Question

Let $A=\begin{bmatrix}2& 1& -2\\ 1& 1& -1\\ 1& 0& 2\end{bmatrix}$ and if $B=|A|adj(A)$ Then $|B|$ is equal to

• A. 3
• B. 9
• C. 81
• D. 243

Hint:

The key formula here is $|k \cdot adj(A)| = k^n |A|^{n-1}$. For a $3 \times 3$ matrix, this becomes $k^3 |A|^2$. If $k = |A|$, the result is $|A|^3 |A|^2 = |A|^5$.

Answer 1

The Correct Option is C

We need to find the determinant of matrix B, which is defined as a scalar multiple of the adjoint of A.

Step 1: \color{redCalculate the Determinant of A
$|A| = 2(1(2) - 0(-1)) - 1(1(2) - 1(-1)) + (-2)(1(0) - 1(1))$
$|A| = 2(2) - 1(2 + 1) - 2(-1)$
$|A| = 4 - 3 + 2 = 3$.
Since $|A| = 3$, the matrix is invertible.

Step 2: \color{redApply Properties of Determinants to B
$B = |A| adj(A)$.
We need $|B| = ||A| adj(A)|$.
For any $n \times n$ matrix M and scalar $k$, $|kM| = k^n |M|$.
Here $n = 3$, so $|B| = |A|^3 |adj(A)|$.

Step 3: \color{redUse the Adjoint Determinant Property
We know that $|adj(A)| = |A|^{n-1}$.
For $n = 3$, $|adj(A)| = |A|^{3-1} = |A|^2$.

Step 4: \color{redCompute the Final Value
Substitute the property from Step 3 into the equation from
Step 2:
$|B| = |A|^3 \cdot |A|^2 = |A|^5$.
Substitute $|A| = 3$:
$|B| = 3^5 = 243$.
re-checking the question and logic.
The options are 3, 9, 81, 243.
My calculation gives 243. Let's re-verify the determinant of A.
Row 1: $2, 1, -2$
$2(1(2)-0) - 1(1(2)-(-1)) - 2(1(0)-1) = 4 - 3 + 2 = 3$. Correct.
$B = |A| adj(A) \implies |B| = |A|^3 |adj(A)| = |A|^3 |A|^2 = |A|^5$.
$3^5 = 243$.
The correct option is (4).