Class 12

Math

Algebra

Determinants

Suppose $A$ and $B$ are two non singular matrices such that $B=I,A_{6}=I$ and $AB_{2}=BA$. Find the least value of $k$ for $B_{k}=I$.

- 63

Step 1: Given:

$∣A∣=∣B∣=0$

$A_{6}=I$

$AB_{2}=BA$

$B_{k}=I$

Step 2: To find:

Value of $k$ for which $B_{k}=I$

Step 3: Formula used:

$AA_{−1}=A_{−1}A=I$

Step 4: Solution:

Consider the given expression

$AB_{2}=BA$ ... (1)Step 5: Aim to make RHS as independent of $A$

Pre-multiply (1) from both sides by $A_{5}$, we get

$A_{6}B_{2}=A_{5}BA$

Post-multiply above equation from both sides by $A_{5}$, we get

$A_{6}B_{2}A_{5}=A_{5}BA_{6}$

Since, it is given that $A_{6}=I$

Expression (3) reduces to

$B_{2}A_{5}=A_{5}B$

Pre-multiply this on both sides by $A$, we get

$AB_{2}A_{5}=A_{6}B$

$⇒AB_{2}A_{5}=B$ $(A_{6}=I)$ .....(2)

Step 8:

Now substituting the value of $B$ from (2) in LHS of (2) itself, we get

$A(AB_{2}A_{5})_{2}A_{5}=B$

$⇒AAB_{2}A_{5}AB_{2}A_{5}A_{5}=B$

$⇒A_{2}B_{2}A_{6}B_{2}A_{10}=B$

Since, we know $A_{6}=I$

$⇒A_{2}B_{4}A_{6}A_{4}=B$

$⇒A_{2}B_{4}A_{4}=B$ ... (3)

Step 9:

Now substituting the value of $B$ from (2) in LHS of (3), we get

$A_{2}(AB_{2}A_{5})_{4}A_{4}=B$

$⇒A_{2}AB_{2}A_{5}AB_{2}A_{5}AB_{2}A_{5}AB_{2}A_{5}A_{4}=B$

$⇒A_{3}B_{2}A_{6}B_{2}A_{6}B_{2}A_{6}B_{2}A_{5}A_{4}=B$

Since, we know $A_{6}=I$

$⇒A_{3}B_{8}A_{3}=B$ ...(4)

Step 10:

Looking at the pattern of (2), (3) and (4), we can say that

$A_{p}B_{q}A_{r}=B$

$p$ is increasing by $1$,

$q$ is forming a GP as $2,4,8,...$, and

$r$ is decreasing by $1$

$∴$ In the same manner, we find that when $p=6$, we get $q=64$ and $r=0$

So, the expression becomes

$A_{6}B_{64}A_{0}=B$

Since, we know $A_{6}=I$

$⇒B_{64}=B$ ...(5)

Step 11:

Now, we have $B_{64}=B$

Post-Multiplying this by $B_{−1}$ both sides, we get

$B_{63}BB_{−1}=BB_{−1}$ ... (6)

Since, $BB_{−1}=I$, (6) reduces to

$B_{63} =I$ .... (7)

Step 12:

Comparing (7) with the given condition where $B_{k}=I$, we find that

$B_{63}=B_{k}$

$⇒k=63$

Step 13: Result:

$k=63$