Determinants And Matrices Set-3

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Example -7

Let M be a 3 × 3 matrix such that \({M^T}M - I\;{\rm{and}}\;|M| = 1.\) Find the value of \(|M - I|.\)

Solution:

\[\begin{align}|M - I| = |{(M - I)^T}| &  = \;|{M^T}I|\\ &= \;|{M^T} - {M^T}M|\\ &  = \;|{M^T}(I - M)|\\ &  = \;|{M^T}|I - M|\\ &  = \;1 \cdot ( - |M - I|)\\ \Rightarrow  \qquad \qquad |M - I|\;  &= 0\end{align}\]

Example -8 

Let A and B be two invertible matrices of the same order such that \({(AB)^n} = {A^n}{B^n}\) holds for three consecutive positive integers. Show that AB = BA.

Solution: Note that since A and B are invertible, so are \({A^n}\;{\rm{and}}\;{B^n}.\) Now for some positive integer k, we have.

\[\begin{align}{(AB)^k}&= {A^k}{B^k}\\{(AB)^{k + 1}}&= {A^{k + 1}}{B^{k + 1}}\\{(AB)^{k + 2}}&= {A^{k + 2}}{B^{k + 2}}\end{align}\]

The second relation can be written as

\[{(AB)^k}(AB) = {A^k}(A{B^k})B\]

Now using the first relation in the LHS, we have

\[{A^k}\left( {{B^k}A} \right)B = {A^k}\left( {A{B^k}} \right)B\]

Since \({A^k}\;{\rm{and}}\;B\) are invertible, we have

\[{B^k}A = A{B^k}\]

Similarly,

\[\begin{align}{B^{k + 1}}A &= A{B^{k + 1}}\\ \Rightarrow  \qquad B({B^k}A) &= A{B^{k + 1}}\\ \Rightarrow  \qquad (BA){B^k} &= (AB){B^k}\end{align}\]

Since \({B^k}\) is invertible, we have

\[BA = AB\]

Example -9

Let A be a a skew-symmetric matrix. Let B be a matrix of the same order such that \(B(I - A) = I + A\) . Show that B is orthogonal.

Solution: We have to show that \({B^T}B = B{B^T} = I\)

Note that since A is skew-symmetric, \({A^T} =  - A\)

We are given that

\[B(I - A) = I + A \qquad\qquad \dots(1)\]

Taking the transpose on both sides, we have

\[\begin{align}  {(I - A)^T}{B^T} &= I + {A^T}\\ \Rightarrow  \qquad (I + A){B^T} &= I - A\end{align} \qquad \qquad \dots(2)\]

From (1) × (2), we have

\[\begin{align} (I + A){B^T}B(I - A) & = (I - A)(I + A)\\ &  = (I + A)(I - A) \qquad \qquad \qquad \text{(How?)}\\\ \Rightarrow  \qquad \qquad \qquad \qquad {B^T}B  &= I \end{align}\]

By (1) × (2) again in another order, we have

\[\begin{align}  &\qquad \;\;B(I - A)(I + A){B^T} = {I^2} - {A^2}\\ &\Rightarrow \quad   B({I^2} - {A^2}){B^T}B = ({I^2} - {A^2})B \qquad \qquad \qquad \dots (3)\end{align}\]

Since \({B^T}B = I,\) we have

\[B({I^2} - {A^2}) = ({I^2} - {A^2})B \qquad \qquad \dots(4)\]

Using (4) in (3), we have

\[\begin{align} &({I^2} - {A^2})B{B^T} = {I^2} - {A^2}\\ \Rightarrow \qquad & B{B^T} = I\end{align}\]

Thus, \(B{B^T} = {B^T}B = I\)

Example -10 

Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of the these entries are 1 and four of them are 0.

(a) Find the number of matrices in A

(b) Find the number of matrices A in A for which the system of linear equations

\[A\left[ {\;\begin{gathered}x\\y\\z\end{gathered}\;} \right] = \left[ {\;\begin{gathered}
1\\0\\0\end{gathered}\;} \right]\]

has a unique solution

(c) Find the number of matrices A in A for which this system of linear equations is inconsistent.

Solution: (a) Since all the matrices are symmetric, this means that on the diagonal, there can be either one ‘1’ or three ‘1’s. There cannot be two ‘1’s.

If all entries on the  diagonal are 1:

 In this case, 3 matrices are possible, since the other two ‘1’s can be placed in 3 possible ways

If one entry on the diagonal is 1 :

 In this case,

3 × 3 = 9

matrices are possible, since the ‘1’ on the can be placed in 3 ways, and the other two pairs of ‘1’s can be placed in three ways. diagonal is

\(\Rightarrow \)       The number of matrices in A is 12.

(b) For a unique solution, | A | should be non-zero. We count such cases explicitly:

(i)  All diagonal element are

\[\left[ {\;\begin{align}1 \quad x \quad y\\x\quad1\quad z\\y\quad z \quad 1\end{align}\;} \right] \qquad \text {One of }x, y, z \; \text{is 1, the other two are 0.}\]

If \(x = y = 0,\;{\rm{then}}\;|A| = 0\). Verify this explicitly. Similarly, if \(y = z = 0\;{\rm{or}}\;{\rm{if}}\;x = z = 0\), then also \(|A| = 0\). So in this case, there are no matrices A for which a unique solution exists

(ii)   One diagonal element is 1, the other two are 0

\[\left[ {\;\begin{array}{*{20}{c}}1&x&y\\x&0&z\\y&z&0\end{array}\;} \right]\;{\rm{or}}\;\left[ {\;\begin{array}{*{20}{c}}0&x&y\\x&1&z\\y&z&0\end{array}\;} \right]\;{\rm{or}}\;\left[ {\;\begin{array}{*{20}{c}}0&x&y\\x&0&z\\y&z&1\end{array}\;} \right]\]

in each case, one of  \(x,y,z\) is 0, the other two are 1. In the first case, if  \(x = 0,\;y = z = 1,\) then \(|A| \ne 0\) . If y = 0, \(x = z = 1,\) then \(\left| A \right| \ne 0\) If \(z = 0,\;x = y = 1,\) then \(|A| = 0\) . Thus, there are 2 matrices from the first case for which \(|A|\; \ne 0\). Similarly, each of the other two cases give rise to two matrices with non-zero determinants.

\( \Rightarrow \)       There are 6 matrices A for which \(|A|\; \ne 0,\) For which the system has a unique solution.

(c) There are 6 cases where | A | = 0. Out of these, in 2 cases, namely, \[\left| {\;\begin{array}{*{20}{c}}1&1&1\\1&0&0\\1&0&0\end{array}\;} \right|\;\;{\rm{and}}\;\;\left| {\;\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&1&1\end{array}\;} \right|\]

the number of solutions is infinite. The other 4 cases are inconsistent systems.

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