Question

For any \[2 \times 2\] matrix A, if\[A({\mathop{\rm adj}\nolimits} .A)\left| { = \left[ {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right]} \right|\] then \[|A| = \]
A. 0
B. 10
C. 20
D. 100

Answer 1

Hint:
To answer the question, we must first define a matrix and an adjoint of a matrix. The determinant of A and adjoint of A in terms of |A| are then found by \[\left| {adj\left( A \right)} \right| = {\left| A \right|^{n - 1}}\] and the determinant of the LHS and RHS of the equation given in the question will then be calculated. In this section, we shall express the values of A's determinants and adjoint in terms of |A|. We will next determine the value of |A|, which will be our needed answer.



FORMULA USED:
\[\left| {adj\left( X \right)} \right| = {\left| X \right|^{n - 1}}\]
Where: n is the order of the matrix.



Complete Step-by-Step Answer:
We have been provided in the question that,
For any \[2 \times 2\] matrix A,
\[A({\mathop{\rm adj}\nolimits} .A)\left| { = \left[ {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right]} \right|\]
In our case, the value of n = 2.
So, we have,
\[\left| {adj\left( A \right)} \right| = {\left| A \right|^{2 - 1}}\]
Let us consider the above as
\[ \Rightarrow |adj(A)| = |A|\]
Now, let us take determinant on the either sides of the above expression, we get
\[A[{\mathop{\rm adj}\nolimits} (A)] = \left| {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right|\]
Now, we have to solve the above using the property
\[\left| {XY} \right| = \left| X \right|\left| Y \right|\]
From that, we will obtain
\[\left| A \right|[{\mathop{\rm adj}\nolimits} (A)] = \left| {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right|\]------ (2)
Now, we have to substitute the value of \[\left| {adj{\rm{ }}A} \right|\] from the equation (1) to equation (2),we obtain
\[\left| A \right|\left| A \right| = \left| {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right|\]
The above can be restructured as,
\[ \Rightarrow {\left| A \right|^2} = \left| {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right|\]
The determinant \[\left| {\begin{array}{*{20}{c}}a&c\\b&d\end{array}} \right|\] now has the value \[ad-bc\]
 As a result, we have,
\[ \Rightarrow {\left| A \right|^2} = 10 \times 10 - 0 \times 0\]
On simplifying the RHS of the above equation we get
\[ \Rightarrow {\left| A \right|^2} = 100\]
Now, let’s square on both sides of the equation,
\[ \Rightarrow \left| A \right| = \sqrt {100} \]
On taking roots, we get
\[ \Rightarrow |A| = 10\]
Therefore, if\[A({\mathop{\rm adj}\nolimits} .A)\left| { = \left[ {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right]} \right|\] then \[|A| = 10\]
Hence, the option B is correct
NOTE:
Students must be careful in solving these types of problems, because it has determinants. Students most likely makes mistake in calculating determinants with signs. The provided question can also be solved in an alternate manner, as illustrated. We know that if A is a square matrix, then the following relationship exists \[A\;adj(A) = |A|{I_n}\].