Find the inverse of each of the following matrices-i 123231312ii 1251-1-123-1iii 2-11-12-11-12iv 20-1510013v 01-14-343-34vi 00-1345-2-4-7vii 1000 cos-sin- 0 sin-cos-

I #160;A=123231312Now,C11=3112=5, #160;C12= 2132= 1 #160;and #160;C13=2331= 7C21= 2312= 1, #160;C22=1332= 7 #160;and #160;C23= 1231=5C31=2331= # ...

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Byju's Answer

Standard XII

Mathematics

Associative Law of Binary Operation

Find the inve...

Question

Find the inverse of each of the following matrices.

(i) $[\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix}]$

(ii) $[\begin{matrix} 1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1 \end{matrix}]$

(iii) $[\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}]$

(iv) $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

(v) $[\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}]$

(vi) $[\begin{matrix} 0 & 0 & - 1 \\ 3 & 4 & 5 \\ - 2 & - 4 & - 7 \end{matrix}]$

(vii) $[\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

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Solution

$(i) A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 3 & 1 \\ 1 & 2 \end{matrix}| = 5, C_{12} = - |\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 2 & 3 \\ 3 & 1 \end{matrix}| = - 7 C_{21} = - |\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 1 & 3 \\ 3 & 2 \end{matrix}| = - 7 and C_{23} = - |\begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix}| = 5 C_{31} = |\begin{matrix} 2 & 3 \\ 3 & 1 \end{matrix}| = - 7, C_{32} = - |\begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix}| = 5 and C_{33} = |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = - 1 adj A = {[\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}] and |A| = - 18 ∴ A^{- 1} = - \frac{1}{18} [\begin{matrix} 5 & - 1 & - 7 \\ - 1 & - 7 & 5 \\ - 7 & 5 & - 1 \end{matrix}] (ii) B = [\begin{matrix} 1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1 \end{matrix}] Now, C_{11} = |\begin{matrix} - 1 & - 1 \\ 3 & - 1 \end{matrix}| = 4, C_{12} = - |\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 1 & - 1 \\ 2 & 3 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 2 & 5 \\ 3 & - 1 \end{matrix}| = 17, C_{22} = |\begin{matrix} 1 & 5 \\ 2 & - 1 \end{matrix}| = - 11 and C_{23} = - |\begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix}| = 1 C_{31} = |\begin{matrix} 2 & 5 \\ - 1 & - 1 \end{matrix}| = 3, C_{32} = - |\begin{matrix} 1 & 5 \\ 1 & - 1 \end{matrix}| = 6 and C_{33} = |\begin{matrix} 1 & 2 \\ 1 & - 1 \end{matrix}| = - 3 adj B = {[\begin{matrix} 4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3 \end{matrix}]}^{T} = [\begin{matrix} 4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3 \end{matrix}] and |B| = 27 ∴ B^{- 1} = \frac{1}{27} [\begin{matrix} 4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3 \end{matrix}] (iii) C = [\begin{matrix} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}| = 3, C_{12} = - |\begin{matrix} - 1 & - 1 \\ 1 & 2 \end{matrix}| = 1 and C_{13} = |\begin{matrix} - 1 & 2 \\ 1 & - 1 \end{matrix}| = - 1 C_{21} = - |\begin{matrix} - 1 & 1 \\ - 1 & 2 \end{matrix}| = 1, C_{22} = |\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}| = 3 and C_{23} = - |\begin{matrix} 2 & - 1 \\ 1 & - 1 \end{matrix}| = 1 C_{31} = |\begin{matrix} - 1 & 1 \\ 2 & - 1 \end{matrix}| = - 1, C_{32} = - |\begin{matrix} 2 & 1 \\ - 1 & - 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}| = 3 adj C = {[\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}]}^{T} = [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] and |C| = 4 ∴ C^{- 1} = \frac{1}{4} [\begin{matrix} 3 & 1 & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{matrix}] (iv) D = [\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}] Now, C_{11} = |\begin{matrix} 1 & 0 \\ 1 & 3 \end{matrix}| = 3, C_{12} = - |\begin{matrix} 5 & 0 \\ 0 & 3 \end{matrix}| = - 15 and