Let Mij and Cij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.
Also, Cij = (–1)i+j × Mij
\( M_{11}=-12\)
\( M_{21}=-16\)
M31 = –3 × 2 – (–1) × 2
M31 = –4
C11 = (–1)1+1 × M11
= 1 × –12
= –12
C21 = (–1)2+1 × M21
= –1 × –16
= 16
C31 = (–1)3+1 × M31
= 1 × –4
= –4
Now expanding along the first column we get
|A| = a11 × C11 + a21× C21+ a31× C31
= 1× (–12) + 4 × 16 + 3× (–4)
= –12 + 64 –12
= 40
Answered by Sakshi | 1 year agoEvaluate
\(\begin{vmatrix}2&3&-5 \\[0.3em] 7& 1&-2\\[0.3em] -3&4&1\end{vmatrix}\)
Show that
\(\begin{vmatrix}sin10°&-cos10° \\[0.3em] sin80°& cos80°\\[0.3em] \end{vmatrix}\)
Evaluate the determinants:
\(\begin{vmatrix}2&3&7 \\[0.3em] 13& 17&5\\[0.3em] 15&20&12\end{vmatrix}^2\)
Evaluate the determinants:
\(\begin{bmatrix}a+ib &c+id \\[0.3em] -c+id & a-ib \\[0.3em] \end{bmatrix}\)
Evaluate the determinants:
\(\begin{bmatrix}cos 15° &-sin 15° \\[0.3em] sin75° & cos75° \\[0.3em] \end{bmatrix}\)