Q.Find the minors and cofactors of every element of $\begin{pmatrix}5&20\\0&-1\end{pmatrix}$.

A minor is defined for each element of the determinant, and a $3\times3$ determinant has 9 elements.

For a determinant of order $n$, a minor $M_{ij}$ is defined for every element $a_{ij}$, obtained by deleting its row and column. Since $A$ is $3\times3$, it has $3\times3=9$ elements, and hence 9 minors (one per element).

The cofactor of the entry -3 (at row 3, column 2) is computed as C32=(-1)^5 times its minor, giving 2 - not any of the listed options.

The entry $-3$ sits at position $a_{32}$ (row 3, column 2) in $\begin{vmatrix}0&1&-2\\-1&0&2\\2&-3&0\end{vmatrix}$.

Minor $M_{32}$ = determinant left after deleting row 3 and column 2:

$M_{32}=\begin{vmatrix}0&-2\\-1&2\end{vmatrix} = (0)(2)-(-2)(-1) = 0-2=-2$

Cofactor $C_{32}=(-1)^{3+2}M_{32}=(-1)(-2)=2$.

Since 2 is not among options (a) 1,

(b) -1,

(c) 0, the correct choice is 'none of these'.

Expansion of a determinant along a column, using cofactors of that column's entries.

The value of a determinant can be obtained by expanding along any row or column, using the cofactors of that row/column's entries. Expanding $\Delta$ along the first column ($a_{11},a_{21},a_{31}$) using their respective cofactors $A_{11},A_{21},A_{31}$:

$\Delta = a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$

This is the standard cofactor (Laplace) expansion along the first column. (Note: expanding along the first row would instead give $a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}$ — a different but equally valid expression not among the options.)

Standard definition from the Determinants chapter.

Minor of an element: For a determinant, the minor $M_{ij}$ of the element $a_{ij}$ is the determinant obtained after deleting the $i$-th row and the $j$-th column, i.e. the row and column in which $a_{ij}$ lies. It is a determinant of order one less than the order of the original determinant.

For example, in a $3\times3$ determinant, $M_{11}$ is the $2\times2$ determinant left after deleting the 1st row and 1st column.

Cofactor $C_{ij}=(-1)^{i+j}M_{ij}$; delete row 3 and column 1, then evaluate the $2\times2$ minor.

$\begin{vmatrix}1 & 3 & 5\\ -1 & 2 & 3\\ 2 & 6 & -7\end{vmatrix}$

To find $C_{31}$, delete row 3 and column 1:

$M_{31} = \begin{vmatrix}3 & 5\\ 2 & 3\end{vmatrix} = 3(3)-5(2) = 9-10=-1$

$C_{31} = (-1)^{3+1}M_{31} = (+1)(-1) = -1$

Element 8 is at row 1, column 1; its cofactor is $(-1)^{1+1}$ times its minor, giving $+14$.

For the determinant $\begin{vmatrix} 8 & 10 \\ 12 & 14 \end{vmatrix}$, the element $8$ lies in the first row and first column, so $i=1,\ j=1$.

Step 1 — the minor $M_{11}$ is obtained by deleting the first row and first column, leaving the single element $14$:

$M_{11}=14$

Step 2 — the cofactor applies the sign $(-1)^{i+j}$:

$A_{11}=(-1)^{1+1}\,M_{11}=(+1)(14)=14$

The minor of an element is the determinant left after deleting its row and column.

In $\begin{vmatrix}1 & 3 & 2\\ 8 & 6 & 3\\ 9 & 5 & 4\end{vmatrix}$, the element 6 is in row 2, column 2.

Deleting row 2 and column 2 leaves $\begin{vmatrix}1 & 2\\ 9 & 4\end{vmatrix} = (1)(4)-(2)(9) = 4-18=-14$.

State the definition of a determinant.

To every square matrix $A=[a_{ij}]$ of order $n$, we can associate a number (real or complex), called the determinant of the square matrix $A$, denoted by $\det(A)$ or $|A|$.

For example, for a $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$,

$|A|=ad-bc$

This mapping which associates each square matrix with a unique number is known as determinant.

The minor of an element is the determinant left after deleting that element's row and column; for a $2\times2$ matrix each minor is just the single opposite-corner entry.

Given $\begin{vmatrix}2 & -4\\ 0 & 3\end{vmatrix}$, the minor $M_{ij}$ is obtained by deleting row $i$ and column $j$ and taking the determinant of what remains (here, a single number).

$M_{11}$ (delete row 1, col 1): remaining entry is $3$, so $M_{11}=3$

$M_{12}$ (delete row 1, col 2): remaining entry is $0$, so $M_{12}=0$

$M_{21}$ (delete row 2, col 1): remaining entry is $-4$, so $M_{21}=-4$

$M_{22}$ (delete row 2, col 2): remaining entry is $2$, so $M_{22}=2$