# Free Printable Worksheet on Matrix | Matrices Worksheet with Answers

The Worksheet on Matrix helps students to understand the concept deeply. First, try to solve the problems given on this Matrices Worksheet for Grade 10 PDF and gain knowledge easily. Get to know different problems related to the Matrix concept with the help of the Matrix Worksheet. Practice our Matrices Worksheet with Answers as many times as you wish to get a grip on all the problems. Check out all the 10th Grade Math articles and worksheets on our website for free.

### Free Matrix Worksheet with Answers PDF

Problem 1: If matrix A has 6 number elements, then determine the order of the matrix.

Solution:

Given that, the number of elements is 8.
Let’s write all the possible factors of the number 6.
i.e., 6 = 1 x 6
6 = 3 x 2
6 = 2 x 3
6 = 6 x 1
we can get the number 6 in four ways.
Therefore, there are four possible ways or orders of the 6 number matrix 6 elements are 1 x 6, 2 x 3, 3 x 2, and 6 x 1.

Problem 2: What is the order of a matrix given below? The matrix $$A = \left[ \begin{matrix} 1 & 3 & 7\cr 14 & 5 & 6\cr \end{matrix} \right]$$

Solution:

Given that, the matrix $$A = \left[ \begin{matrix} 1 & 3 & 7\cr 14 & 5 & 6\cr \end{matrix} \right]$$
The number of rows in a given matrix is A = 2.
The number of columns in a given matrix is A = 3
Thus, the order of matrix A is 2 x 3.

Problem 3: Prove that the additive identity of $$A = \left[ \begin{matrix} 2 & 3 & 4\cr 6 & 9& 7\cr \end{matrix} \right]$$ is a null matrix.

Solution:

Given that, the non zero matrix is $$A = \left[ \begin{matrix} 2 & 3 & 4\cr 6 & 9& 7\cr \end{matrix} \right]$$
To prove that the additive identity of a given non-zero matrix is a null matrix, we need to add the given matrix with the null matrix.
$$A = \left[ \begin{matrix} 2 & 3 & 4\cr 6 & 9& 7\cr \end{matrix} \right]$$ + $$A = \left[ \begin{matrix} 0 & 0 & 0\cr 0 & 0& 0\cr \end{matrix} \right]$$
The value of $$A = \left[ \begin{matrix} 2 & 3 & 4\cr 6 & 9& 7\cr \end{matrix} \right]$$
Thus, the addition of a null matrix to the given matrix also gives the same matrix.
Therefore, the null matrix is the additive identity of the given matrix.

Problem 4: Determine if the given matrix is a unit matrix. The value of $$A = \left[ \begin{matrix} 1 & 1\cr 2 & 1\cr \end{matrix} \right]$$

Solution:

The given matrix is $$A = \left[ \begin{matrix} 1 & 1\cr 2 & 1\cr \end{matrix} \right]$$
The Identity Matrix consists of all the diagonal elements as 1 and other elements as zero. In the given matrix A, all the diagonal matrices are 1 but the remaining elements are not zero.
Therefore, the given matrix $$A = \left[ \begin{matrix} 1 & 1\cr 2 & 1\cr \end{matrix} \right]$$ is not an identity matrix.

Problem 5: Find the values of ‘m’ and ‘n’ in the given matrix A such that A is a strictly upper triangular matrix. The matrix $$A = \left[ \begin{matrix} 3m & 4\cr n & 5\cr \end{matrix} \right]$$

Solution:

Given matrix is $$A = \left[ \begin{matrix} 3m & 4\cr n & 5\cr \end{matrix} \right]$$
For the strictly upper triangular matrix, the elements below the diagonal are zero and the elements of the main diagonal are zero.
Therefore, we must have 3m = 0 that is m = 0 and n = 0.
Hence, the values of m and n are 0.

