}\left( \text{A} \right) \right)\text{=}\left| \text{A} \right|\cdot \text{I}$$ Taking determinants on both sides, we get $$\text{ }\\!\\!|\\!\\!\text{ A}\left( \text{adj}\left( \text{A} \right) \right)\text{ }\\!\\!|\\!\\!\text{ = }\\!\\!|\\!\\!\text{ }\left| \text{A} \right|\cdot \text{I }\\!\\!|\\!\\!\text{ }$$ Since det(AB) = det(A)det(B) $$\text{ }\\!\\!|\\!\\!\text{ A }\\!\\!|\\!\\!\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ = }\\!\\!|\\!\\!\text{ }\left| \text{A} \right|\cdot \text{I }\\!\\!|\\!\\!\text{ }$$ For nxn matrix, det(kA) = kndet(A), where k is constant number $$\text{ }\\!\\!|\\!\\!\text{ A }\\!\\!|\\!\\!\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ =}{{\left| \text{A} \right|}^{n}}\text{ }\\!\\!|\\!\\!\text{ I }\\!\\!|\\!\\!\text{ }$$ Since the determinant of the identity matrix is 1. $$\text{ }\\!\\!|\\!\\!\text{ A }\\!\\!|\\!\\!\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ =}{{\left| \text{A} \right|}^{n}}\cdot 1$$ $$\text{ }\\!\\!|\\!\\!\text{ A }\\!\\!|\\!\\!\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ =}{{\left| \text{A} \right|}^{n}}$$ By cancelling |A| on both sides, we get $$\text{ }\\!\\!|\\!\\!\text{ adj}\left( \text{A} \right)\text{ }\\!\\!|\\!\\!\text{ =}{{\left| \text{A} \right|}^{n-1}}$$