The inverse of the matrix [1&0&0 3&3&0 5&2&-1] is

Question

The inverse of the matrix [1&0&0 3&3&0 5&2&-1] is (A)  - (1/3) [ - 3 &0&0 3&1&0 9 &2& - 3 ] (B)  - (1/3) [ - 3 &0&0 3& - 1&0 - 9 - 2& 3 ] (C)  - (1/3) [ 3 &0&0 3& - 1&0 - 9 - 2& 3 ] (D)  - (1/3) [ - 3 &0&0 - 3& 1&0 - 9 - 2& 3 ]

Tardigrade Admin · Accepted Answer

A=[1 0 0 3 3 0 5 2 -1] | A |=-3 ≠0 ∴ A -1 exist A -1=(1/| A |) Adj A textAdj( A )= Transpose of co-factor of A M 11=(-1)1+1(-3-0)=-3, M 12=(-1)1+2(-3-0)=3, M 13=(-1)1+3(6-15)=-9 M 21=(-1)2+1(0-0)=0, M 22=(-1)2+2(-1+0)=-1, M 23=(-1)2+3(2-0)=-2 M 31=(-1)3+1(0-0)=0, M 32=(-1)3+2(0-0)=0, M 33=(-1)3+3(3-0)=3 textAdj(A)=[-3 0 0 3 -1 0 -9 -2 3] ∴ A -1=-(1/3)[-3 0 0 3 -1 0 -9 -2 3]

Q. The inverse of the matrix $⎣ ⎡ 135032 00 - 1 ⎦ ⎤$ is

$- \frac{1}{3} ⎣ ⎡ - 3 39 012 00 - 3 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ - 3 3 - 9 0 - 1 - 2 003 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ 33 - 9 0 - 1 - 2 003 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ - 3 - 3 - 9 01 - 2 003 ⎦ ⎤$

Q. The inverse of the matrix ⎣⎡​135​032​00−1​⎦⎤​ is

−31​⎣⎡​−339​012​00−3​⎦⎤​

−31​⎣⎡​−33−9​0−1−2​003​⎦⎤​

−31​⎣⎡​33−9​0−1−2​003​⎦⎤​

−31​⎣⎡​−3−3−9​01−2​003​⎦⎤​

Q. The inverse of the matrix $⎣ ⎡ 135032 00 - 1 ⎦ ⎤$ is

$- \frac{1}{3} ⎣ ⎡ - 3 39 012 00 - 3 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ - 3 3 - 9 0 - 1 - 2 003 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ 33 - 9 0 - 1 - 2 003 ⎦ ⎤$

$- \frac{1}{3} ⎣ ⎡ - 3 - 3 - 9 01 - 2 003 ⎦ ⎤$