Problem 6: The matrix $$A = \left[ \begin{matrix} 12 & 5\cr 6 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 4 & 3\cr 2 & 5\cr \end{matrix} \right]$$. What is the added value of Matrices A and B?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 12 & 5\cr 6 & 4\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 4 & 3\cr 2 & 5\cr \end{matrix} \right]$$
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is $$A +B= \left[ \begin{matrix} 12+4 & 5+\3cr 6+2 & 4+5\cr \end{matrix} \right]$$
Therefore, the addition matrices value of A +B is $$\left[ \begin{matrix} 16 & 8\cr 8 & 9\cr \end{matrix} \right]$$

Project 7: Subtract X and Y where
$$X = \left[ \begin{matrix} 12 & 8\cr 5 & -4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -5 & -4\cr 2 & 1\cr \end{matrix} \right]$$

Solution:

Given matrices are $$X = \left[ \begin{matrix} 12 & 8\cr 5 & -4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} -5 & -4\cr 2 & 1\cr \end{matrix} \right]$$
Both the matrices have the same order of 2 × 2. So, we will subtract the elements of the first matrix with the respective elements of the second matrix.
So, the matrices X-Y is $$= \left[ \begin{matrix} 12 & 8\cr 5 & -4\cr \end{matrix} \right]$$ – $$\left[ \begin{matrix} -5 & -4\cr 2 & 1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 12-(-5) & 8-(-4) \cr 5-2 & -4-1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 17 & 12\cr 3 & -5\cr \end{matrix} \right]$$
Thus, the subtraction of X and Y is $$\left[ \begin{matrix} 17 & 12\cr 3 & -5\cr \end{matrix} \right]$$

Problem 8: The matrix is,
$$A = \left[ \begin{matrix} 2 & 0\cr 3 & 5\cr \end{matrix} \right]$$  and $$B = \left[ \begin{matrix} 6 & 12\cr 4 & 5\cr \end{matrix} \right]$$. What is the final product value of two matrices?

Solution:

As given in the question,
The matrix A is $$A = \left[ \begin{matrix} 2 & 0\cr 3 & 5\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} 6 & 12\cr 4 & 5\cr \end{matrix} \right]$$
Now, we need to find out the final product value of two matrices.
After the multiplication of two matrices is  $$\left[ \begin{matrix} (2*6) + (0*4) & (2*12) + (0*5) \cr (3*6) + (5*4) & (3*12) + (5*5) \cr \end{matrix}\right]$$
$$\left[ \begin{matrix} 12+0 & 24+0\cr 18+20 & 36+25\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 12 & 24\cr 38 & 61\cr \end{matrix} \right]$$
Therefore, the final product value of the two matrices is $$\left[ \begin{matrix} 12 & 24\cr 38 & 61\cr \end{matrix} \right]$$.

Problem 9: Find the transpose of a column matrix $$A = \left[ \begin{matrix} 8\cr 26\cr 12\cr \end{matrix} \right]$$

Solution:

The given matrix is $$A = \left[ \begin{matrix} 8\cr 26 \cr 12\cr \end{matrix} \right]$$
To find the transpose of the given column matrix, the elements of the column matrix must be written as row elements.
A=[8 26 12]
Therefore, the transpose of the given column matrix is A=[8 26 12].

Problem 10: Find 3B where
$$B= \left[ \begin{matrix} 3 & 7\cr 2 & 13\cr \end{matrix} \right]$$?

Solution:

The given matrix is $$B = \left[ \begin{matrix} 3 & 7\cr 2 & 13\cr \end{matrix} \right]$$
To find the value of 3B, we need to multiply the value of 3 by all the elements of matrix B.
i.e., 3B= $$\left[ \begin{matrix} 3*3 & 3*7\cr 3*2 & 3*13\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 9 & 21\cr 6 & 39\cr \end{matrix} \right]$$
Therefore, the required matrix is 3C $$=\left[ \begin{matrix} 9 & 21\cr 6 & 39\cr \end{matrix} \right]$$

Problem 11: Let  $$A =\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ and $$B= \left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$. Prove that AB = BA = A?