C_{13} = |\begin{matrix} 5 & 1 \\ 0 & 1 \end{matrix}| = 5 C_{21} = - |\begin{matrix} 0 & - 1 \\ 1 & 3 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 2 & - 1 \\ 0 & 3 \end{matrix}| = 6 and C_{23} = - |\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}| = - 2 C_{31} = |\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 2 & - 1 \\ 5 & 0 \end{matrix}| = - 5 and C_{33} = |\begin{matrix} 2 & 0 \\ 5 & 1 \end{matrix}| = 2 adj D = {[\begin{matrix} 3 & - 15 & 5 \\ - 1 & 6 & - 2 \\ 1 & - 5 & 2 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] and |D| = 1 ∴ D^{- 1} = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}] (v) E = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] Now, C_{11} = |\begin{matrix} - 3 & 4 \\ - 3 & 4 \end{matrix}| = 0, C_{12} = - |\begin{matrix} 4 & 4 \\ 3 & 4 \end{matrix}| = - 4 and C_{13} = |\begin{matrix} 4 & - 3 \\ 3 & - 3 \end{matrix}| = - 3 C_{21} = - |\begin{matrix} 1 & - 1 \\ - 3 & 4 \end{matrix}| = - 1, C_{22} = |\begin{matrix} 0 & - 1 \\ 3 & 4 \end{matrix}| = 3 and C_{23} = - |\begin{matrix} 0 & 1 \\ 3 & - 3 \end{matrix}| = 3 C_{31} = |\begin{matrix} 1 & - 1 \\ - 3 & 4 \end{matrix}| = 1, C_{32} = - |\begin{matrix} 0 & - 1 \\ 4 & 4 \end{matrix}| = - 4 and C_{33} = |\begin{matrix} 0 & 1 \\ 4 & - 3 \end{matrix}| = - 4 adj E = {[\begin{matrix} 0 & - 4 & - 3 \\ - 1 & 3 & 3 \\ 1 & - 4 & - 4 \end{matrix}]}^{T} = [\begin{matrix} 0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4 \end{matrix}] and |E| = - 1 ∴ E^{- 1} = - 1 [\begin{matrix} 0 & - 1 & 1 \\ - 4 & 3 & - 4 \\ - 3 & 3 & - 4 \end{matrix}] = [\begin{matrix} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{matrix}] (vi) F = [\begin{matrix} 0 & 0 & - 1 \\ 3 & 4 & 5 \\ - 2 & - 4 & - 7 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 5 \\ - 4 & - 7 \end{matrix}| = - 8, C_{12} = - |\begin{matrix} 3 & 5 \\ - 2 & - 7 \end{matrix}| = 11 and C_{13} = |\begin{matrix} 3 & 4 \\ - 2 & - 4 \end{matrix}| = - 4 C_{21} = - |\begin{matrix} 0 & - 1 \\ - 4 & - 7 \end{matrix}| = 4, C_{22} = |\begin{matrix} 0 & - 1 \\ - 2 & - 7 \end{matrix}| = - 2 and C_{23} = - |\begin{matrix} 0 & 0 \\ - 2 & - 4 \end{matrix}| = 0 C_{31} = |\begin{matrix} 0 & - 1 \\ 4 & 5 \end{matrix}| = 4, C_{32} = - |\begin{matrix} 0 & - 1 \\ 3 & 5 \end{matrix}| = - 3 and C_{33} = |\begin{matrix} 0 & 0 \\ 3 & 4 \end{matrix}| = 0 adj F = {[\begin{matrix} - 8 & 11 & - 4 \\ 4 & - 2 & 0 \\ 4 & - 3 & 0 \end{matrix}]}^{T} = [\begin{matrix} - 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0 \end{matrix}] and |F| = 4 ∴ F^{- 1} = \frac{1}{4} [\begin{matrix} - 8 & 4 & 4 \\ 11 & - 2 & - 3 \\ - 4 & 0 & 0 \end{matrix}] (vii) G = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}] Now, C_{11} = |\begin{matrix} \cos α & \sin α \\ \sin α & - \cos α \end{matrix}| = - 1, C_{12} = - |\begin{matrix} 0 & \sin α \\ 0 & - \cos α \end{matrix}| = 0 and C_{13} = |\begin{matrix} 0 & \cos α \\ 0 & \sin α \end{matrix}| = 0 C_{21} = - |\begin{matrix} 0 & 0 \\ \sin α & - \cos α \end{matrix}| = 0, C_{22} = |\begin{matrix} 1 & 0 \\ 0 & - \cos α \end{matrix}| = - \cos α and C_{23} = - |\begin{matrix} 1 & 0 \\ 0 & \sin α \end{matrix}| = - \sin α C_{31} = |\begin{matrix} 0 & 0 \\ \cos α & \sin α \end{matrix}| = 0, C_{32} = - |\begin{matrix} 1 & 0 \\ 0 & \sin α \end{matrix}| = - \sin α and C_{33} = |\begin{matrix} 1 & 0 \\ 0 & \cos α \end{matrix}| = \cos α adj F = {[\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}]}^{T} = [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}] and |F| = - 1 ∴ F^{- 1} = - 1 [\begin{matrix} - 1 & 0 & 0 \\ 0 & - \cos α & - \sin α \\ 0 & - \sin α & \cos α \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & \cos α & \sin α \\ 0 & \sin α & - \cos α \end{matrix}]$

Suggest Corrections

Find the inverse of each of the following matrices. (i) 123231312 (ii) 1251-1-123-1 (iii) 2-11-12-11-12 (iv) 20-1510013 (v) 01-14-343-34 (vi) 00-1345-2-4-7 (vii) 1000cos αsin α0sin α-cos α