If A and B are square matrices of size n × n, then which of the following is not true?

Explanation:

det (A + B) = det(A) + det(B) is not correct.

Taking an example:

Considering A = I (I is an identity matrix of order 2 × 2)

Let B = -A

So, B  = -I

∴ det(A + B)  = 0

But, det(A) + det(B) = 2

Additional Information Properties of determinants:

• The value of a determinant does not change when rows and columns are interchanged. i.e. \(\left|A^T \right|~=~\left|A \right|\).
• If any row or column of a matrix A is completely zero, then \(\left|A \right|~=~0\), such a row or column is called zero rows or column.
• Also if two rows or columns of a matrix A are identical then \(\left|A \right|~=~0\).
• If any two rows or two columns of a determinant are interchanged, the value of the determinant is multiplied by -1.
• If all elements of one row (or one column) of the determinant are multiplied by the same number k, the value of the determinant is k times the value of the given determinant.
• If A is an n-rowed square row matrix, and k be any scalar, then \(\left|kA \right|~=~k^n\left|A \right|\)
• In a determinant, the sum of the products of the elements of any row (or column) with the cofactors of corresponding elements of any row or column is equal to the determinant value.
• In a determinant, the sum of the products of the elements of any row (or column) with the cofactors of ​some other row or column is zero.

Example:

Δ = \(\begin{vmatrix} a_1 & b_1 &c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 &c_3 \\ \end{vmatrix}\)

Then,

\(a_1 A_1~+~b_1 B_1~+~c_1 C_1~=~\Delta\)

\(a_1 A_2~+~b_1 B_2~+~c_1 C_2~=~0\)

\(a_1 A_3~+~b_1 B_3~+~c_1 C_3~=~0\)

\(a_2 A_2~+~b_2 B_2~+~c_2 C_2~=~\Delta\)

\(a_2 A_1~+~b_2 B_1~+~c_2 C_1~=~0\)

where A₁, B₁, and C₁ be the cofactors of the elements a₁, b₁, and c₁ in D.

• If the elements of a row or column of a determinant are added m times the corresponding elements of another row or column, the value of the determinant thus obtained is equal to the value of the original determinant:

i.e. \(A~\xrightarrow[]{R_i ~+~mR_j}~B~then~\left|A \right|~=~\left|B \right|\)

And \(A~\xrightarrow[]{C_i ~+~mC_j}~B~then~\left|A \right|~=~\left|B \right|\)

• \(\left|AB \right|~=~\left|A \right|~×~\left|B \right|\) and based on this we can prove the following

(a). \(\left|A^n \right|~=~(\left|A \right|)^n\)

(b). \(\left|A^{-1} \right|~=~\frac{1}{\left|A \right|}\)

• Using the fact that \(A.Adj~A~=~\left|A \right|~.I\) the following can be obtained for A_{n × n}.

(a). \(\left|Adj~A \right|~=~{\left|A \right|}^{n~-~1}\)

(b). \(\left|Adj~(Adj~(A)) \right|~=~{\left|A \right|}^{(n~-~1)^2}\)