Solution:

Given that, the A and B matrices values.
The matrix A is the order of 2 × 2 and the matrix B is the order of 2 × 2.
So, the number of columns in matrix A equals the number of rows in matrix B. Hence, AB can be found. Also, the number of columns in B equals the number of rows in A. So, BA can also be found.
Now, find the value of AB,
AB = $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ $$\left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8*1 + 5*0 & 8*0+ 5*1\cr -4*1 + 3*0 & -4*0+ 3*1\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ = A
Next, find the value of the BA.
$$BA= \left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$ $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 1*8 + 0*(-4) & 1*5+ 0*3\cr 0*8+ 1*(-4) & 0*5+ 1*3\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ = A
Clearly, $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 8 & 5\cr -4 & 3\cr \end{matrix} \right]$$
Therefore, the value of AB and BA is equal. Hence its proved AB = BA = A.

Problem 12: If $$A = \left[ \begin{matrix} cosθ & 0\cr 0 & cosθ\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} sinθ & 0\cr 0 & sinθ\cr \end{matrix} \right]$$ then show that A2 +B2 = I.

Solution:

In the given question, the matrices are $$A = \left[ \begin{matrix} cosθ & 0\cr 0 & cosθ\cr \end{matrix} \right]$$ and $$B = \left[ \begin{matrix} sinθ & 0\cr 0 & sinθ\cr \end{matrix} \right]$$
Now, we will prove that the value of A2 +B2 = I.
First, find the value of A2
So, A2 = A x A = $$\left[ \begin{matrix} cosθ & 0\cr 0 & cosθ\cr \end{matrix} \right]$$ x $$\left[ \begin{matrix} cosθ & 0\cr 0 & cosθ\cr \end{matrix} \right]$$
=  $$\left[ \begin{matrix} cos^2θ+0 &(0+0)\cr (0+0) & cos^2θ+0\cr \end{matrix} \right]$$
The value of $$A^2 = \left[ \begin{matrix} cos^2θ & 0\cr 0 & cos^2θ\cr \end{matrix} \right]$$
Now, find the value of B^2
So, the value is B x B = $$\left[ \begin{matrix} Sinθ & 0\cr 0 & Sinθ\cr \end{matrix} \right]$$ x $$\left[ \begin{matrix} Sinθ & 0\cr 0 & Sinθ\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} sin^2θ+0 &(0+0)\cr (0+0) & sin^2θ+0\cr \end{matrix} \right]$$
The value of $$B^2= \left[ \begin{matrix} sin^2θ & 0\cr 0 & sin^2θ\cr \end{matrix} \right]$$
Now, adding the corresponding terms. We get,
A^2+B^2 = $$\left[ \begin{matrix} cos^2θ+sin^2θ & 0 \cr 0 & cos^2θ+sin^2θ \cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$ = I.

Problem 13: Verify that X2 = I when $$X = \left[ \begin{matrix} 5 & -4\cr 6 & -5\cr \end{matrix} \right]$$ ?

Solution:

As given in the question, the value of X is $$\left[ \begin{matrix} 5 & -4\cr 6 & -5\cr \end{matrix} \right]$$
Now, we will verify that X2 = I.
First, we will find the value of the X2.
The value of X2 is  X x X = $$\left[ \begin{matrix} 5 & -4\cr 6 & -5\cr \end{matrix} \right]$$ x $$\left[ \begin{matrix} 5 & -4\cr 6 & -5\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (25-24) & (20-20)\cr (30-30) & (-24+25)\cr \end{matrix} \right]$$
So, the final value is $$\left[ \begin{matrix} 1 & 0\cr 0 & 1\cr \end{matrix} \right]$$
Hence, proved the value of X2 is I.

Problem 14:  Prove that A(B + C) = AB + AC. The matrix of $$A=\left[ \begin{matrix} 1 & 3\cr 5 & -1\cr \end{matrix} \right]$$, $$B= \left[ \begin{matrix} 1 & -1 & 2\cr 3 & 5 & 2\cr \end{matrix} \right]$$, and $$C = \left[ \begin{matrix} 1 & 3 & 2\cr -4 & 1 & 3\cr \end{matrix} \right]$$ ?

