Question

Problem 3. Determine (with proof) whether each of the following statements is true or false. (a) For every m xn matrix A, det(AAT) = det(ATA) (b) Let A be an invertible n xn matrix, and suppose that B, C, and D are n x n matrices [det(A) |det(C) det (B) CA-1B. Then the 2 x 2 matrix is not invertible satisfying D (c) If A is an invertible n x n matrix such that A = A-1 then det(A) = +1.t (d) If A E Rnxn is invertible and all entries of A and A-1 are integers, then det(A) = det(A-1)

Answer 1

(a) Let us consider a 2x1 matrix A = $\begin{bmatrix} a\\ b \end{bmatrix}$ .

Then, A^TA = $\begin{bmatrix} a & b \end{bmatrix}$ $\begin{bmatrix} a\\ b \end{bmatrix}$ = $\begin{bmatrix} a^2+b^2\end{bmatrix}$

And, AA^T = $\begin{bmatrix} a\\ b \end{bmatrix}$ $\begin{bmatrix} a & b \end{bmatrix}$ =

2 Tab ab 6²

Now, det(A^TA) =
Ta² +62
= $a^2+b^2$

and, det(AA^T) = $\begin{vmatrix} a^2 & ab\\ ab & b^2 \end{vmatrix}$ = $a^2.b^2-(ab).(ab)$ = $a^2b^2-a^2b^2$ =

Since $a^2+b^2\neq0$ , then det(A^TA) $\neq$ det(AA^T) for all A.

Therefore, the given statement is false.

(b) Given, $D=CA^{-1}B$

i.e., $\det(D)=\det(CA^{-1}B)$

i.e., $\det(D)=\det(C)*\det(A^{-1})*\det(B)$

i.e., $\det(D)=\det(C)*\frac{1}{\det(A)}*\det(B)$

i.e.,

det (D) * det(A) = det(C) * det (B)

i.e., $\det(D)*\det(A)-\det(C)*\det(B)=0$

This shows that the determinant of $\begin{bmatrix} \det(A) & \det(B)\\ \det(C) & \det(D) \end{bmatrix}$ is equal to 0.

We know that if determinant of any matrix is equal to 0, then inverse of that matrix does not exist ,i.e, the matrix is not invertible.

Therefore, $\begin{bmatrix} \det(A) & \det(B)\\ \det(C) & \det(D) \end{bmatrix}$ is not invertible.

Hence, the given statement is true.

(c) Given, $A=A^{-1}$

i.e., $\det(A)=\det(A^{-1})$

i.e., $\det(A)=\frac{1}{\det(A)}$

i.e., $[\det(A)]^2=1$

i.e., $\det(A)=\pm\sqrt{1}$

i.e., $\det(A)=\pm1$

Therefore, the given statement is true.

(d) Let A = $\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ be a 2x2 invertible matrix.

Then, A^-1 = $\frac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$

i.e., A^-1 = $\begin{bmatrix} d/(ad-bc) & -b/(ad-bc)\\ -c/(ad-bc) & a/(ad-bc) \end{bmatrix}$

Now, det(A) = $\begin{vmatrix} a & b\\ c & d \end{vmatrix}$

i.e., det(A) =

ad - be

and, det(A^-1) = $\begin{vmatrix} d/(ad-bc) & -b/(ad-bc)\\ -c/(ad-bc) & a/(ad-bc) \end{vmatrix}$

i.e., det(A^-1) = $\left ( \frac{d}{ad-bc} \right )*\left ( \frac{a}{ad-bc} \right )-\left ( -\frac{b}{ad-bc} \right )\left ( -\frac{c}{ad-bc} \right )$

i.e., det(A^-1) = $\frac{ad}{(ad-bc)^2}-\frac{bc}{(ad-bc)^2}$

i.e., det(A^-1) = $\frac{ad-bc}{(ad-bc)^2}$

i.e., det(A^-1) = $\frac{1}{ad-bc}$

Since
ad - be
$\neq$$\frac{1}{ad-bc}$, therefore det(A) $\neq$ det(A^-1).

If $ad-bc=1$, only then det(A) = det(A^-1).

Therefore, the given statement is false.