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Let A be an mxn matrix and B an nxm matrix.

It is given that determinant ( Im + AB ) = determinant ( In + BA ) , where Ik is the k×k identity matrix. Using the above property, the determinant of the matrix given below is

$\begin{bmatrix} 2& 1& 1& 1\\ 1& 2& 1& 1\\ 1& 1& 2& 1\\ 1& 1& 1& 2 \end{bmatrix}$

A) 2

B) 5

C) 8

D) 16
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m not getting this question @Bikram sir
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Im is the Identity matrix with m * m

In is the identity matrix with n * n

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3 Answers

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$\begin{bmatrix} 2 &1 &1 &1 \\ 1 &2 &1 &1\\ 1 &1 &2 &1\\ 1 &1 &1 &2 \end{bmatrix} = \begin{bmatrix} 1 &1 &1 &1 \\ 1 &1 &1 &1\\ 1 &1 &1 &1\\ 1 &1 &1 &1 \end{bmatrix} + \begin{bmatrix} 1 &0 &0 &0 \\ 0 &1 &0 &0\\ 0 &0 &1 &0\\ 0 &0 &0 &1 \end{bmatrix}$

Let AB = $\begin{bmatrix} 1 &1 &1 &1 \\ 1 &1 &1 &1\\ 1 &1 &1 &1\\ 1 &1 &1 &1 \end{bmatrix}$

Now, AB = $\begin{bmatrix}1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ * $\begin{bmatrix} 1 &1 & 1 & 1 \end{bmatrix}$

and I4 = $\begin{bmatrix} 1 &0 &0 &0 \\ 0 &1 &0 &0\\ 0 &0 &1 &0\\ 0 &0 &0 &1 \end{bmatrix}$

Now use the given property,

determinant ( Im + AB ) = determinant ( In + BA )

So, BA = $\begin{bmatrix} 1 &1 & 1 & 1 \end{bmatrix}$ * $\begin{bmatrix}1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ = $\begin{bmatrix} 4 \end{bmatrix}$

$So, determinant ( In + BA ) = $$\begin{vmatrix} \begin{bmatrix} 4 \end{bmatrix} +\begin{bmatrix} In \end{bmatrix} \end{vmatrix}$

= $\begin{vmatrix} 4 \end{vmatrix} + \begin{vmatrix} 1 \end{vmatrix}$ ($\because$ determinant of an Identity matrix is 1)

= 5

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4 Comments

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This is not |4| + |1| but here | mat(4) + mat(1)| = |mat(5)| = 5
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very nice explanation
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4 votes
4 votes

Elimination method.

 

$\begin{vmatrix} 2 & 1 & 1& 1\\ 1& 2& 1 &1 \\ 1 & 1& 2& 1\\ 1 & 1& 1& 2 \end{vmatrix}$

$R_1 \leftarrow R_1+R_2+R_3+R_4$

$\begin{vmatrix} 5 & 5 & 5& 5\\ 1& 2& 1 &1 \\ 1 & 1& 2& 1\\ 1 & 1& 1& 2 \end{vmatrix}$

Take $5$ out from the first row.

$5*\begin{vmatrix} 1 & 1 & 1& 1\\ 1& 2& 1 &1 \\ 1 & 1& 2& 1\\ 1 & 1& 1& 2 \end{vmatrix}$

So, this means that determinant should be a multiple of $5$

Since there is only one option having multiple of $5$  i.e. option $B.$ so all other options are eliminated.

$\therefore$ Option $B.$ is correct choice

3 votes
3 votes

It is based on one of the exercise problems from Gilbert Strang's text.

We'll Solve a general case which will automatically provide an answer to this.

Problem : Suppose the 4 by 4 matrix m has four equal rows all containing a,b,c,d. We know that Det(M)=0. The problem is to find the det(I+M) by any method

det(I+M)=$\begin{vmatrix} 1+a &b &c &d \\ a & 1+b &c &d\\ a &b & 1+c &d \\ a& b & c & 1+d \end{vmatrix}$

Now from the above determinant, Subtract row 4 from rows 1,2,3

$\begin{vmatrix} 1 &0 &0 &-1 \\ 0&1 &0 &-1 \\ 0 & 0 & 1 & -1\\ a & b &c & 1+d \end{vmatrix}$

Now, perform a(Row1)+b(Row2)+c(Row3) and subtract this from Row 4

$\begin{vmatrix} 1 &0 &0 &-1 \\ 0&1 &0 &-1 \\ 0 & 0 & 1 & -1\\ 0 & 0 &0 & 1+a+b+c+d \end{vmatrix}$

And this is an upper triangular matrix form hence determinant=1+a+b+c+d.

In our GATE question case,a=b=c=d=1 , and hence our det=1+1+1+1+1=5.

Answer-(B)

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@Ayush Upadhyaya Sir, it is really good explanation. ty

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