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Q : 18       If A is an invertible matrix of order 2, then det  \small (A^-^1) is equal to

                (A) \small det(A)       (B)  \small \frac{1}{det (A)}       (C)  \small 1       (D) \small 0

Given that the matrix is invertible hence  exists and  Let us assume a matrix of the order of 2; . Then .     and   Now,     Taking determinant both sides; Therefore we get; Hence the correct answer is B.  

Q : 17        Let A be a nonsingular square matrix of order \small 3\times 3. Then \small |adjA| is equal to

                 (A) \small |A|      (B) \small |A|^2      (C) \small |A|^3      (D) \small 3|A|

We know the identity  Hence we can determine the value of . Taking both sides determinant value we get,      or    or taking R.H.S., or, we have then      Therefore  Hence the correct answer is B.  

Q : 16      If   \small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix} , verify that \small A^3-6A^2+9A-4I=O. Hence find \small A^-^1.

Given matrix: ; To show:  Finding each term: So now we have,  Now finding the inverse of A; Post-multiplying by  as,                                          ...................(1) Now, From equation (1) we get; Hence inverse of A is :            

Q : 15     For the matrix  \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}   Show that \small A^3-6A^2+5A+11I=O     Hence, find \small A^-^1.
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Given matrix: ; To show:  Finding each term: So now we have,  Now finding the inverse of A; Post-multiplying by  as,                                          ...................(1) Now, From equation (1) we get;        

Q : 14      For the matrix \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} , find the numbers \small a and \small b such that A^2+aA+bI=0.

Given  then we have the relation   So, calculating each term; therefore  ; So, we have equations;     and  We get .  

Q : 13            If   \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}​ , show that  A^2-5A+7I=O. Hence find \small A^-^1

Given  then we have to show the relation  So, calculating each term; therefore  ; Hence .   [Post multiplying by , also ]  

Q : 12        Let   \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}   and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix} .  Verify that  \small (AB)^-^1=B^{-1}A^{-1}.

We have  and . then calculating; Finding the inverse of AB. Calculating the cofactors fo AB:                Then we have adj(AB): and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2 Therefore we have inverse:                                                        .....................................(1) Now, calculating inverses of A and B. |A| = 15-14 = 1   and |B| = 54- 56 = -2      and ...

Q : 11       Find the inverse of each of the matrices (if it exists).

                 \small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : Now, calculating the cofactors terms and then adjA.                                                                                       So, we have  Therefore inverse of A will be:  

Q : 10         Find the inverse of each of the matrices (if it exists).

                   \small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : Now, calculating the cofactors terms and then adjA.                                                                                       So, we have  Therefore inverse of A will be:

Q : 9         Find the inverse of each of the matrices (if it exists).

                 \small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : Now, calculating the cofactors terms and then adjA.                                                                                       So, we have  Therefore inverse of A will be:  

Q : 8       Find the inverse of each of the matrices (if it exists).

               \small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : Now, calculating the cofactors terms and then adjA.                                                                                       So, we have  Therefore inverse of A will be:

Q : 7       Find the inverse of each of the matrices (if it exists).

               \small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : Now, calculating the cofactors terms and then adjA.                                                                                       So, we have  Therefore inverse of A will be:

Q : 6         Find the inverse of each of the matrices (if it exists).

                 \small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix}

Given the matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : |A| = (-2+15) = 13 Now, calculating the cofactors terms and then adjA. So, we have  Therefore inverse of A will be:

Q : 5          Find the inverse of each of the matrices (if it exists).

                   \small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}

Given matrix :   To find the inverse we have to first find adjA then as we know the relation: So, calculating |A| : |A| = (6+8) = 14 Now, calculating the cofactors terms and then adjA. So, we have  Therefore inverse of A will be:  

Q : 4          Verify \small A (adj A)=(adjA)A=|A| I.

                  \small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}

Given matrix:  Let   Calculating the cofactors;            Hence,  Now,   also,   Now, calculating |A|;    So,  Hence we get,  .

Q : 3           Verify \small A (adj A)=(adj A)A=|A|I.

                    \small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Given the matrix:  Let   Calculating the cofactors;     Hence,  Now,   aslo,   Now, calculating |A|;    So,  Hence we get  

Q : 2     Find adjoint of each of the matrices

             \small \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}

Given the matrix:  Then we have, Hence we get:

Q : 1       Find adjoint of each of the matrices.

               \small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}

Given matrix:  Then we have, Hence we get:  
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