Solution:

In the given questions are, the matrices are $$A=\left[ \begin{matrix} 1 & 3\cr 5 & -1\cr \end{matrix} \right]$$, $$B= \left[ \begin{matrix} 1 & -1 & 2\cr 3 & 5 & 2\cr \end{matrix} \right]$$, and $$C = \left[ \begin{matrix} 1 & 3 & 2\cr -4 & 1 & 3\cr \end{matrix} \right]$$
Now, we will prove that A(B + C) = AB + AC.
First, take the L.H.S part that is A(B + C).
(B+C) = $$\left[ \begin{matrix} 1 & -1 & 2\cr 3 & 5 & 2\cr \end{matrix} \right]$$ + $$\left[ \begin{matrix} 1 & 3 & 2\cr -4 & 1 & 3\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (1+1) & (-1+3) & (2+2)\cr (3-4) & (5+1) & (2+3)\cr \end{matrix} \right]$$ = $$\left[ \begin{matrix} 2 & 2 & 4\cr -1 & 6 & 5\cr \end{matrix} \right]$$
Now, multiply the B+C value with A, We get,
A(B+C) = $$\left[ \begin{matrix} 1 & 3 \cr 5 & -1 \cr \end{matrix} \right]$$. $$\left[ \begin{matrix} 2 & 2 & 4\cr -1 & 6 & 5\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (2-3) & (2+18) & (4+15)\cr (10+1) & (10-6) & (20-5)\cr \end{matrix} \right]$$
A(B+C) = $$\left[ \begin{matrix} -1 & 20 & 19\cr 11 & 4 & 15\cr \end{matrix} \right]$$
Next, find the value of the R.H.S that is AB+AC
First take AB= $$\left[ \begin{matrix} 1 & 3\cr 5 & -1\cr \end{matrix} \right]$$ x $$\left[ \begin{matrix} 1 & -1 & 2\cr 3 & 5 & 2\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (1+9) & (-1+15) & (2+6)\cr (5-3) & (-5-5) & (10-2)\cr \end{matrix} \right]$$
$$AB=\left[ \begin{matrix} 10 & 14 & 8\cr 2 & -10 & 8\cr \end{matrix} \right]$$
Now, find the value of AC,
AC = $$\left[ \begin{matrix} 1 & 3\cr 5 & -1\cr \end{matrix} \right]$$ x $$\left[ \begin{matrix} 1 & 3 & 2\cr -4 & 1 & 3\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (1-12) & (3+3) & (2+19)\cr (5+4) & (15-1) & (10-3)\cr \end{matrix} \right]$$
So, the value of AC is $$\left[ \begin{matrix} -11 & 6 & 11\cr 9 & 14 & 7\cr \end{matrix} \right]$$
Now, let us add AB+ AC.
AB+AC = $$\left[ \begin{matrix} 10 & 14 & 8\cr 2 & -10 & 8\cr \end{matrix} \right]$$ + $$\left[ \begin{matrix} -11 & 6 & 11\cr 9 & 14 & 7\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} (10-11) & (14+6) & (8+11)\cr (2+9) & (-10+14) & (8+7)\cr \end{matrix} \right]$$
= $$\left[ \begin{matrix} -1 & 20 & 19\cr 11 & 4 & 15\cr \end{matrix} \right]$$
The value of A(B+C) and AB+AC both are Equal.
Hence, prove that L.H.S = R.H.S.

Problem 15: The matrix $$X = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$. What is the added value of Matrices X and Y?

Solution:

As given in the question,
The matrix A is $$X = \left[ \begin{matrix} 11 & 2\cr 3 & 4\cr \end{matrix} \right]$$ and $$Y = \left[ \begin{matrix} 7 & 8\cr 6 & 5\cr \end{matrix} \right]$$
Now, we will find the addition value of A and B.
Here, both the matrices have the same order of 2 × 2. Now, we can add the elements of the first matrix with the respective elements of the second matrix.
After, addition the matrices is $$X+Y= \left[ \begin{matrix} 11+7 & 2+8\cr 3+6 & 4+5\cr \end{matrix} \right]$$
Therefore, the addition matrices value of X+Y is $$\left[ \begin{matrix} 18 & 10\cr 9 & 9\cr \end{matrix} \right]$$

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