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Q1. Ravi purchases 1 pen, 4 pencils and 1 box in Rs. 34. From the same store Neeraj purchases 2 pens, 2 pencils in Rs. 26 and Neetu purchases 1 pen, 3 pencils and 2 boxes in Rs. 43. Express this problem into matrix form.

Solution

Let price of 1 pen = Rs.x, Price of 1 pencil = Rs. y and Price of 1 box = Rs. z   From the given conditions, X + 4y + z = 34    …(1), 2x + 2y = 26        …(2) and x + 3y + 2z = 43  …(3)   The system of linear equations can be written as:

Q2. Ravi purchases 1 pen, 4 pencils and 1 box in Rs. 34. From the same store Neeraj purchases 2 pens, 2 pencils in Rs. 26 and Neetu purchases 1 pen, 3 pencils and 2 boxes in Rs. 43. Express this problem into matrix form.

Solution

Let price of 1 pen = Rs.x, Price of 1 pencil = Rs. y and Price of 1 box = Rs. z   From the given conditions, X + 4y + z = 34    …(1), 2x + 2y = 26        …(2) and x + 3y + 2z = 43  …(3)   The system of linear equations can be written as:

Q3. If , then =?

Solution

                                  

Q4. Solve the system of linear equations by matrix method: x + y + z = 3, 2x – y + z = 2, x – 2y + 3z = 2.

Solution

The system of linear equations can be written as AX = B,

Q5. Show that =0.

Solution

, Since, If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.   Since, if each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.   Since, if any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero.

Q6. If , verify that (AB)^–1 = B^–1A^–1.

Solution

Here, , Co-factors of elements of AB,

Q7. Write the given system of linear equations in matrix form. 2x – 3y = 1, x + 3z = 11 and x + 2y + z = 7.

Solution

The given system of linear equation can be written as:

Q8. If , find.

Solution

Q9. Show that if any two columns of a determinant are interchanged, then sign of the determinant changes.

Solution

        On interchanging the 1^st and the 2^nd column. i.e.               Thus, the value of the determinant changes on interchanging two columns.

Q10. If A =, find A^–1.  Using A^–1 solve the system of equations X + 2y – 2z = 5, – x + 3y = 2 and – 2y + z = 7.

Solution

Now, the system of linear equations can be written as AX = B,

Q11. Find the inverse of A, if

Solution

Q12. If , calculate the value of x.

Solution

Q13. Find whether the system of linear equation has a unique solution or not. 3x + 2y = 5 and 6x + 5y = 10.

Solution

The given system can be written as AX = B, where Thus, the system of linear equations has a unique solution.

Q14. Given  and . find

Solution

Given  and              

Q15.       If and . find .

Solution

Given: and ⇒AB is a zero matrix.   Thus, =0

Q16. Evaluate

Solution

                         Here C₁ and C₂ are identical.

Q17. Write the given system of linear equations in matrix form. 2x – 3y = 1, x + 3z = 11 and x + 2y + z = 7.

Solution

The given system of linear equation can be written as:

Q18. Find the value of the determinant,

Solution

Q19. Find the cofactors of all the elements of the determinant .

Solution

Co-factor of element aij, A_ij = (–1)^i+j M_ij Co-factor of element 3 = (–1)¹⁺¹.1 = 1, Co-factor of element – 2 = (–1)¹⁺².7 = – 7, Co-factor of element 7 = (–1)²⁺¹.(– 2) = 2 and Co-factor of element 1 = (–1)²⁺².3 = 3,

Q20. If A =, find A^–1.  Using A^–1 solve the system of equations X + 2y – 2z = 5, – x + 3y = 2 and – 2y + z = 7.

Solution

Now, the system of linear equations can be written as AX = B,

Q21. Define adjoint of a matrix.

Solution

Adjoint of a Matrix: If A = [a_ij] be a square matrix of order n, then adjoint of A is defined as the transpose of the matrix [Aij]_{n x n}, where A_ij is the co-factor of the element a_ij. Adjoint of A is denoted by Adj A.

Q22. Show that A =  satisfy the equation x² – 5x – 14 = 0.

Solution

A will satisfy the equation x² – 5x – 14 = 0, if A² – 5A – 14I = O.

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