If A - - 1 1- 2 2 - B - - -1 1- 1 -1 - then detAB -

AB = [ 1 amp; 1; 2 amp; 2 #160; ].[ 1 amp; 1; 1 amp; 1 #160; ]=[ 1+1 amp; 1 1; 2+2 amp; 2 2 ]=[ 0 amp; 0; 0 ...

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Equality of Matrices

If A = [ 1 ...

Question

If $A = [\begin{matrix} 1 & 1 2 & 2 \end{matrix}]$ , $B = [\begin{matrix} - 1 & 1 1 & - 1 \end{matrix}]$ , then $d e t (A B) =$

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A = [\begin{matrix} 1 & 1 2 & 2 \end{matrix}], B = [\begin{matrix} 1 & 1 3 & 4 \end{matrix}]

| A B |

[\begin{matrix} 0 & 1 2 & 4 \end{matrix}]

[\begin{matrix} - 1 & 1 2 & 2 \end{matrix}]

C = [\begin{matrix} 1 & 0 1 & 0 \end{matrix}],

2 A + 3 B - C =

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & - 1 2 & - 3 & 4 3 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦,

B = ⎡ ⎢ ⎣ \begin{matrix} - 1 & - 2 & - 1 6 & 12 & 6 5 & 10 & 5 \end{matrix} ⎤ ⎥ ⎦

C = ⎡ ⎢ ⎣ \begin{matrix} - 1 & - 1 & 1 2 & 2 & - 2 - 3 & - 3 & 3 \end{matrix} ⎤ ⎥ ⎦,

If A is an invertible matrix of order 2, then det $(A^{- 1})$ is equal to
a) $d e t (A)$
(b) $\frac{1}{d e t (A)}$
c) 1
d) zero

A

B

3 \times 3

det (A B A^{T}) = 8

det (A B^{- 1}) = 8

det (B A^{- 1} B^{T})

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